$\displaystyle

\int_{0}^1 (x^2+e^x-6)\, dx

$

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- Apr 14th 2009, 06:51 AMfoodmmmdefinite integral help
$\displaystyle

\int_{0}^1 (x^2+e^x-6)\, dx

$ - Apr 14th 2009, 06:57 AMCalculus26solution
http://www.mathhelpforum.com/math-he...0a183b23-1.gif = x^3/3 +e^x -6x = 1/3+e -6 -1

- Apr 14th 2009, 07:00 AMapcalculus

Find the antiderivative, without the constant C. You can either do it in one step or separate the function into three terms.

The antiderivative of x^2 is x^3 / 3

The antiderivative of e^x is e^x

The antiderivative of -6 is -6x

Evaluate at 1

Evaluate at 0

Subtract the answer at 0 from the answer at 1.

Good luck! - Apr 14th 2009, 07:00 AMfoodmmmthanks
thank you so much! i appreciate it.

- Apr 14th 2009, 07:03 AMmr fantastic
Read this: Table of Integrals

All these should be in your textbook and/or class notes.