For a function defined by parts study continuity, and differentiability at two points
For the function defined by $$F(x)=begin{cases}displaystyleint_x^{2x}sin t^2,mathrm dt,&xneq0\0,&x=0end{cases}$$ analyze continuity and derivability at the origin. Is $F$ derivable at point $x_0=sqrt{pi/2}$? Justify the answer, and if possible, calculate $F'(x_0)$.
I have been told that I must use the Fundamental Theorem of Integral Calculus but I do not know how to apply it to this case.
For the function to be continuous at the origin, it must happen that $F(0)=lim_{xto0}F(x)$. We know that $F(0)=0$, and $$lim_{xto0}F(x)=lim_{xto0}int_x^{2x}sin t^2,mathrm dt;{bfcolor{red}=}int_0^{2cdot0}sin t^2,mathrm dt=0,$$ so the statement holds, but here I do now how to justify the $bfcolor{red}=$.
To find the derivative at $x_0=0$ I tried the differentiate directly $F(x)$ but it is wrong, so I have been told that I must use the definition. So we have to find $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x.$$ Why we have to bound $left|sin t^2right|leq t^2$? How can we do that?
Finally, I do not know how to use the aforementioned theorem to justify that the function is derivable in $sqrt{pi/2}$. Using the definition again:
begin{align*}
F'left(sqrt{fracpi2}right)&=lim_{xtosqrt{fracpi2}}frac{F(x)-Fleft(sqrt{fracpi2}right)}{x-sqrt{fracpi2}}\
&=lim_{xtosqrt{fracpi2}}frac{int_x^{2x}sin t^2,mathrm dt-int_{sqrt{pi/2}}^{2sqrt{pi/2}}sin t^2,mathrm dt}{x-sqrt{fracpi2}}\
&leqlim_{xtosqrt{fracpi2}}frac{int_x^{2x}t^2,mathrm dt-int_{sqrt{pi/2}}^{2sqrt{pi/2}}t^2,mathrm dt}{x-sqrt{fracpi2}}\
&underbrace=_{A=sqrt{pi/2}}lim_{xto A}frac{1/3((2x)^3-x^3)-1/3((2A)^3-(A^3))}{x-A}\
&=frac73lim_{xto A}frac{x^3-A^3}{x-A}\
&=frac73lim_{xto A}frac{(x-A)(x^2+Ax+A^2)}{x-A}\
&=frac73(A^2+A^2+A^2)\
&=7A^2\
&=frac{7pi}2,
end{align*}
but it is wrong.
How can we solve the statement?
Thanks!
calculus derivatives continuity
add a comment |
For the function defined by $$F(x)=begin{cases}displaystyleint_x^{2x}sin t^2,mathrm dt,&xneq0\0,&x=0end{cases}$$ analyze continuity and derivability at the origin. Is $F$ derivable at point $x_0=sqrt{pi/2}$? Justify the answer, and if possible, calculate $F'(x_0)$.
I have been told that I must use the Fundamental Theorem of Integral Calculus but I do not know how to apply it to this case.
For the function to be continuous at the origin, it must happen that $F(0)=lim_{xto0}F(x)$. We know that $F(0)=0$, and $$lim_{xto0}F(x)=lim_{xto0}int_x^{2x}sin t^2,mathrm dt;{bfcolor{red}=}int_0^{2cdot0}sin t^2,mathrm dt=0,$$ so the statement holds, but here I do now how to justify the $bfcolor{red}=$.
To find the derivative at $x_0=0$ I tried the differentiate directly $F(x)$ but it is wrong, so I have been told that I must use the definition. So we have to find $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x.$$ Why we have to bound $left|sin t^2right|leq t^2$? How can we do that?
Finally, I do not know how to use the aforementioned theorem to justify that the function is derivable in $sqrt{pi/2}$. Using the definition again:
begin{align*}
F'left(sqrt{fracpi2}right)&=lim_{xtosqrt{fracpi2}}frac{F(x)-Fleft(sqrt{fracpi2}right)}{x-sqrt{fracpi2}}\
&=lim_{xtosqrt{fracpi2}}frac{int_x^{2x}sin t^2,mathrm dt-int_{sqrt{pi/2}}^{2sqrt{pi/2}}sin t^2,mathrm dt}{x-sqrt{fracpi2}}\
&leqlim_{xtosqrt{fracpi2}}frac{int_x^{2x}t^2,mathrm dt-int_{sqrt{pi/2}}^{2sqrt{pi/2}}t^2,mathrm dt}{x-sqrt{fracpi2}}\
&underbrace=_{A=sqrt{pi/2}}lim_{xto A}frac{1/3((2x)^3-x^3)-1/3((2A)^3-(A^3))}{x-A}\
&=frac73lim_{xto A}frac{x^3-A^3}{x-A}\
&=frac73lim_{xto A}frac{(x-A)(x^2+Ax+A^2)}{x-A}\
&=frac73(A^2+A^2+A^2)\
&=7A^2\
&=frac{7pi}2,
end{align*}
but it is wrong.
How can we solve the statement?
Thanks!
calculus derivatives continuity
1
Would it help if you write $int_x^{2x}=int_0^{2x}-int_0^x$?
– A.Γ.
Jan 3 at 23:59
@A.Γ. probably. Can we separate it into two integrals because the limits of integrations are continuous?
– manooooh
Jan 4 at 0:01
It is because the integral is additive. Or if you wish $int_a^b=F(b)-F(a)$ where $F(y)$ is antiderivative, which is e.g. $int_0^y$.
– A.Γ.
Jan 4 at 0:05
@A.Γ. how do you know that $0in[x,2x]$ for all $xneq0$? Because for example if $xin(0,infty)$ then $0notin[x,2x]$.
– manooooh
Jan 4 at 0:29
1
One does not need $0in[x,2x]$. Integral is additive and if $f$ is Riemann integrable on some closed interval containing $a, b, c$ then $$int_{a} ^{b} f(x) , dx+int_{b} ^{c} f(x) , dx=int_{a} ^{c} f(x) , dx$$ irrespective of the linear order of $a, b, c$. This assumes that we have by definition $int_{a} ^{a}f(x),dx=0$ and $int_{b} ^{a} f(x) , dx=-int_{a} ^{b} f(x) , dx$.
– Paramanand Singh
Jan 4 at 7:23
add a comment |
For the function defined by $$F(x)=begin{cases}displaystyleint_x^{2x}sin t^2,mathrm dt,&xneq0\0,&x=0end{cases}$$ analyze continuity and derivability at the origin. Is $F$ derivable at point $x_0=sqrt{pi/2}$? Justify the answer, and if possible, calculate $F'(x_0)$.
I have been told that I must use the Fundamental Theorem of Integral Calculus but I do not know how to apply it to this case.
For the function to be continuous at the origin, it must happen that $F(0)=lim_{xto0}F(x)$. We know that $F(0)=0$, and $$lim_{xto0}F(x)=lim_{xto0}int_x^{2x}sin t^2,mathrm dt;{bfcolor{red}=}int_0^{2cdot0}sin t^2,mathrm dt=0,$$ so the statement holds, but here I do now how to justify the $bfcolor{red}=$.
To find the derivative at $x_0=0$ I tried the differentiate directly $F(x)$ but it is wrong, so I have been told that I must use the definition. So we have to find $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x.$$ Why we have to bound $left|sin t^2right|leq t^2$? How can we do that?
Finally, I do not know how to use the aforementioned theorem to justify that the function is derivable in $sqrt{pi/2}$. Using the definition again:
begin{align*}
F'left(sqrt{fracpi2}right)&=lim_{xtosqrt{fracpi2}}frac{F(x)-Fleft(sqrt{fracpi2}right)}{x-sqrt{fracpi2}}\
&=lim_{xtosqrt{fracpi2}}frac{int_x^{2x}sin t^2,mathrm dt-int_{sqrt{pi/2}}^{2sqrt{pi/2}}sin t^2,mathrm dt}{x-sqrt{fracpi2}}\
&leqlim_{xtosqrt{fracpi2}}frac{int_x^{2x}t^2,mathrm dt-int_{sqrt{pi/2}}^{2sqrt{pi/2}}t^2,mathrm dt}{x-sqrt{fracpi2}}\
&underbrace=_{A=sqrt{pi/2}}lim_{xto A}frac{1/3((2x)^3-x^3)-1/3((2A)^3-(A^3))}{x-A}\
&=frac73lim_{xto A}frac{x^3-A^3}{x-A}\
&=frac73lim_{xto A}frac{(x-A)(x^2+Ax+A^2)}{x-A}\
&=frac73(A^2+A^2+A^2)\
&=7A^2\
&=frac{7pi}2,
end{align*}
but it is wrong.
