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Integral of a Power

Date: 9/8/96 at 21:51:56
From: Chris Cummings
Subject: Integral of a Power

I've forgotten a lot about calculus. Could you explain why

integral from -1 to 2 of x^2/3   =   x^3/9 |from -1 to 2


Date: 9/9/96 at 20:37:32
From: Doctor Tom
Subject: Re: Integral of a Power

Well, I clearly can't teach you a course on integration, but here's
basically what you need to do:

1) Find the indefinite integral of what's inside the integration.
The indefinite integral is some function whose derivative is equal
to the stuff inside.

In this case, x^3/9 works, since the derivative of x^3/9 is x^2/3.

2) Plug in the value at the top of the integral sign and the value
at the bottom of the integral sign into the indefinite integral and
subtract the lower from the upper.  If I plug in 2, I get 2^3/9 = 8/9
and if I plug in -1, I get (-1)^3/9 = -1/9

So the answer (the area under the curve of x^2/3 from -1 to 3) is
given by 8/9 - (-1/9) = 9/9 = 1.

-Doctor Tom,  The Math Forum
 Check out our web site!   

Date: 9/9/96 at 22:12:23
From: Chris Cummings
Subject: Re: Integral of a Power

Thanks, Doc Tom, for the quick integration lesson... it helped a lot!
Associated Topics:
High School Calculus

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