Integration by PartsDate: 15 Feb 1995 17:08:26 -0500 From: Courtois Subject: Help!!! Since most people did calculus in high school, I assume this question you folks will be able to answer. I've been working all day on it, and no luck. I need the integral from 0 to 1 of X^4 * exp(-x/2). Can you help? Stuck in Seattle Date: 15 Feb 1995 22:10:07 -0500 From: Dr. Ken Subject: Re: Help!!! Hello there! This is a classic example of a problem that can be solved by using the technique of Integration by Parts. But first, let's make the problem a little bit nicer by getting rid of the -x/2. Let's let y = -x/2 (i.e. x = -2y), and then we have dx = -2dy. So we have the integral from 0 to -1/2 of 16y^4 * e^y dy. We can pull the 16 outside of the integral. So now we've got 16 times the integral from 0 to -1/2 of y^4 * e^y dy. Using Integration by Parts, let u = y^4 and let dv = e^y dy. So then du = 4y^3 dy, and v = e^y. Remember the formula for integration by parts? It's Integral(u dv) = uv - Integral(v du). So we get 16 * Integral(y^4 e^y dy) = 16*y^4 e^y - 16*Integral(4y^3 e^y dy) = 16/(16Sqrt{e}) - 64*Integral(y^3 e^y dy)) = 1/Sqrt{e} - 64*Integral(y^3 e^y dy). ***note: all integrals are between 0 and -1/2*** Notice that we've knocked the power down on the y^4 so that it became y^3. You can apply this method again and knock the power down another notch, and another, until you get the integral of e^y. Which you can do, because it's e^y. I hope this makes sense to you. -Ken "Dr." Math |
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