How can we solve the statement?
Thanks!
calculus derivatives continuity
For the function defined by $$F(x)=begin{cases}displaystyleint_x^{2x}sin t^2,mathrm dt,&xneq0\0,&x=0end{cases}$$ analyze continuity and derivability at the origin. Is $F$ derivable at point $x_0=sqrt{pi/2}$? Justify the answer, and if possible, calculate $F'(x_0)$.
I have been told that I must use the Fundamental Theorem of Integral Calculus but I do not know how to apply it to this case.
For the function to be continuous at the origin, it must happen that $F(0)=lim_{xto0}F(x)$. We know that $F(0)=0$, and $$lim_{xto0}F(x)=lim_{xto0}int_x^{2x}sin t^2,mathrm dt;{bfcolor{red}=}int_0^{2cdot0}sin t^2,mathrm dt=0,$$ so the statement holds, but here I do now how to justify the $bfcolor{red}=$.
To find the derivative at $x_0=0$ I tried the differentiate directly $F(x)$ but it is wrong, so I have been told that I must use the definition. So we have to find $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x.$$ Why we have to bound $left|sin t^2right|leq t^2$? How can we do that?
Finally, I do not know how to use the aforementioned theorem to justify that the function is derivable in $sqrt{pi/2}$. Using the definition again:
begin{align*}
F'left(sqrt{fracpi2}right)&=lim_{xtosqrt{fracpi2}}frac{F(x)-Fleft(sqrt{fracpi2}right)}{x-sqrt{fracpi2}}\
&=lim_{xtosqrt{fracpi2}}frac{int_x^{2x}sin t^2,mathrm dt-int_{sqrt{pi/2}}^{2sqrt{pi/2}}sin t^2,mathrm dt}{x-sqrt{fracpi2}}\
&leqlim_{xtosqrt{fracpi2}}frac{int_x^{2x}t^2,mathrm dt-int_{sqrt{pi/2}}^{2sqrt{pi/2}}t^2,mathrm dt}{x-sqrt{fracpi2}}\
&underbrace=_{A=sqrt{pi/2}}lim_{xto A}frac{1/3((2x)^3-x^3)-1/3((2A)^3-(A^3))}{x-A}\
&=frac73lim_{xto A}frac{x^3-A^3}{x-A}\
&=frac73lim_{xto A}frac{(x-A)(x^2+Ax+A^2)}{x-A}\
&=frac73(A^2+A^2+A^2)\
&=7A^2\
&=frac{7pi}2,
end{align*}
but it is wrong.
How can we solve the statement?
Thanks!
calculus derivatives continuity
calculus derivatives continuity
asked Jan 3 at 23:43
manooooh
5441417
5441417
1
Would it help if you write $int_x^{2x}=int_0^{2x}-int_0^x$?
– A.Γ.
Jan 3 at 23:59
@A.Γ. probably. Can we separate it into two integrals because the limits of integrations are continuous?
– manooooh
Jan 4 at 0:01
It is because the integral is additive. Or if you wish $int_a^b=F(b)-F(a)$ where $F(y)$ is antiderivative, which is e.g. $int_0^y$.
– A.Γ.
Jan 4 at 0:05
@A.Γ. how do you know that $0in[x,2x]$ for all $xneq0$? Because for example if $xin(0,infty)$ then $0notin[x,2x]$.
– manooooh
Jan 4 at 0:29
1
One does not need $0in[x,2x]$. Integral is additive and if $f$ is Riemann integrable on some closed interval containing $a, b, c$ then $$int_{a} ^{b} f(x) , dx+int_{b} ^{c} f(x) , dx=int_{a} ^{c} f(x) , dx$$ irrespective of the linear order of $a, b, c$. This assumes that we have by definition $int_{a} ^{a}f(x),dx=0$ and $int_{b} ^{a} f(x) , dx=-int_{a} ^{b} f(x) , dx$.
– Paramanand Singh
Jan 4 at 7:23
add a comment |
1
Would it help if you write $int_x^{2x}=int_0^{2x}-int_0^x$?
– A.Γ.
Jan 3 at 23:59
@A.Γ. probably. Can we separate it into two integrals because the limits of integrations are continuous?
– manooooh
Jan 4 at 0:01
It is because the integral is additive. Or if you wish $int_a^b=F(b)-F(a)$ where $F(y)$ is antiderivative, which is e.g. $int_0^y$.
– A.Γ.
Jan 4 at 0:05
@A.Γ. how do you know that $0in[x,2x]$ for all $xneq0$? Because for example if $xin(0,infty)$ then $0notin[x,2x]$.
– manooooh
Jan 4 at 0:29
1
One does not need $0in[x,2x]$. Integral is additive and if $f$ is Riemann integrable on some closed interval containing $a, b, c$ then $$int_{a} ^{b} f(x) , dx+int_{b} ^{c} f(x) , dx=int_{a} ^{c} f(x) , dx$$ irrespective of the linear order of $a, b, c$. This assumes that we have by definition $int_{a} ^{a}f(x),dx=0$ and $int_{b} ^{a} f(x) , dx=-int_{a} ^{b} f(x) , dx$.
– Paramanand Singh
Jan 4 at 7:23
1
1
Would it help if you write $int_x^{2x}=int_0^{2x}-int_0^x$?
– A.Γ.
Jan 3 at 23:59
Would it help if you write $int_x^{2x}=int_0^{2x}-int_0^x$?
– A.Γ.
Jan 3 at 23:59
@A.Γ. probably. Can we separate it into two integrals because the limits of integrations are continuous?
– manooooh
Jan 4 at 0:01
@A.Γ. probably. Can we separate it into two integrals because the limits of integrations are continuous?
– manooooh
Jan 4 at 0:01
It is because the integral is additive. Or if you wish $int_a^b=F(b)-F(a)$ where $F(y)$ is antiderivative, which is e.g. $int_0^y$.
– A.Γ.
Jan 4 at 0:05
It is because the integral is additive. Or if you wish $int_a^b=F(b)-F(a)$ where $F(y)$ is antiderivative, which is e.g. $int_0^y$.
– A.Γ.
Jan 4 at 0:05
@A.Γ. how do you know that $0in[x,2x]$ for all $xneq0$? Because for example if $xin(0,infty)$ then $0notin[x,2x]$.
– manooooh
Jan 4 at 0:29
@A.Γ. how do you know that $0in[x,2x]$ for all $xneq0$? Because for example if $xin(0,infty)$ then $0notin[x,2x]$.
– manooooh
Jan 4 at 0:29
1
1
One does not need $0in[x,2x]$. Integral is additive and if $f$ is Riemann integrable on some closed interval containing $a, b, c$ then $$int_{a} ^{b} f(x) , dx+int_{b} ^{c} f(x) , dx=int_{a} ^{c} f(x) , dx$$ irrespective of the linear order of $a, b, c$. This assumes that we have by definition $int_{a} ^{a}f(x),dx=0$ and $int_{b} ^{a} f(x) , dx=-int_{a} ^{b} f(x) , dx$.
– Paramanand Singh
Jan 4 at 7:23
One does not need $0in[x,2x]$. Integral is additive and if $f$ is Riemann integrable on some closed interval containing $a, b, c$ then $$int_{a} ^{b} f(x) , dx+int_{b} ^{c} f(x) , dx=int_{a} ^{c} f(x) , dx$$ irrespective of the linear order of $a, b, c$. This assumes that we have by definition $int_{a} ^{a}f(x),dx=0$ and $int_{b} ^{a} f(x) , dx=-int_{a} ^{b} f(x) , dx$.
