# definite integral

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• May 4th 2010, 11:04 AM
JGaraffa
definite integral
a little confused on this one

"solve the definite integral"

$\displaystyle \int^2_0 2x^2e^{x^3+5}$
• May 4th 2010, 11:08 AM
Failure
Quote:

Originally Posted by JGaraffa
a little confused on this one

"solve the definite integral"

$\displaystyle \int^2_0 2x^2e^{x^3+5}$

Well, again, use substitution $\displaystyle z := x^3+5$, which gives that $\displaystyle dz=3x^2\,dx$ and hence you have

$\displaystyle \int 2x^2e^{x^3+5}\,dx=\tfrac{2}{3}\cdot\int e^{x^3+5}\cdot 3x^2\, dx=\tfrac{2}{3}\cdot\int z\,dz=\ldots$
• May 4th 2010, 11:10 AM
piglet
Quote:

Originally Posted by JGaraffa
a little confused on this one

"solve the definite integral"
$\displaystyle \int^2_0 2x^2e^{x^3+5}$

What exactly are you confused about? Have you made an attempt at it?

Also the integral should be $\displaystyle \int^2_0 2x^2e^{x^3+5}dx$

I could tell you how to do it but i think i it's only fair that an attempt is made first

Hint: use substitution
• May 4th 2010, 11:11 AM
JGaraffa
ah perfect! i couldn't figure out how i could change that $\displaystyle 2x^2$ to a $\displaystyle 3x^2$ . thanks!
• May 4th 2010, 11:31 AM
JGaraffa
actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.

$\displaystyle \int^2_0 2x^2e^{x^3+5}$

u= $\displaystyle x^3 + 5$
h(u) = $\displaystyle 3x^2$

$\displaystyle \frac{2}{3} \int^2_0 3x^2 e^{x^3+5}$

when x = 0, u = 5
when x = 2, u = 13

$\displaystyle \frac{2}{3} \int^{13}_5 e^u du$
$\displaystyle \frac{2}{3} e^{13} - \frac{2}{3} e^5$

i'm getting 294,843.32 which makes me think i'm doing something wrong.
• May 4th 2010, 11:43 AM
Failure
Quote:

Originally Posted by JGaraffa
actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.

$\displaystyle \int^2_0 2x^2e^{x^3+5}$

u= $\displaystyle x^3 + 5$
h(u) = $\displaystyle 3x^2$

$\displaystyle \frac{2}{3} \int^2_0 3x^2 e^{x^3+5}$

when x = 0, u = 5
when x = 2, u = 13

$\displaystyle \frac{2}{3} \int^{13}_5 e^u du$
$\displaystyle \frac{2}{3} e^{13} - \frac{2}{3} e^5$

i'm getting 294,843.32 which makes me think i'm doing something wrong.

No, that result is ok. Of course, $\displaystyle e^{x^3+5}$ is a fast growing function.
• May 4th 2010, 11:47 AM
surjective
Quote:

Originally Posted by JGaraffa
actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.

$\displaystyle \int^2_0 2x^2e^{x^3+5}$

u= $\displaystyle x^3 + 5$
h(u) = $\displaystyle 3x^2$

$\displaystyle \frac{2}{3} \int^2_0 3x^{2} e^{x^3+5}$

when x = 0, u = 5
when x = 2, u = 13

$\displaystyle \frac{2}{3} \int^{13}_5 e^u du$
$\displaystyle \frac{2}{3} e^{13} - \frac{2}{3} e^5$

i'm getting 294,843.32 which makes me think i'm doing something wrong.

Apart from [tex]3x^{2} it lookes fine to me.
• May 4th 2010, 11:27 PM
CaptainBlack
Quote:

Originally Posted by JGaraffa
a little confused on this one

"solve the definite integral"

$\displaystyle \int^2_0 2x^2e^{x^3+5}$

Again observe that $\displaystyle 3x^2$ is the derivative of $\displaystyle x^3+5$, so consider:

$\displaystyle \frac{d}{dx}e^{x^3+5}$

CB