(e^x/10 - x^2 - 1)dx {5

{-5

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- Jul 1st 2008, 09:04 PM #1

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- Jul 1st 2008, 09:07 PM #2
With my super amazing mind-reading ablilities...this reads

$\displaystyle \int_{-5}^{5}e^{\frac{x}{10}}-x^2-1~~dx$

Well I will compute the antiderivative and you do the definite part

$\displaystyle \int{e^{\frac{x}{10}}~dx}=10\int{u}~du=10e^{\frac{ x}{10}}$

Now what substitution did I make?

$\displaystyle \int{x^2}~dx=\frac{x^3}{3}\longleftarrow\text{no surprises here}$

$\displaystyle \int~dx=x\longleftarrow\text{none here either}$

- Jul 1st 2008, 09:10 PM #3

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- Jul 1st 2008, 09:18 PM #4

- Jul 1st 2008, 09:23 PM #5

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- Jul 1st 2008, 09:26 PM #6

- Jul 1st 2008, 10:51 PM #7

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