– Paramanand Singh
Jan 4 at 7:23
add a comment |
3 Answers
3
active
oldest
votes
You just need to use the fundamental theorem of calculus. Since the integrand $sin(t^2)$ is continuous everywhere we can write $$F(x) =int_{0}^{2x}sin t^2,dt-int_{0}^{x}sin t^2,dt$$ Use substitution $z=t/2$ in first integral on right to get $$F(x) =2int_{0}^{x}sin (4z^2),dz-int_{0}^{x}sin t^2,dt$$ and by FTC we can see that $F$ is continuous and differentiable everywhere with derivative $$F'(x) =2sin (4x^2)-sin x^2$$ for all $xinmathbb {R} $.
For reference I mention FTC explicitly :
Fundamental Theorem of Calculus Part 1: Let the function $f:[a, b] tomathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $cin[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.
Using the above theorem it can be proved that if a function $f:mathbb {R} tomathbb {R} $ is Riemann integrable on every bounded and closed interval then the function $F:mathbb {R} tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ for some $ainmathbb {R} $ is continuous everywhere and if $f$ is continuous at some point $cin mathbb {R} $ then $F$ is differentiable at $c$ with $F'(c) =f(c) $.
1
@manooooh: if you carefully note the FTC you will see that it deals with integrals where upper limit of integral is $x$ and lower limit is a constant. The substitution in first integral is done to change upper limit $2x$ to $x$. The second integral is already having $x$ as upper limit. If we apply substitution in second integral also then the upper limit changes from $x$ to $x/2$ and FTC can't be applied directly. Continued in next comment.
– Paramanand Singh
2 days ago
1
@manooooh: Using FTC and chain rule for derivatives it can be proved that if $$F(x) =int_{a} ^{g(x)} f(t) , dt$$ then $F'(x) =f(g(x)) g'(x) $. This handles the case when the upper limit of integral is not $x$ but rather some complicated function $g(x) $. Regarding your second doubt, you can check using this formula that the answer remains same even if we apply substitution in second integral.
– Paramanand Singh
2 days ago
1
@manooooh: you need to revisit substitution in definite integrals in your textbook and observe very carefully the given examples. During substitution in a definite integral the function as well as the limit of integral change according to the substitution used. Thus if we use $z=t/2, t=2z,dt=2,dz$ in second integral the limits $0$ and $x$ change to $0$ and $x/2$ to give $$int_{0}^{x}sin t^2,dt=2int_{0}^{x/2}sin (4z^2),dz$$
– Paramanand Singh
2 days ago
1
@manooooh: the limits of integral indicate the range of values being taken by the variable of integration. Thus in $int_{0}^{x}sin t^2,dt$ the variable $t$ varies from $0$ to $x$. If $z=t/2$ then $z$ should vary from $0$ to $x/2$ (as $z$ is half of $t$).
– Paramanand Singh
2 days ago
1
@manooooh: irrespective of how proficient you are in mathematics, your comments indicate that you have a sincere desire to learn. That's what matters here and nothing else. +1 for your question.
– Paramanand Singh
2 days ago
|
show 5 more comments
If you want to evaluate the limit:
$$displaystylelim_{xto 0}F(x)=lim_{xto 0}int_{x}^{2x}sin(t^2)dt$$
you can observe that $forall x>0$ (the case $x<0$ is the same), $f(t)=sin(t^2)$ is continuous in $[x,2x]$ so for the mean value theorem, exists $xi_{x}in (x,2x)$ such that
$$int_{x}^{2x}sin(t^2)dt=sin(xi_{x}^2)(2x-x)implies F(x)=sin(xi_{x}^2)x$$
Now $xi_{x}to 0$ for $xto 0^{+}$ so:
$$lim_{xto 0^{+}}F(x)=lim_{xto 0}sin(xi_{x}^2)x=[sin(0)cdot 0]=0$$
Note that this argument can be used to show that $F(x)$ is derivable for $x=0$, infact:
$$lim_{xto 0^{+}}frac{F(x)-F(0)}{x-0}=lim_{xto 0}frac{sin(xi_{x}^2)x}{x}=lim_{xto 0}sin(xi_{x}^2)=0$$
For $x_0=sqrt{frac{pi}{2}}$, the derivative of $F(x)$ can be found using the Fundamental Theorem of Integral Calculus.
$F'(x)=2sin(4x^2)-sin(x^2)implies F'left(sqrt{frac{pi}{2}}right)=2sin(4cdotfrac{pi}{2})-sin(4cdotfrac{pi}{2})=-1$
Thanks for the answer! How do you know that $F(x)=f(xi_{x})(2x-x)$? From Mean Value Theorem, we have that exists $xi_{x}in (x,2x)$ such that $F(x)=frac{F(2x)-F(x)}{2x-x}$.
– manooooh
Jan 4 at 0:49
I applied the mean value theorem for the function $f(t)=sin(t^2)$ over the intervall $[x,2x]$. So exists $xi_xin(x,2x)$ such that $int_{x}^{2x}f(t)dt=f(xi_x)(2x-x)$.
– Ixion
Jan 4 at 0:52
I do not know how do you get $f(xi_x)(2x-x)$. Since $f(x)$ is continuous at $[x,2x]$ and $f(x)$ is differentiable at $(x,2x)$ then exists $xi_xin(x,2x)$ such that $require{cancel}f'(xi_x)=dfrac{f(2x)-f(x)}{2x-x}=dfrac{sin(4x^2)-sin(2x)}xcancelimplies f(xi_x)(2x-x)$.
– manooooh
Jan 4 at 5:31
1
The mean value theorem states that: if $f(t)$ is continuous in $[a,b]$ then it exists $xiin (a,b)$ such that: $$int_{a}^{b}f(t)dt=f(xi)(b-a)$$ In this case $f(t)=sin(t^2), a=x$ and $b=2x.$
– Ixion
2 days ago
add a comment |
Since $frac {sin, x} xto1 $as $x to 0$ we can find $delta >0$ such that $frac 1 2 t^{2} leqsin(t^{2})leq 2t^{2}$ for $|t| <delta$. This gives $frac 7 6 x^{3} leq F(x) leq frac {14} 3x^{3}$ for $0<x<sqrt {delta}$ and it follows easily from the definition that the right hand derivative of $F$ at $0$ is $0$. Make the substitution $s=-t$ to see that the left hand derivative is also $0$. Hence $F'(0)=0$. For $x>0$ we have $F(x)=int_0^{2x}sin(t^{2}), dt -int_0^{x}sin(t^{2}), dt$ from which it follows (by Fundamental Theorem of Calculus) that $F'(x)=2sin(4x^{2})-sin(x^{2})$. At the given point $x_0$ the derivative is $-1$.
Thanks for the answer! Why do you use "Since $frac {sin, x} xto1$ as $xto0$ (...)"?
– manooooh
Jan 4 at 6:35
1
@manooooh it makes it easy to see that $frac {F(x)} x to 0$ as $xto 0$. I can use just the definition of derivative instead of using MVT, etc.
– Kavi Rama Murthy
Jan 4 at 6:40
Ok. Could you give me the guidelines on how to use the derivative definition at a point, please? From $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x$$ I should use the fact that $left|sin t^2right|leq t^2$ so now the limit becomes $lim_{xto0}left|frac{int_x^{2x}t^2,mathrm dt}xright|$?
– manooooh
Jan 4 at 6:44
1
@manooooh Now calculate he integral of $t^{2}$ from $x$ to $2x$. You will get $7x^{3} /3$. You now see that the limit is $0$, right?
– Kavi Rama Murthy
Jan 4 at 7:17
1
@manooooh You said you have been asked to use the Fundamental Theorem of Calculus. That is what I have done in my answer. DO not try to find the derivative at $x_0$ using the definition of derivative. That is messy!
– Kavi Rama Murthy
Jan 4 at 7:30
|
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You just need to use the fundamental theorem of calculus. Since the integrand $sin(t^2)$ is continuous everywhere we can write $$F(x) =int_{0}^{2x}sin t^2,dt-int_{0}^{x}sin t^2,dt$$ Use substitution $z=t/2$ in first integral on right to get $$F(x) =2int_{0}^{x}sin (4z^2),dz-int_{0}^{x}sin t^2,dt$$ and by FTC we can see that $F$ is continuous and differentiable everywhere with derivative $$F'(x) =2sin (4x^2)-sin x^2$$ for all $xinmathbb {R} $.
