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SOLUTIONS TO INTEGRATION OF EXPONENTIAL FUNCTIONS



SOLUTION 6 : Integrate $ \displaystyle{ \int { e^x (1+2e^x)^4 } \,dx } $ . Use u-substitution. Let

u = 1+2ex

so that

du = 2ex dx ,

or

(1/2) du = ex dx .

Substitute into the original problem, replacing all forms of x, getting

$ \displaystyle{ \int { e^x (1+2e^x)^4 } \,dx } = \displaystyle{ \int { (1+2e^x)^4 } \,e^x dx } $

$ = \displaystyle{ \int { u^4 } \, (1/2) du } $

$ = \displaystyle{ (1/2) \int { u^4 } \, du } $

$ = \displaystyle{ (1/2) { u^5 \over 5 } } + C $

$ = \displaystyle{ (1/10) u^5 } + C $

$ = \displaystyle{ (1/10) (1+2e^x)^5 } + C $ .

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SOLUTION 7 : Integrate $ \displaystyle{ \int ( e^{4x} - e^{-4x} )^2 \,dx } $. First, square the exponential function, recalling that (A-B)2 = A2 - 2AB + B2 . The result is

$ \displaystyle{ \int ( e^{4x} - e^{-4x} )^2 \,dx } = \displaystyle{ \int ( e^{4x} e^{4x} - 2 e^{4x} e^{-4x} + e^{-4x}e^{-4x} ) \,dx } $

(Recall that $ \displaystyle{ R^M R^N = R^{M+N}} $ .)

$ = \displaystyle{ \int ( e^{4x+4x} - 2 e^{4x-4x} + e^{-4x-4x} ) \,dx } $

$ = \displaystyle{ \int ( e^{8x} - 2 e^{0} + e^{-8x} ) \,dx } $

$ = \displaystyle{ \int ( e^{8x} - 2(1) + e^{-8x} ) \,dx } $

(Use the properties of integrals.)

$ = \displaystyle{ \int e^{8x} \,dx } - \displaystyle{ \int 2 \,dx } + \displaystyle{ \int e^{-8x} \,dx } $

(Use formula 3 from the introduction to this section on integrating exponential functions.)

$ = \displaystyle{ { e^{8x} \over 8 } } - 2x + \displaystyle{ { e^{-8x} \over -8 } } + C $

$ = \displaystyle{ { e^{8x} \over 8 } } - 2x - \displaystyle{ { e^{-8x} \over 8 } } + C $ .

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SOLUTION 8 : Integrate $ \displaystyle{ \int e^x(1-e^x)(1+e^x)^{10} \,dx } $ . Use u-substitution. Let

u = 1+ex

so that

du = ex dx .

In addition, we can "back substitute" with

ex = u-1 .

Substitute into the original problem, replacing all forms of x, getting

$ \displaystyle{ \int e^x(1-e^x)(1+e^x)^{10} \,dx } = \displaystyle{ \int (1-e^x)(1+e^x)^{10} \,e^x dx }$

$ = \displaystyle{ \int (1-(u-1))(u)^{10} \, du }$

$ = \displaystyle{ \int (1-u+1)u^{10} \, du }$

$ = \displaystyle{ \int (2-u)u^{10} \, du }$

$ = \displaystyle{ \int (2u^{10}-uu^{10}) \, du }$

$ = \displaystyle{ \int ( 2u^{10}-u^{11} ) \, du }$

$ = \displaystyle{ 2 { u^{11} \over 11 } - { u^{12} \over 12 } } + C $

$ = \displaystyle{ (2/11) (1+e^x)^{11} - (1/12) (1+e^x)^{12} } + C $ .

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SOLUTION 9 :Integrate $ \displaystyle{ \int_{0}^{1} { 3^x+4^x \over 5^x } \,dx } $ . First, split the function into two parts, getting

$ \displaystyle{ \int_{0}^{1} { 3^x+4^x \over 5^x } \,dx } = \displaystyle{ \int_{0}^{1} \Big\{ {3^x \over 5^x } + {4^x \over 5^x} \Big\} \,dx } $

(Recall that $ \displaystyle{ { A^B \over C^B } = \Big( { A \over C }\Big)^B } $ .)

$ = \displaystyle{ \int_{0}^{1} \Big\{ \Big({3 \over 5 }\Big)^x + \Big({4 \over 5}\Big)^x \Big\} \,dx } $

(Use formula 2 from the introduction to this section on integrating exponential functions.)

$ = \displaystyle{ \Bigg\{ { \Big(\displaystyle{3 \over 5 }\Big)^x \over \ln(3/5... ...(\displaystyle{4 \over 5}\Big)^x \over \ln(4/5) } \Bigg\} \Bigg\vert_{0}^{1} } $

$ = \displaystyle{ \Bigg\{ { \Big(\displaystyle{3 \over 5 }\Big)^{(1)} \over \ln... .../5) }+ { \Big(\displaystyle{4 \over 5}\Big)^{(0)} \over \ln(4/5) } \Bigg\} } $

$ = \displaystyle{ \Bigg\{ { \displaystyle{3 \over 5 } \over \ln(3/5) }+ { \di... ... \displaystyle{ \Bigg\{ { 1 \over \ln(3/5) }+ { 1 \over \ln(4/5) } \Bigg\} } $

$ = \displaystyle{ { \displaystyle{3 \over 5 } - 1 \over \ln(3/5) }+ { \displaystyle{4 \over 5} - 1 \over \ln(4/5) } } $

$ = \displaystyle{ { -2/5 \over \ln(3/5) }+ { -1/5 \over \ln(4/5) } } $ .

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Duane Kouba
1999-05-19