21b_integration

21b_integration - 21B INTEGR ATION The first step in all...

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21B INTEGRATION The first step in all integration problems is to determine which type of problem it is… algebra (no exponentials (e's), logs or trig), trig or 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. Algebra Trig Mixture * Can you make it an exponents problem? * Does the derivative of one part = the other part? * Does the derivative of one part = the other part? * Does the derivative of one part equal the other part? * Tricks… * Integration by Parts (LATE) * Tricks… Long division Partial Fractions Complete the square Trig substitution Algebraic Substitution I. Algebra A. Can you make it an exponent problem? [Any necessary quantities in parentheses are raised to a positive whole number power.] By an exponent problem, I mean any expression which may be written as the sums and/or differences of terms of the form ax n where a and n are constants. [e.g. 10 x 1 2 ! 3 x ! 2 + 4 x ! 2 3 ] For example: (2x+3) 2 = 4 x 2 + 12 x + 9 3 x 2 + 1 2 x 2 = 3 x 2 + 1 ( ) ! 1 2 x " 2 = 3 2 + 1 2 x " 2 are each written in the desired form. Note that any terms in parentheses are raised to a whole number power. These functions may now be integrated by the power rule. 3 x 2 + 1 2 x 2 ! dx = 3 2 + 1 2 x " 2 ! dx = 3 2 x " 1 2 x " 1 + C Integrands like 3 x + 1 = (3 x + 1) 1 2 and 1 ( x 2 + 3) 2 = x 2 + 3 ( ) ! 2 require parentheses with powers that are not positive whole numbers and therefore cannot be written as exponent problems. B. Does the derivative of one part = the other part? [ Substitution - The degree of one part is one more 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 derivative of one part is to equal the other part, it is clear that the powers must differ by one. For example the degree of 4x 2 + 2x +13 and that of 8x + 2 differ by one (2 and 1 respectively) so in this case try letting u = 4x 2 + 2x - 13, the one with the larger power. Since du = (8x + 2) dx we are in luck.
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8 x + 2 (4 x 2 + 2 x + 13) 3 ! dx = du u 3 ! = so. .. u ! 3 " du = ! 1 2 u ! 2 + C = ! 1
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21b_integration - 21B INTEGR ATION The first step in all...

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