For reference I mention FTC explicitly :
Fundamental Theorem of Calculus Part 1: Let the function $f:[a, b] tomathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $cin[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.
Using the above theorem it can be proved that if a function $f:mathbb {R} tomathbb {R} $ is Riemann integrable on every bounded and closed interval then the function $F:mathbb {R} tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ for some $ainmathbb {R} $ is continuous everywhere and if $f$ is continuous at some point $cin mathbb {R} $ then $F$ is differentiable at $c$ with $F'(c) =f(c) $.
1
@manooooh: if you carefully note the FTC you will see that it deals with integrals where upper limit of integral is $x$ and lower limit is a constant. The substitution in first integral is done to change upper limit $2x$ to $x$. The second integral is already having $x$ as upper limit. If we apply substitution in second integral also then the upper limit changes from $x$ to $x/2$ and FTC can't be applied directly. Continued in next comment.
– Paramanand Singh
2 days ago
1
@manooooh: Using FTC and chain rule for derivatives it can be proved that if $$F(x) =int_{a} ^{g(x)} f(t) , dt$$ then $F'(x) =f(g(x)) g'(x) $. This handles the case when the upper limit of integral is not $x$ but rather some complicated function $g(x) $. Regarding your second doubt, you can check using this formula that the answer remains same even if we apply substitution in second integral.
– Paramanand Singh
2 days ago
1
@manooooh: you need to revisit substitution in definite integrals in your textbook and observe very carefully the given examples. During substitution in a definite integral the function as well as the limit of integral change according to the substitution used. Thus if we use $z=t/2, t=2z,dt=2,dz$ in second integral the limits $0$ and $x$ change to $0$ and $x/2$ to give $$int_{0}^{x}sin t^2,dt=2int_{0}^{x/2}sin (4z^2),dz$$
– Paramanand Singh
2 days ago
1
@manooooh: the limits of integral indicate the range of values being taken by the variable of integration. Thus in $int_{0}^{x}sin t^2,dt$ the variable $t$ varies from $0$ to $x$. If $z=t/2$ then $z$ should vary from $0$ to $x/2$ (as $z$ is half of $t$).
– Paramanand Singh
2 days ago
1
@manooooh: irrespective of how proficient you are in mathematics, your comments indicate that you have a sincere desire to learn. That's what matters here and nothing else. +1 for your question.
– Paramanand Singh
2 days ago
|
show 5 more comments
You just need to use the fundamental theorem of calculus. Since the integrand $sin(t^2)$ is continuous everywhere we can write $$F(x) =int_{0}^{2x}sin t^2,dt-int_{0}^{x}sin t^2,dt$$ Use substitution $z=t/2$ in first integral on right to get $$F(x) =2int_{0}^{x}sin (4z^2),dz-int_{0}^{x}sin t^2,dt$$ and by FTC we can see that $F$ is continuous and differentiable everywhere with derivative $$F'(x) =2sin (4x^2)-sin x^2$$ for all $xinmathbb {R} $.
For reference I mention FTC explicitly :
Fundamental Theorem of Calculus Part 1: Let the function $f:[a, b] tomathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $cin[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.
Using the above theorem it can be proved that if a function $f:mathbb {R} tomathbb {R} $ is Riemann integrable on every bounded and closed interval then the function $F:mathbb {R} tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ for some $ainmathbb {R} $ is continuous everywhere and if $f$ is continuous at some point $cin mathbb {R} $ then $F$ is differentiable at $c$ with $F'(c) =f(c) $.
1
@manooooh: if you carefully note the FTC you will see that it deals with integrals where upper limit of integral is $x$ and lower limit is a constant. The substitution in first integral is done to change upper limit $2x$ to $x$. The second integral is already having $x$ as upper limit. If we apply substitution in second integral also then the upper limit changes from $x$ to $x/2$ and FTC can't be applied directly. Continued in next comment.
– Paramanand Singh
2 days ago
1
@manooooh: Using FTC and chain rule for derivatives it can be proved that if $$F(x) =int_{a} ^{g(x)} f(t) , dt$$ then $F'(x) =f(g(x)) g'(x) $. This handles the case when the upper limit of integral is not $x$ but rather some complicated function $g(x) $. Regarding your second doubt, you can check using this formula that the answer remains same even if we apply substitution in second integral.
– Paramanand Singh
2 days ago
1
@manooooh: you need to revisit substitution in definite integrals in your textbook and observe very carefully the given examples. During substitution in a definite integral the function as well as the limit of integral change according to the substitution used. Thus if we use $z=t/2, t=2z,dt=2,dz$ in second integral the limits $0$ and $x$ change to $0$ and $x/2$ to give $$int_{0}^{x}sin t^2,dt=2int_{0}^{x/2}sin (4z^2),dz$$
– Paramanand Singh
2 days ago
1
@manooooh: the limits of integral indicate the range of values being taken by the variable of integration. Thus in $int_{0}^{x}sin t^2,dt$ the variable $t$ varies from $0$ to $x$. If $z=t/2$ then $z$ should vary from $0$ to $x/2$ (as $z$ is half of $t$).
– Paramanand Singh
2 days ago
1
@manooooh: irrespective of how proficient you are in mathematics, your comments indicate that you have a sincere desire to learn. That's what matters here and nothing else. +1 for your question.
– Paramanand Singh
2 days ago
|
show 5 more comments
You just need to use the fundamental theorem of calculus. Since the integrand $sin(t^2)$ is continuous everywhere we can write $$F(x) =int_{0}^{2x}sin t^2,dt-int_{0}^{x}sin t^2,dt$$ Use substitution $z=t/2$ in first integral on right to get $$F(x) =2int_{0}^{x}sin (4z^2),dz-int_{0}^{x}sin t^2,dt$$ and by FTC we can see that $F$ is continuous and differentiable everywhere with derivative $$F'(x) =2sin (4x^2)-sin x^2$$ for all $xinmathbb {R} $.
For reference I mention FTC explicitly :
Fundamental Theorem of Calculus Part 1: Let the function $f:[a, b] tomathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $cin[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.
Using the above theorem it can be proved that if a function $f:mathbb {R} tomathbb {R} $ is Riemann integrable on every bounded and closed interval then the function $F:mathbb {R} tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ for some $ainmathbb {R} $ is continuous everywhere and if $f$ is continuous at some point $cin mathbb {R} $ then $F$ is differentiable at $c$ with $F'(c) =f(c) $.
You just need to use the fundamental theorem of calculus. Since the integrand $sin(t^2)$ is continuous everywhere we can write $$F(x) =int_{0}^{2x}sin t^2,dt-int_{0}^{x}sin t^2,dt$$ Use substitution $z=t/2$ in first integral on right to get $$F(x) =2int_{0}^{x}sin (4z^2),dz-int_{0}^{x}sin t^2,dt$$ and by FTC we can see that $F$ is continuous and differentiable everywhere with derivative $$F'(x) =2sin (4x^2)-sin x^2$$ for all $xinmathbb {R} $.
For reference I mention FTC explicitly :
Fundamental Theorem of Calculus Part 1: Let the function $f:[a, b] tomathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $cin[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.
Using the above theorem it can be proved that if a function $f:mathbb {R} tomathbb {R} $ is Riemann integrable on every bounded and closed interval then the function $F:mathbb {R} tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ for some $ainmathbb {R} $ is continuous everywhere and if $f$ is continuous at some point $cin mathbb {R} $ then $F$ is differentiable at $c$ with $F'(c) =f(c) $.
edited Jan 4 at 8:14
answered Jan 4 at 7:28
Paramanand Singh
49k555161
49k555161
1
@manooooh: if you carefully note the FTC you will see that it deals with integrals where upper limit of integral is $x$ and lower limit is a constant. The substitution in first integral is done to change upper limit $2x$ to $x$. The second integral is already having $x$ as upper limit. If we apply substitution in second integral also then the upper limit changes from $x$ to $x/2$ and FTC can't be applied directly. Continued in next comment.
