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Unformatted text preview: x for “u” and x 2 dx for “dv”. The derivative for e x is e x , so this is an acceptable choice for “u”. This leaves us with x 2 dx for “dv”. Next we integrate x 2 dx we end up with 1/3 x 3 dx. When we proceed with our problem by following the formula of: udv = uv vdu. We find that the power of x has increased, therefore making our problem more difficult than it originally was. This is a poor choice for “dv”. If we choose x 2 for “u” and e x dx for “dv” then the derivative of x 2 is 2x and the integral of e x dx is e x . So when we integrate by parts, our problem is easier after we apply the formula....
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This note was uploaded on 09/20/2008 for the course MATH 151 taught by Professor Bray during the Spring '07 term at Mesa CC.
 Spring '07
 Bray
 Integration By Parts

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