Comments on: Calculus extra credit problem using the rules only of substitution by parts and substitution? http://www.safedebthelp.com/blog/calculus-extra-credit-problem-using-the-rules-only-of-substitution-by-parts-and-substitution/ Fri, 18 May 2012 01:10:53 +0000 http://wordpress.org/?v=2.7 hourly 1 By: jdrawz4 http://www.safedebthelp.com/blog/calculus-extra-credit-problem-using-the-rules-only-of-substitution-by-parts-and-substitution/comment-page-1/#comment-3094 jdrawz4 Tue, 20 Jul 2010 00:30:00 +0000 http://www.safedebthelp.com/blog/calculus-extra-credit-problem-using-the-rules-only-of-substitution-by-parts-and-substitution/#comment-3094 Hi. I assume when you say subsitution by parts you are obviously referring to what my AP Calculus BC class calls integration by parts which is in the form of uv - ∫v du. For your problem --> ∫ x^3(x^2-1)^10 Set u = x^2 where du = 2x dx And dv = x (x^2-1)^10 where you integrate to get v = [½ (x^2-1)^11]/11 or v = [(x^2-1)^11]/22 *Notice one power of x from the term x^3 of the original equation is separated and placed in the expression dv rather than u Now set up your integration by parts as uv - ∫v du where: x^2 [(x^2-1)^11]/22 - ∫ [(x^2-1)^11]/22 * 2xdx (Integrate the second expression now by using u = x^2 and du = 2xdx) So ∫ x^3(x^2-1)10 = x^2 [(x^2-1)^11]/22 - [(x^2-1)^12]/264<a href="http://www.resumeminers.com/online-degree/major-education-schools.htm"> jdrawz4</a> Hi. I assume when you say subsitution by parts you are obviously referring to what my AP Calculus BC class calls integration by parts which is in the form of uv - ∫v du.

For your problem –> ∫ x^3(x^2-1)^10

Set u = x^2 where du = 2x dx
And dv = x (x^2-1)^10 where you integrate to get v = [½ (x^2-1)^11]/11 or v = [(x^2-1)^11]/22

*Notice one power of x from the term x^3 of the original equation is separated and placed in the expression dv rather than u

Now set up your integration by parts as uv - ∫v du where:

x^2 [(x^2-1)^11]/22 - ∫ [(x^2-1)^11]/22 * 2xdx (Integrate the second expression now by using u = x^2 and du = 2xdx)

So ∫ x^3(x^2-1)10 = x^2 [(x^2-1)^11]/22 - [(x^2-1)^12]/264 jdrawz4

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