– Paramanand Singh
2 days ago
1
@manooooh: Using FTC and chain rule for derivatives it can be proved that if $$F(x) =int_{a} ^{g(x)} f(t) , dt$$ then $F'(x) =f(g(x)) g'(x) $. This handles the case when the upper limit of integral is not $x$ but rather some complicated function $g(x) $. Regarding your second doubt, you can check using this formula that the answer remains same even if we apply substitution in second integral.
– Paramanand Singh
2 days ago
1
@manooooh: you need to revisit substitution in definite integrals in your textbook and observe very carefully the given examples. During substitution in a definite integral the function as well as the limit of integral change according to the substitution used. Thus if we use $z=t/2, t=2z,dt=2,dz$ in second integral the limits $0$ and $x$ change to $0$ and $x/2$ to give $$int_{0}^{x}sin t^2,dt=2int_{0}^{x/2}sin (4z^2),dz$$
– Paramanand Singh
2 days ago
1
@manooooh: the limits of integral indicate the range of values being taken by the variable of integration. Thus in $int_{0}^{x}sin t^2,dt$ the variable $t$ varies from $0$ to $x$. If $z=t/2$ then $z$ should vary from $0$ to $x/2$ (as $z$ is half of $t$).
– Paramanand Singh
2 days ago
1
@manooooh: irrespective of how proficient you are in mathematics, your comments indicate that you have a sincere desire to learn. That's what matters here and nothing else. +1 for your question.
– Paramanand Singh
2 days ago
|
show 5 more comments
1
@manooooh: if you carefully note the FTC you will see that it deals with integrals where upper limit of integral is $x$ and lower limit is a constant. The substitution in first integral is done to change upper limit $2x$ to $x$. The second integral is already having $x$ as upper limit. If we apply substitution in second integral also then the upper limit changes from $x$ to $x/2$ and FTC can't be applied directly. Continued in next comment.
– Paramanand Singh
2 days ago
1
@manooooh: Using FTC and chain rule for derivatives it can be proved that if $$F(x) =int_{a} ^{g(x)} f(t) , dt$$ then $F'(x) =f(g(x)) g'(x) $. This handles the case when the upper limit of integral is not $x$ but rather some complicated function $g(x) $. Regarding your second doubt, you can check using this formula that the answer remains same even if we apply substitution in second integral.
– Paramanand Singh
2 days ago
1
@manooooh: you need to revisit substitution in definite integrals in your textbook and observe very carefully the given examples. During substitution in a definite integral the function as well as the limit of integral change according to the substitution used. Thus if we use $z=t/2, t=2z,dt=2,dz$ in second integral the limits $0$ and $x$ change to $0$ and $x/2$ to give $$int_{0}^{x}sin t^2,dt=2int_{0}^{x/2}sin (4z^2),dz$$
– Paramanand Singh
2 days ago
1
@manooooh: the limits of integral indicate the range of values being taken by the variable of integration. Thus in $int_{0}^{x}sin t^2,dt$ the variable $t$ varies from $0$ to $x$. If $z=t/2$ then $z$ should vary from $0$ to $x/2$ (as $z$ is half of $t$).
– Paramanand Singh
2 days ago
1
@manooooh: irrespective of how proficient you are in mathematics, your comments indicate that you have a sincere desire to learn. That's what matters here and nothing else. +1 for your question.
– Paramanand Singh
2 days ago
1
1
@manooooh: if you carefully note the FTC you will see that it deals with integrals where upper limit of integral is $x$ and lower limit is a constant. The substitution in first integral is done to change upper limit $2x$ to $x$. The second integral is already having $x$ as upper limit. If we apply substitution in second integral also then the upper limit changes from $x$ to $x/2$ and FTC can't be applied directly. Continued in next comment.
– Paramanand Singh
2 days ago
@manooooh: if you carefully note the FTC you will see that it deals with integrals where upper limit of integral is $x$ and lower limit is a constant. The substitution in first integral is done to change upper limit $2x$ to $x$. The second integral is already having $x$ as upper limit. If we apply substitution in second integral also then the upper limit changes from $x$ to $x/2$ and FTC can't be applied directly. Continued in next comment.
– Paramanand Singh
2 days ago
1
1
@manooooh: Using FTC and chain rule for derivatives it can be proved that if $$F(x) =int_{a} ^{g(x)} f(t) , dt$$ then $F'(x) =f(g(x)) g'(x) $. This handles the case when the upper limit of integral is not $x$ but rather some complicated function $g(x) $. Regarding your second doubt, you can check using this formula that the answer remains same even if we apply substitution in second integral.
– Paramanand Singh
2 days ago
@manooooh: Using FTC and chain rule for derivatives it can be proved that if $$F(x) =int_{a} ^{g(x)} f(t) , dt$$ then $F'(x) =f(g(x)) g'(x) $. This handles the case when the upper limit of integral is not $x$ but rather some complicated function $g(x) $. Regarding your second doubt, you can check using this formula that the answer remains same even if we apply substitution in second integral.
– Paramanand Singh
2 days ago
1
1
@manooooh: you need to revisit substitution in definite integrals in your textbook and observe very carefully the given examples. During substitution in a definite integral the function as well as the limit of integral change according to the substitution used. Thus if we use $z=t/2, t=2z,dt=2,dz$ in second integral the limits $0$ and $x$ change to $0$ and $x/2$ to give $$int_{0}^{x}sin t^2,dt=2int_{0}^{x/2}sin (4z^2),dz$$
– Paramanand Singh
2 days ago
@manooooh: you need to revisit substitution in definite integrals in your textbook and observe very carefully the given examples. During substitution in a definite integral the function as well as the limit of integral change according to the substitution used. Thus if we use $z=t/2, t=2z,dt=2,dz$ in second integral the limits $0$ and $x$ change to $0$ and $x/2$ to give $$int_{0}^{x}sin t^2,dt=2int_{0}^{x/2}sin (4z^2),dz$$
– Paramanand Singh
2 days ago
1
1
@manooooh: the limits of integral indicate the range of values being taken by the variable of integration. Thus in $int_{0}^{x}sin t^2,dt$ the variable $t$ varies from $0$ to $x$. If $z=t/2$ then $z$ should vary from $0$ to $x/2$ (as $z$ is half of $t$).
– Paramanand Singh
2 days ago
@manooooh: the limits of integral indicate the range of values being taken by the variable of integration. Thus in $int_{0}^{x}sin t^2,dt$ the variable $t$ varies from $0$ to $x$. If $z=t/2$ then $z$ should vary from $0$ to $x/2$ (as $z$ is half of $t$).
– Paramanand Singh
2 days ago
1
1
@manooooh: irrespective of how proficient you are in mathematics, your comments indicate that you have a sincere desire to learn. That's what matters here and nothing else. +1 for your question.
– Paramanand Singh
2 days ago
@manooooh: irrespective of how proficient you are in mathematics, your comments indicate that you have a sincere desire to learn. That's what matters here and nothing else. +1 for your question.
– Paramanand Singh
2 days ago
|
show 5 more comments
If you want to evaluate the limit:
$$displaystylelim_{xto 0}F(x)=lim_{xto 0}int_{x}^{2x}sin(t^2)dt$$
you can observe that $forall x>0$ (the case $x<0$ is the same), $f(t)=sin(t^2)$ is continuous in $[x,2x]$ so for the mean value theorem, exists $xi_{x}in (x,2x)$ such that
$$int_{x}^{2x}sin(t^2)dt=sin(xi_{x}^2)(2x-x)implies F(x)=sin(xi_{x}^2)x$$
Now $xi_{x}to 0$ for $xto 0^{+}$ so:
$$lim_{xto 0^{+}}F(x)=lim_{xto 0}sin(xi_{x}^2)x=[sin(0)cdot 0]=0$$
Note that this argument can be used to show that $F(x)$ is derivable for $x=0$, infact:
$$lim_{xto 0^{+}}frac{F(x)-F(0)}{x-0}=lim_{xto 0}frac{sin(xi_{x}^2)x}{x}=lim_{xto 0}sin(xi_{x}^2)=0$$
For $x_0=sqrt{frac{pi}{2}}$, the derivative of $F(x)$ can be found using the Fundamental Theorem of Integral Calculus.
