# definite integral

• May 4th 2010, 12:04 PM
JGaraffa
definite integral
a little confused on this one

"solve the definite integral"

$\int^2_0 2x^2e^{x^3+5}$
• May 4th 2010, 12:08 PM
Failure
Quote:

Originally Posted by JGaraffa
a little confused on this one

"solve the definite integral"

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

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

$\int 2x^2e^{x^3+5}\,dx=\tfrac{2}{3}\cdot\int e^{x^3+5}\cdot 3x^2\, dx=\tfrac{2}{3}\cdot\int z\,dz=\ldots$
• May 4th 2010, 12:10 PM
piglet
Quote:

Originally Posted by JGaraffa
a little confused on this one

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

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

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

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, 12:11 PM
JGaraffa
ah perfect! i couldn't figure out how i could change that $2x^2$ to a $3x^2$ . thanks!
• May 4th 2010, 12:31 PM
JGaraffa
actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.

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

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

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

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

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

\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, 12:43 PM
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.

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

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

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

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

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

\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, $e^{x^3+5}$ is a fast growing function.
• May 4th 2010, 12:47 PM
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.

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

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

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

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

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

\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 5th 2010, 12:27 AM
CaptainBlack
Quote:

Originally Posted by JGaraffa
a little confused on this one

"solve the definite integral"

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

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

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

CB