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Section 6.2: Integration by substitution

Section 6.2: Integration by substitution. When Substitution is Just Easier. 1. Find: ∫ 16(2x+5) 3 dx. You could FOIL it out, but why waste your time doing all that work?. 0. u =2x+5 Determine an expression for u. du = 2dx Differentiate u.

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Section 6.2: Integration by substitution

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  1. Section 6.2: Integration by substitution

  2. When Substitution is Just Easier 1 Find: ∫ 16(2x+5)3 dx You could FOIL it out, but why waste your time doing all that work? 0 u=2x+5 Determine an expression for u. du = 2dx Differentiate u. ∫ 8u3 du Write the integral in terms of u. 2u4 Find the antiderivative. Notice that for the sake of keeping the problem simple, we temporarily neglected the bounds of integration. However, our last step requires us to evaluate on these bounds because this is a definite integral. Because we integrated du and not dx, we need to put our bounds of integration in terms of u. Alternatively, we could make the final equation in terms of x by substitution. u(1)=2(1)+5=7 u(0)=2(0)+5=5 2(7)4 - 2(5)4= 3552 Evaluate and simplify.

  3. Trigonometry Find: ∫cos(x)sin2(x) dx u= sin x Determine an expression for u. du = cos x dx Differentiate u. ∫ u2 du Write the integral in terms of u. (1/3)u3 Find the antiderivative. For an indefinite integral, the final equation must be put in terms of x, not u. (1/3) sin3 (x) Use substitution and put equation in terms of x. (1/3) sin3 (x) + CAdd an integration constant.

  4. Exponential Functions 1 Find: ∫ 4xe2x2dx 0 u= 2x2 Determine an expression for u. du = 4x dx Differentiate u. ∫ eu du Write the integral in terms of u. eu Find the antiderivative. u(1)=2(1)2=2 Find u-values for bounds • u(0)=2(0)2=0 of integration. e2 – 1 Evaluate.

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