Finding the lower and upper Riemann Sums
I'm having a hard time figuring this questions out. I've looked on google and on the book and so far I haven't gotten a good explanation for this questions. I know they aren't hard and are probably easy but I've haven't gotten a good explanation.
Find the lower and Upper Riemann Sums $L(P)$ and $U(P)$ for the function $f(x)=x^2$ on the interval $[0,1]$ using the partition $P={0,frac12,
frac34,1}$Given that $int_1^3 f(x)~dx=4$ and $int_1^5 f(x)~dx=7$,
find $int_3^5 f(x)~dx$.Given that $int_1^3 f(x)~dx=4$ and $int_1^3 g(x)~dx=2$,
find $int_1^3 3f(x)-g(x)~dx$.
definite-integrals riemann-sum
edited Jul 8 '16 at 1:34
Frentos
2,5151622
asked Nov 29 '15 at 6:15
Atif ShahAtif Shah
12
add a comment |
I'm having a hard time figuring this questions out. I've looked on google and on the book and so far I haven't gotten a good explanation for this questions. I know they aren't hard and are probably easy but I've haven't gotten a good explanation.
Find the lower and Upper Riemann Sums $L(P)$ and $U(P)$ for the function $f(x)=x^2$ on the interval $[0,1]$ using the partition $P={0,frac12,
frac34,1}$Given that $int_1^3 f(x)~dx=4$ and $int_1^5 f(x)~dx=7$,
find $int_3^5 f(x)~dx$.Given that $int_1^3 f(x)~dx=4$ and $int_1^3 g(x)~dx=2$,
find $int_1^3 3f(x)-g(x)~dx$.
definite-integrals riemann-sum
edited Jul 8 '16 at 1:34
Frentos
2,5151622
asked Nov 29 '15 at 6:15
Atif ShahAtif Shah
12
add a comment |
I'm having a hard time figuring this questions out. I've looked on google and on the book and so far I haven't gotten a good explanation for this questions. I know they aren't hard and are probably easy but I've haven't gotten a good explanation.
Find the lower and Upper Riemann Sums $L(P)$ and $U(P)$ for the function $f(x)=x^2$ on the interval $[0,1]$ using the partition $P={0,frac12,
frac34,1}$Given that $int_1^3 f(x)~dx=4$ and $int_1^5 f(x)~dx=7$,
find $int_3^5 f(x)~dx$.Given that $int_1^3 f(x)~dx=4$ and $int_1^3 g(x)~dx=2$,
find $int_1^3 3f(x)-g(x)~dx$.
definite-integrals riemann-sum
edited Jul 8 '16 at 1:34
Frentos
2,5151622
asked Nov 29 '15 at 6:15
Atif ShahAtif Shah
12
I'm having a hard time figuring this questions out. I've looked on google and on the book and so far I haven't gotten a good explanation for this questions. I know they aren't hard and are probably easy but I've haven't gotten a good explanation.
Find the lower and Upper Riemann Sums $L(P)$ and $U(P)$ for the function $f(x)=x^2$ on the interval $[0,1]$ using the partition $P={0,frac12,
frac34,1}$Given that $int_1^3 f(x)~dx=4$ and $int_1^5 f(x)~dx=7$,
find $int_3^5 f(x)~dx$.Given that $int_1^3 f(x)~dx=4$ and $int_1^3 g(x)~dx=2$,
find $int_1^3 3f(x)-g(x)~dx$.
definite-integrals riemann-sum
definite-integrals riemann-sum
edited Jul 8 '16 at 1:34
Frentos
2,5151622
asked Nov 29 '15 at 6:15
Atif ShahAtif Shah
12
edited Jul 8 '16 at 1:34
Frentos
2,5151622
asked Nov 29 '15 at 6:15
Atif ShahAtif Shah
12
edited Jul 8 '16 at 1:34
Frentos
2,5151622
edited Jul 8 '16 at 1:34
Frentos
2,5151622
edited Jul 8 '16 at 1:34
Frentos
2,5151622
2,5151622
asked Nov 29 '15 at 6:15
Atif ShahAtif Shah
12
asked Nov 29 '15 at 6:15
Atif ShahAtif Shah
12
asked Nov 29 '15 at 6:15
Atif ShahAtif Shah
12
12
add a comment |
add a comment |
1 Answer
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Just so this isn't hanging around unanswered forever.
