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 Thanks!
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! http://mathforum.org/dr.math/
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!
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