a little confused on this one

"solve the definite integral"

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

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- May 4th 2010, 11:04 AMJGaraffadefinite 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 AMFailure
- May 4th 2010, 11:10 AMpiglet
- May 4th 2010, 11:11 AMJGaraffa
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 AMJGaraffa
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 AMFailure
- May 4th 2010, 11:47 AMsurjective
- May 4th 2010, 11:27 PMCaptainBlack