Choose the minimum function value over each interval for $L(P)$, and the maximum for $U(P)$: $L(P)=0^2(frac12)+(frac12)^2(frac14)+(frac34)^2(frac14) = frac{13}{64}. U(P)=(frac12)^2(frac12)+(frac34)^2(frac14)+1^2(frac14)=frac{33}{64}$
$int_3^5 f(x)~dx = int_1^5f(x)~dx - int_1^3 f(x)~dx = 7 - 4 = 3$
$int_1^3 3f(x)-g(x)~dx =int_1^3 3f(x)-int_1^3g(x)~dx =3int_1^3 f(x)-int_1^3g(x)~dx = 3(4)-2=10$.
answered Jul 8 '16 at 1:47
FrentosFrentos
2,5151622
add a comment |
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1 Answer
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1 Answer
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Just so this isn't hanging around unanswered forever.
Choose the minimum function value over each interval for $L(P)$, and the maximum for $U(P)$: $L(P)=0^2(frac12)+(frac12)^2(frac14)+(frac34)^2(frac14) = frac{13}{64}. U(P)=(frac12)^2(frac12)+(frac34)^2(frac14)+1^2(frac14)=frac{33}{64}$
$int_3^5 f(x)~dx = int_1^5f(x)~dx - int_1^3 f(x)~dx = 7 - 4 = 3$
$int_1^3 3f(x)-g(x)~dx =int_1^3 3f(x)-int_1^3g(x)~dx =3int_1^3 f(x)-int_1^3g(x)~dx = 3(4)-2=10$.
answered Jul 8 '16 at 1:47
FrentosFrentos
2,5151622
add a comment |
Just so this isn't hanging around unanswered forever.
Choose the minimum function value over each interval for $L(P)$, and the maximum for $U(P)$: $L(P)=0^2(frac12)+(frac12)^2(frac14)+(frac34)^2(frac14) = frac{13}{64}. U(P)=(frac12)^2(frac12)+(frac34)^2(frac14)+1^2(frac14)=frac{33}{64}$
$int_3^5 f(x)~dx = int_1^5f(x)~dx - int_1^3 f(x)~dx = 7 - 4 = 3$
$int_1^3 3f(x)-g(x)~dx =int_1^3 3f(x)-int_1^3g(x)~dx =3int_1^3 f(x)-int_1^3g(x)~dx = 3(4)-2=10$.
answered Jul 8 '16 at 1:47
FrentosFrentos
2,5151622
add a comment |
Just so this isn't hanging around unanswered forever.
Choose the minimum function value over each interval for $L(P)$, and the maximum for $U(P)$: $L(P)=0^2(frac12)+(frac12)^2(frac14)+(frac34)^2(frac14) = frac{13}{64}. U(P)=(frac12)^2(frac12)+(frac34)^2(frac14)+1^2(frac14)=frac{33}{64}$
$int_3^5 f(x)~dx = int_1^5f(x)~dx - int_1^3 f(x)~dx = 7 - 4 = 3$
$int_1^3 3f(x)-g(x)~dx =int_1^3 3f(x)-int_1^3g(x)~dx =3int_1^3 f(x)-int_1^3g(x)~dx = 3(4)-2=10$.
answered Jul 8 '16 at 1:47
FrentosFrentos
2,5151622
Just so this isn't hanging around unanswered forever.
Choose the minimum function value over each interval for $L(P)$, and the maximum for $U(P)$: $L(P)=0^2(frac12)+(frac12)^2(frac14)+(frac34)^2(frac14) = frac{13}{64}. U(P)=(frac12)^2(frac12)+(frac34)^2(frac14)+1^2(frac14)=frac{33}{64}$
$int_3^5 f(x)~dx = int_1^5f(x)~dx - int_1^3 f(x)~dx = 7 - 4 = 3$
$int_1^3 3f(x)-g(x)~dx =int_1^3 3f(x)-int_1^3g(x)~dx =3int_1^3 f(x)-int_1^3g(x)~dx = 3(4)-2=10$.
answered Jul 8 '16 at 1:47
FrentosFrentos
2,5151622
answered Jul 8 '16 at 1:47
FrentosFrentos
2,5151622
answered Jul 8 '16 at 1:47
FrentosFrentos
2,5151622
answered Jul 8 '16 at 1:47
FrentosFrentos
2,5151622
2,5151622
add a comment |
add a comment |
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