integration by parts revision sheet

By Tammy Nelson,2014-07-05 09:37
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integration by parts revision sheet

Integration by parts is used to integrate a product. The formula (given in formula book): One part of the product (u) should be easy to dvdudifferentiate (and will usually simplify when udxuvvdx,?~~dxdxdv(?differentiated). The other part should be easy ;? dx,?The other part One part is to integrate and should not become too much harder is integrated. differentiated when integrated.

xCommon examination questions Definite integrals (using by parts) 2xedxExample 1: Find . ~ Example: a) Find the points where the graph of xxdxlnExample 1: Find . ~This is a suitable candidate for integration by parts x yxe,?(2) cuts the x and y axes. dvxThis can be found using integration by parts if we xwith : 2anduxe,,yxe,?(2)b) Sketch the graph of . dvdxc) Find the area of the region between the axes and take . uxx,,lnandduxdxyxe,?(2)22ux,,,the graph of . dxdu1 ux,,,lndvxxa) Graph cuts y-axis when x = 0, i.e. at y = 2 ,,,evedxx 2dxGraph cuts the x-axis when y = 0, i.e. when x = 2.. dvx,,,xvSubstitute these into the formula: ydx2xxxxxb) The graph looks like: 322222xedxxeedxxeec,?,?? Substitute these into the formula: ~~ 2.5222 x1x2211x dxxxxdx,?lnxxdxxlnln,?22c) Area is . (2)xedx~~~~1.52xxdxcos2Example 2: Find . x~022111 ,??xxxcln 24du0.5,?,,?21uxdv Here we take : uxx,,andcosdxx0.511.52 dxdvlnxdxExample 2: Find . ??xx~,,,?evedu dxux,,,1???xxx1lnxdxThis can be thought of as and so can be dx(2)(2)()1?,????!?xedxxeedx~So ~~dv,,,cossinxvxdv??xx???(2)()xeedx = integrated by parts with ux,,lnand1~dxdx??xxSubstitute these into the formula: ???(2)()xee = du1xxdxxxxdxxxxccossinsinsin(cos),?,???ux,,,lnNow that we’ve integrated, we substitute in our ~~ dxx limits: dv,??xxxcsincos2,,,1vx2???xxx :?(2)(2)?,???xedxxeedx~;(0Note: Sometimes it is necessary to use the 01??2200 xdxxxdx,?ln1lnlnxdxxx,? ,?????02eeeeintegration by parts formula twice (e.g. with ;;;;~~~x2,?,0.13511.135xxdxsin). xxxcln??~ =

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