$F'(x)=2sin(4x^2)-sin(x^2)implies F'left(sqrt{frac{pi}{2}}right)=2sin(4cdotfrac{pi}{2})-sin(4cdotfrac{pi}{2})=-1$
Thanks for the answer! How do you know that $F(x)=f(xi_{x})(2x-x)$? From Mean Value Theorem, we have that exists $xi_{x}in (x,2x)$ such that $F(x)=frac{F(2x)-F(x)}{2x-x}$.
– manooooh
Jan 4 at 0:49
I applied the mean value theorem for the function $f(t)=sin(t^2)$ over the intervall $[x,2x]$. So exists $xi_xin(x,2x)$ such that $int_{x}^{2x}f(t)dt=f(xi_x)(2x-x)$.
– Ixion
Jan 4 at 0:52
I do not know how do you get $f(xi_x)(2x-x)$. Since $f(x)$ is continuous at $[x,2x]$ and $f(x)$ is differentiable at $(x,2x)$ then exists $xi_xin(x,2x)$ such that $require{cancel}f'(xi_x)=dfrac{f(2x)-f(x)}{2x-x}=dfrac{sin(4x^2)-sin(2x)}xcancelimplies f(xi_x)(2x-x)$.
– manooooh
Jan 4 at 5:31
1
The mean value theorem states that: if $f(t)$ is continuous in $[a,b]$ then it exists $xiin (a,b)$ such that: $$int_{a}^{b}f(t)dt=f(xi)(b-a)$$ In this case $f(t)=sin(t^2), a=x$ and $b=2x.$
– Ixion
2 days ago
add a comment |
If you want to evaluate the limit:
$$displaystylelim_{xto 0}F(x)=lim_{xto 0}int_{x}^{2x}sin(t^2)dt$$
you can observe that $forall x>0$ (the case $x<0$ is the same), $f(t)=sin(t^2)$ is continuous in $[x,2x]$ so for the mean value theorem, exists $xi_{x}in (x,2x)$ such that
$$int_{x}^{2x}sin(t^2)dt=sin(xi_{x}^2)(2x-x)implies F(x)=sin(xi_{x}^2)x$$
Now $xi_{x}to 0$ for $xto 0^{+}$ so:
$$lim_{xto 0^{+}}F(x)=lim_{xto 0}sin(xi_{x}^2)x=[sin(0)cdot 0]=0$$
Note that this argument can be used to show that $F(x)$ is derivable for $x=0$, infact:
$$lim_{xto 0^{+}}frac{F(x)-F(0)}{x-0}=lim_{xto 0}frac{sin(xi_{x}^2)x}{x}=lim_{xto 0}sin(xi_{x}^2)=0$$
For $x_0=sqrt{frac{pi}{2}}$, the derivative of $F(x)$ can be found using the Fundamental Theorem of Integral Calculus.
$F'(x)=2sin(4x^2)-sin(x^2)implies F'left(sqrt{frac{pi}{2}}right)=2sin(4cdotfrac{pi}{2})-sin(4cdotfrac{pi}{2})=-1$
Thanks for the answer! How do you know that $F(x)=f(xi_{x})(2x-x)$? From Mean Value Theorem, we have that exists $xi_{x}in (x,2x)$ such that $F(x)=frac{F(2x)-F(x)}{2x-x}$.
– manooooh
Jan 4 at 0:49
I applied the mean value theorem for the function $f(t)=sin(t^2)$ over the intervall $[x,2x]$. So exists $xi_xin(x,2x)$ such that $int_{x}^{2x}f(t)dt=f(xi_x)(2x-x)$.
– Ixion
Jan 4 at 0:52
I do not know how do you get $f(xi_x)(2x-x)$. Since $f(x)$ is continuous at $[x,2x]$ and $f(x)$ is differentiable at $(x,2x)$ then exists $xi_xin(x,2x)$ such that $require{cancel}f'(xi_x)=dfrac{f(2x)-f(x)}{2x-x}=dfrac{sin(4x^2)-sin(2x)}xcancelimplies f(xi_x)(2x-x)$.
– manooooh
Jan 4 at 5:31
1
The mean value theorem states that: if $f(t)$ is continuous in $[a,b]$ then it exists $xiin (a,b)$ such that: $$int_{a}^{b}f(t)dt=f(xi)(b-a)$$ In this case $f(t)=sin(t^2), a=x$ and $b=2x.$
– Ixion
2 days ago
add a comment |
If you want to evaluate the limit:
$$displaystylelim_{xto 0}F(x)=lim_{xto 0}int_{x}^{2x}sin(t^2)dt$$
you can observe that $forall x>0$ (the case $x<0$ is the same), $f(t)=sin(t^2)$ is continuous in $[x,2x]$ so for the mean value theorem, exists $xi_{x}in (x,2x)$ such that
$$int_{x}^{2x}sin(t^2)dt=sin(xi_{x}^2)(2x-x)implies F(x)=sin(xi_{x}^2)x$$
Now $xi_{x}to 0$ for $xto 0^{+}$ so:
$$lim_{xto 0^{+}}F(x)=lim_{xto 0}sin(xi_{x}^2)x=[sin(0)cdot 0]=0$$
Note that this argument can be used to show that $F(x)$ is derivable for $x=0$, infact:
$$lim_{xto 0^{+}}frac{F(x)-F(0)}{x-0}=lim_{xto 0}frac{sin(xi_{x}^2)x}{x}=lim_{xto 0}sin(xi_{x}^2)=0$$
For $x_0=sqrt{frac{pi}{2}}$, the derivative of $F(x)$ can be found using the Fundamental Theorem of Integral Calculus.
$F'(x)=2sin(4x^2)-sin(x^2)implies F'left(sqrt{frac{pi}{2}}right)=2sin(4cdotfrac{pi}{2})-sin(4cdotfrac{pi}{2})=-1$
If you want to evaluate the limit:
$$displaystylelim_{xto 0}F(x)=lim_{xto 0}int_{x}^{2x}sin(t^2)dt$$
you can observe that $forall x>0$ (the case $x<0$ is the same), $f(t)=sin(t^2)$ is continuous in $[x,2x]$ so for the mean value theorem, exists $xi_{x}in (x,2x)$ such that
$$int_{x}^{2x}sin(t^2)dt=sin(xi_{x}^2)(2x-x)implies F(x)=sin(xi_{x}^2)x$$
Now $xi_{x}to 0$ for $xto 0^{+}$ so:
$$lim_{xto 0^{+}}F(x)=lim_{xto 0}sin(xi_{x}^2)x=[sin(0)cdot 0]=0$$
Note that this argument can be used to show that $F(x)$ is derivable for $x=0$, infact:
$$lim_{xto 0^{+}}frac{F(x)-F(0)}{x-0}=lim_{xto 0}frac{sin(xi_{x}^2)x}{x}=lim_{xto 0}sin(xi_{x}^2)=0$$
For $x_0=sqrt{frac{pi}{2}}$, the derivative of $F(x)$ can be found using the Fundamental Theorem of Integral Calculus.
$F'(x)=2sin(4x^2)-sin(x^2)implies F'left(sqrt{frac{pi}{2}}right)=2sin(4cdotfrac{pi}{2})-sin(4cdotfrac{pi}{2})=-1$
edited Jan 4 at 0:57
answered Jan 4 at 0:41
Ixion
778419
778419
Thanks for the answer! How do you know that $F(x)=f(xi_{x})(2x-x)$? From Mean Value Theorem, we have that exists $xi_{x}in (x,2x)$ such that $F(x)=frac{F(2x)-F(x)}{2x-x}$.
– manooooh
Jan 4 at 0:49
I applied the mean value theorem for the function $f(t)=sin(t^2)$ over the intervall $[x,2x]$. So exists $xi_xin(x,2x)$ such that $int_{x}^{2x}f(t)dt=f(xi_x)(2x-x)$.
– Ixion
Jan 4 at 0:52
I do not know how do you get $f(xi_x)(2x-x)$. Since $f(x)$ is continuous at $[x,2x]$ and $f(x)$ is differentiable at $(x,2x)$ then exists $xi_xin(x,2x)$ such that $require{cancel}f'(xi_x)=dfrac{f(2x)-f(x)}{2x-x}=dfrac{sin(4x^2)-sin(2x)}xcancelimplies f(xi_x)(2x-x)$.
