21bintegralquestions - INTEGRATION MATH 21 B The first step...

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INTEGRATION MATH 21 BThe first step in all integration problems is to determine which type of problem it is . . .algebra(noexponentials (e's), logs or trig), trigor mixture(some combination of algebra, exponentials (e's), logs and trig).The following outline suggests an orderly approach to deciding which integration technique to use.I. AlgebraA. Can you make it an exponent problem? [Any necessary quantities in parentheses are raised toa whole number power.]
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B. Does the derivative of one part = the other part? [Substitution - The degree of one part is onemore than the degree of the other part. Let u = the part with the higher degree.]Taking the derivative of an expression lowers the degree of the expression by one. Therefore, if the derivativeof one part is to equal the other part, it is clear that the powers must differ by one.For example the degree of 4x2+ 2x - 13 and that of 8x + 2 differ by one (2 and 1 respectively) so in this casetry letting u = 4x2+ 2x - 13, the one with the larger power. Since du = (8x + 2)dx we are in luck.8x + 2(4x2+ 2x - 13)3dx = duu3= u-3du = u-2-2+ C = 12(4x2+ 2x - 13) -2+ CIf du can be obtained by multiplying the "other part," by a constant, this still works. In this case, multiply theother part by the necessary constant and multiply the integral by the reciprocal of the constant. This results indu appearing in the integrand.For example: x(x2+ 4)10dx Note that the degree of x and x2+ 4, we can choose to ignore the exponent 10for now, differ by one. Let u = x2+ 4 , then du = 2xdx which is not exactly in the problem. Note that
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x(x2+ 4)10dx 122x(x2+ 4)10dx 12
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