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Unformatted text preview: MTH 122 — Calculus II Essex County College — Division of Mathematics and Physics 1 Lecture Notes #3 — Sakai Web Project Material 1 Integration by Parts Let’s start with an example where we’re asked to verify: Z xe x d x = xe x e x + C. As you know, we basically just have to check that: d d x ( xe x e x + C ) = xe x . Here goes: d d x ( xe x e x + C ) = d d x ( xe x ) d d x ( e x ) + d d x ( C ) = ( xe x + e x ) ( e x ) + (0) = xe x The main problem here is not the checking, but it is the actual process of finding the antideriva tive. So far we’ve only used one real method, that was usubstitution, and although it is an important method it would fail miserably in this particular case. Another technique known as integration by parts uses the product rule . As you recall, the product rule is: d d x ( uv ) = u v + uv , where u and v are functions of x and u and v are the derivatives with respect to x . We can rewrite this product rule as follows: uv = d d x ( uv ) u v, and then integrate both sides,...
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 Spring '10
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 Calculus, Division, Derivative, Integration By Parts, dx

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