– manooooh
Jan 4 at 5:31
1
The mean value theorem states that: if $f(t)$ is continuous in $[a,b]$ then it exists $xiin (a,b)$ such that: $$int_{a}^{b}f(t)dt=f(xi)(b-a)$$ In this case $f(t)=sin(t^2), a=x$ and $b=2x.$
– Ixion
2 days ago
add a comment |
Thanks for the answer! How do you know that $F(x)=f(xi_{x})(2x-x)$? From Mean Value Theorem, we have that exists $xi_{x}in (x,2x)$ such that $F(x)=frac{F(2x)-F(x)}{2x-x}$.
– manooooh
Jan 4 at 0:49
I applied the mean value theorem for the function $f(t)=sin(t^2)$ over the intervall $[x,2x]$. So exists $xi_xin(x,2x)$ such that $int_{x}^{2x}f(t)dt=f(xi_x)(2x-x)$.
– Ixion
Jan 4 at 0:52
I do not know how do you get $f(xi_x)(2x-x)$. Since $f(x)$ is continuous at $[x,2x]$ and $f(x)$ is differentiable at $(x,2x)$ then exists $xi_xin(x,2x)$ such that $require{cancel}f'(xi_x)=dfrac{f(2x)-f(x)}{2x-x}=dfrac{sin(4x^2)-sin(2x)}xcancelimplies f(xi_x)(2x-x)$.
– manooooh
Jan 4 at 5:31
1
The mean value theorem states that: if $f(t)$ is continuous in $[a,b]$ then it exists $xiin (a,b)$ such that: $$int_{a}^{b}f(t)dt=f(xi)(b-a)$$ In this case $f(t)=sin(t^2), a=x$ and $b=2x.$
– Ixion
2 days ago
Thanks for the answer! How do you know that $F(x)=f(xi_{x})(2x-x)$? From Mean Value Theorem, we have that exists $xi_{x}in (x,2x)$ such that $F(x)=frac{F(2x)-F(x)}{2x-x}$.
– manooooh
Jan 4 at 0:49
Thanks for the answer! How do you know that $F(x)=f(xi_{x})(2x-x)$? From Mean Value Theorem, we have that exists $xi_{x}in (x,2x)$ such that $F(x)=frac{F(2x)-F(x)}{2x-x}$.
– manooooh
Jan 4 at 0:49
I applied the mean value theorem for the function $f(t)=sin(t^2)$ over the intervall $[x,2x]$. So exists $xi_xin(x,2x)$ such that $int_{x}^{2x}f(t)dt=f(xi_x)(2x-x)$.
– Ixion
Jan 4 at 0:52
I applied the mean value theorem for the function $f(t)=sin(t^2)$ over the intervall $[x,2x]$. So exists $xi_xin(x,2x)$ such that $int_{x}^{2x}f(t)dt=f(xi_x)(2x-x)$.
– Ixion
Jan 4 at 0:52
I do not know how do you get $f(xi_x)(2x-x)$. Since $f(x)$ is continuous at $[x,2x]$ and $f(x)$ is differentiable at $(x,2x)$ then exists $xi_xin(x,2x)$ such that $require{cancel}f'(xi_x)=dfrac{f(2x)-f(x)}{2x-x}=dfrac{sin(4x^2)-sin(2x)}xcancelimplies f(xi_x)(2x-x)$.
– manooooh
Jan 4 at 5:31
I do not know how do you get $f(xi_x)(2x-x)$. Since $f(x)$ is continuous at $[x,2x]$ and $f(x)$ is differentiable at $(x,2x)$ then exists $xi_xin(x,2x)$ such that $require{cancel}f'(xi_x)=dfrac{f(2x)-f(x)}{2x-x}=dfrac{sin(4x^2)-sin(2x)}xcancelimplies f(xi_x)(2x-x)$.
– manooooh
Jan 4 at 5:31
1
1
The mean value theorem states that: if $f(t)$ is continuous in $[a,b]$ then it exists $xiin (a,b)$ such that: $$int_{a}^{b}f(t)dt=f(xi)(b-a)$$ In this case $f(t)=sin(t^2), a=x$ and $b=2x.$
– Ixion
2 days ago
The mean value theorem states that: if $f(t)$ is continuous in $[a,b]$ then it exists $xiin (a,b)$ such that: $$int_{a}^{b}f(t)dt=f(xi)(b-a)$$ In this case $f(t)=sin(t^2), a=x$ and $b=2x.$
– Ixion
2 days ago
add a comment |
Since $frac {sin, x} xto1 $as $x to 0$ we can find $delta >0$ such that $frac 1 2 t^{2} leqsin(t^{2})leq 2t^{2}$ for $|t| <delta$. This gives $frac 7 6 x^{3} leq F(x) leq frac {14} 3x^{3}$ for $0<x<sqrt {delta}$ and it follows easily from the definition that the right hand derivative of $F$ at $0$ is $0$. Make the substitution $s=-t$ to see that the left hand derivative is also $0$. Hence $F'(0)=0$. For $x>0$ we have $F(x)=int_0^{2x}sin(t^{2}), dt -int_0^{x}sin(t^{2}), dt$ from which it follows (by Fundamental Theorem of Calculus) that $F'(x)=2sin(4x^{2})-sin(x^{2})$. At the given point $x_0$ the derivative is $-1$.
Thanks for the answer! Why do you use "Since $frac {sin, x} xto1$ as $xto0$ (...)"?
– manooooh
Jan 4 at 6:35
1
@manooooh it makes it easy to see that $frac {F(x)} x to 0$ as $xto 0$. I can use just the definition of derivative instead of using MVT, etc.
– Kavi Rama Murthy
Jan 4 at 6:40
Ok. Could you give me the guidelines on how to use the derivative definition at a point, please? From $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x$$ I should use the fact that $left|sin t^2right|leq t^2$ so now the limit becomes $lim_{xto0}left|frac{int_x^{2x}t^2,mathrm dt}xright|$?
– manooooh
Jan 4 at 6:44
1
@manooooh Now calculate he integral of $t^{2}$ from $x$ to $2x$. You will get $7x^{3} /3$. You now see that the limit is $0$, right?
– Kavi Rama Murthy
Jan 4 at 7:17
1
@manooooh You said you have been asked to use the Fundamental Theorem of Calculus. That is what I have done in my answer. DO not try to find the derivative at $x_0$ using the definition of derivative. That is messy!
– Kavi Rama Murthy
Jan 4 at 7:30
|
show 4 more comments
Since $frac {sin, x} xto1 $as $x to 0$ we can find $delta >0$ such that $frac 1 2 t^{2} leqsin(t^{2})leq 2t^{2}$ for $|t| <delta$. This gives $frac 7 6 x^{3} leq F(x) leq frac {14} 3x^{3}$ for $0<x<sqrt {delta}$ and it follows easily from the definition that the right hand derivative of $F$ at $0$ is $0$. Make the substitution $s=-t$ to see that the left hand derivative is also $0$. Hence $F'(0)=0$. For $x>0$ we have $F(x)=int_0^{2x}sin(t^{2}), dt -int_0^{x}sin(t^{2}), dt$ from which it follows (by Fundamental Theorem of Calculus) that $F'(x)=2sin(4x^{2})-sin(x^{2})$. At the given point $x_0$ the derivative is $-1$.
Thanks for the answer! Why do you use "Since $frac {sin, x} xto1$ as $xto0$ (...)"?
– manooooh
Jan 4 at 6:35
1
@manooooh it makes it easy to see that $frac {F(x)} x to 0$ as $xto 0$. I can use just the definition of derivative instead of using MVT, etc.
– Kavi Rama Murthy
Jan 4 at 6:40
Ok. Could you give me the guidelines on how to use the derivative definition at a point, please? From $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x$$ I should use the fact that $left|sin t^2right|leq t^2$ so now the limit becomes $lim_{xto0}left|frac{int_x^{2x}t^2,mathrm dt}xright|$?
– manooooh
Jan 4 at 6:44
1
@manooooh Now calculate he integral of $t^{2}$ from $x$ to $2x$. You will get $7x^{3} /3$. You now see that the limit is $0$, right?
– Kavi Rama Murthy
Jan 4 at 7:17
1
@manooooh You said you have been asked to use the Fundamental Theorem of Calculus. That is what I have done in my answer. DO not try to find the derivative at $x_0$ using the definition of derivative. That is messy!
– Kavi Rama Murthy
Jan 4 at 7:30
|
show 4 more comments
Since $frac {sin, x} xto1 $as $x to 0$ we can find $delta >0$ such that $frac 1 2 t^{2} leqsin(t^{2})leq 2t^{2}$ for $|t| <delta$. This gives $frac 7 6 x^{3} leq F(x) leq frac {14} 3x^{3}$ for $0<x<sqrt {delta}$ and it follows easily from the definition that the right hand derivative of $F$ at $0$ is $0$. Make the substitution $s=-t$ to see that the left hand derivative is also $0$. Hence $F'(0)=0$. For $x>0$ we have $F(x)=int_0^{2x}sin(t^{2}), dt -int_0^{x}sin(t^{2}), dt$ from which it follows (by Fundamental Theorem of Calculus) that $F'(x)=2sin(4x^{2})-sin(x^{2})$. At the given point $x_0$ the derivative is $-1$.
Since $frac {sin, x} xto1 $as $x to 0$ we can find $delta >0$ such that $frac 1 2 t^{2} leqsin(t^{2})leq 2t^{2}$ for $|t| <delta$. This gives $frac 7 6 x^{3} leq F(x) leq frac {14} 3x^{3}$ for $0<x<sqrt {delta}$ and it follows easily from the definition that the right hand derivative of $F$ at $0$ is $0$. Make the substitution $s=-t$ to see that the left hand derivative is also $0$. Hence $F'(0)=0$. For $x>0$ we have $F(x)=int_0^{2x}sin(t^{2}), dt -int_0^{x}sin(t^{2}), dt$ from which it follows (by Fundamental Theorem of Calculus) that $F'(x)=2sin(4x^{2})-sin(x^{2})$. At the given point $x_0$ the derivative is $-1$.
edited Jan 4 at 7:31
answered Jan 4 at 6:14
Kavi Rama Murthy
51.4k31855
51.4k31855
Thanks for the answer! Why do you use "Since $frac {sin, x} xto1$ as $xto0$ (...)"?
– manooooh
Jan 4 at 6:35
1
@manooooh it makes it easy to see that $frac {F(x)} x to 0$ as $xto 0$. I can use just the definition of derivative instead of using MVT, etc.
– Kavi Rama Murthy
Jan 4 at 6:40
Ok. Could you give me the guidelines on how to use the derivative definition at a point, please? From $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x$$ I should use the fact that $left|sin t^2right|leq t^2$ so now the limit becomes $lim_{xto0}left|frac{int_x^{2x}t^2,mathrm dt}xright|$?
– manooooh
Jan 4 at 6:44
1
@manooooh Now calculate he integral of $t^{2}$ from $x$ to $2x$. You will get $7x^{3} /3$. You now see that the limit is $0$, right?
– Kavi Rama Murthy
Jan 4 at 7:17
1
@manooooh You said you have been asked to use the Fundamental Theorem of Calculus. That is what I have done in my answer. DO not try to find the derivative at $x_0$ using the definition of derivative. That is messy!
– Kavi Rama Murthy
Jan 4 at 7:30
|
show 4 more comments
Thanks for the answer! Why do you use "Since $frac {sin, x} xto1$ as $xto0$ (...)"?
– manooooh
Jan 4 at 6:35
1
@manooooh it makes it easy to see that $frac {F(x)} x to 0$ as $xto 0$. I can use just the definition of derivative instead of using MVT, etc.
– Kavi Rama Murthy
Jan 4 at 6:40
Ok. Could you give me the guidelines on how to use the derivative definition at a point, please? From $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x$$ I should use the fact that $left|sin t^2right|leq t^2$ so now the limit becomes $lim_{xto0}left|frac{int_x^{2x}t^2,mathrm dt}xright|$?
– manooooh
Jan 4 at 6:44
1
@manooooh Now calculate he integral of $t^{2}$ from $x$ to $2x$. You will get $7x^{3} /3$. You now see that the limit is $0$, right?
– Kavi Rama Murthy
Jan 4 at 7:17
1
@manooooh You said you have been asked to use the Fundamental Theorem of Calculus. That is what I have done in my answer. DO not try to find the derivative at $x_0$ using the definition of derivative. That is messy!
– Kavi Rama Murthy
Jan 4 at 7:30
Thanks for the answer! Why do you use "Since $frac {sin, x} xto1$ as $xto0$ (...)"?
– manooooh
Jan 4 at 6:35
Thanks for the answer! Why do you use "Since $frac {sin, x} xto1$ as $xto0$ (...)"?
– manooooh
Jan 4 at 6:35
1
1
@manooooh it makes it easy to see that $frac {F(x)} x to 0$ as $xto 0$. I can use just the definition of derivative instead of using MVT, etc.
– Kavi Rama Murthy
Jan 4 at 6:40
@manooooh it makes it easy to see that $frac {F(x)} x to 0$ as $xto 0$. I can use just the definition of derivative instead of using MVT, etc.
– Kavi Rama Murthy
Jan 4 at 6:40
Ok. Could you give me the guidelines on how to use the derivative definition at a point, please? From $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x$$ I should use the fact that $left|sin t^2right|leq t^2$ so now the limit becomes $lim_{xto0}left|frac{int_x^{2x}t^2,mathrm dt}xright|$?
– manooooh
Jan 4 at 6:44
Ok. Could you give me the guidelines on how to use the derivative definition at a point, please? From $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x$$ I should use the fact that $left|sin t^2right|leq t^2$ so now the limit becomes $lim_{xto0}left|frac{int_x^{2x}t^2,mathrm dt}xright|$?
– manooooh
Jan 4 at 6:44
1
1
@manooooh Now calculate he integral of $t^{2}$ from $x$ to $2x$. You will get $7x^{3} /3$. You now see that the limit is $0$, right?
– Kavi Rama Murthy
Jan 4 at 7:17
@manooooh Now calculate he integral of $t^{2}$ from $x$ to $2x$. You will get $7x^{3} /3$. You now see that the limit is $0$, right?
– Kavi Rama Murthy
Jan 4 at 7:17
1
1
@manooooh You said you have been asked to use the Fundamental Theorem of Calculus. That is what I have done in my answer. DO not try to find the derivative at $x_0$ using the definition of derivative. That is messy!
– Kavi Rama Murthy
Jan 4 at 7:30
@manooooh You said you have been asked to use the Fundamental Theorem of Calculus. That is what I have done in my answer. DO not try to find the derivative at $x_0$ using the definition of derivative. That is messy!
– Kavi Rama Murthy
Jan 4 at 7:30
|
show 4 more comments
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1
Would it help if you write $int_x^{2x}=int_0^{2x}-int_0^x$?
– A.Γ.
Jan 3 at 23:59
@A.Γ. probably. Can we separate it into two integrals because the limits of integrations are continuous?
– manooooh
Jan 4 at 0:01
It is because the integral is additive. Or if you wish $int_a^b=F(b)-F(a)$ where $F(y)$ is antiderivative, which is e.g. $int_0^y$.
– A.Γ.
Jan 4 at 0:05
@A.Γ. how do you know that $0in[x,2x]$ for all $xneq0$? Because for example if $xin(0,infty)$ then $0notin[x,2x]$.
– manooooh
Jan 4 at 0:29
1
One does not need $0in[x,2x]$. Integral is additive and if $f$ is Riemann integrable on some closed interval containing $a, b, c$ then $$int_{a} ^{b} f(x) , dx+int_{b} ^{c} f(x) , dx=int_{a} ^{c} f(x) , dx$$ irrespective of the linear order of $a, b, c$. This assumes that we have by definition $int_{a} ^{a}f(x),dx=0$ and $int_{b} ^{a} f(x) , dx=-int_{a} ^{b} f(x) , dx$.
– Paramanand Singh
Jan 4 at 7:23