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SOLUTIONS TO INTEGRATION OF EXPONENTIAL FUNCTIONS



SOLUTION 1 : Integrate $ \displaystyle{ \int { 5 e^x} \,dx } $ . By formula 1 from the introduction to this section on integrating exponential functions and properties of integrals we get that

$ \displaystyle{ \int { 5 e^x} \,dx } = \displaystyle{ 5 \int { e^x} \,dx }$

$ = \displaystyle{ 5 e^x} + C $ .

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SOLUTION 2 : Integrate $ \displaystyle{ \int ( 2-3e^x ) \,dx } $ . By formula 1 from the introduction to this section on integrating exponential functions and properties of integrals we get that

$ \displaystyle{ \int ( 2-3e^x ) \,dx } = \displaystyle{ \int 2 \,dx } - \displaystyle{ \int 3e^x \,dx } $

$ = \displaystyle{ \int 2 \,dx } - \displaystyle{ 3 \int e^x \,dx } $

$ = \displaystyle{ 2x - 3 e^x} + C $ .

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SOLUTION 3 : Integrate $ \displaystyle{ \int_{0}^{ (1/7) \ln 2 } 14e^{7x} \,dx } $ . Use u-substitution. Let

u = 7x

so that

du = 7 dx ,

or

(1/7) du = dx .

In addition, the range of x-values is

$ x : 0 \longrightarrow (1/7) \ln 2 $ ,

so that the range of u-values is

$ u : 7(0) \longrightarrow 7(1/7) \ln 2 $ ,

or

$ u : 0 \longrightarrow \ln 2 $ .

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

$ \displaystyle{ \int_{0}^{ (1/7) \ln 2 } 14e^{7x} \,dx } = \displaystyle{ 14 \int_{0}^{ (1/7) \ln 2 } e^{7x} \, dx } $

$ = \displaystyle{ 14 \int_{0}^{ \ln 2 } e^{u} \, (1/7) du } $

$ = \displaystyle{ 14 (1/7)\int_{0}^{ \ln 2 } e^{u} \, du } $

$ = \displaystyle{ 2 \int_{0}^{ \ln 2 } e^{u} \, du } $

$ = \displaystyle{ 2 e^u\Big\vert_{0}^{\ln 2} } $

$ = \displaystyle{ 2 e^{\ln 2 } - 2 e^{0} } $

(Recall that $ \displaystyle{ e^{ \ln A} = A } $ .)

$ = \displaystyle{ 2 (2) - 2(1) } $

= 4 - 2

= 2 .

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SOLUTION 4 : Integrate $ \displaystyle{ \int 7^{2x+3} \,dx } $ . Use u-substitution. Let

u = 2x+3

so that

du = 2 dx ,

or

(1/2) du = dx .

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

$ \displaystyle{ \int 7^{2x+3} \, dx } = \displaystyle{ \int 7^u \, (1/2) du } $

$ = \displaystyle{ (1/2) \int 7^u \, du } $

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

$ = (1/2) \displaystyle{ { 7^u \over \ln 7 } + C } $

$ = \displaystyle{ { 7^{2x+3} \over 2 \ln 7 } + C } $

(Recall that $ A \ln B = \ln B^A $ .)

$ = \displaystyle{ { 7^{2x+3} \over \ln 7^2 } + C } $

$ = \displaystyle{ { 7^{2x+3} \over \ln 49 } + C } $ .

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SOLUTION 5 : Integrate $ \displaystyle{ \int e^{5x} \Big( { e^{2x} \over 7 } + { 3 \over e^{3x} } \Big) \,dx } $ . First, multiply the exponential functions together. The result is

$ \displaystyle{ \int e^{5x} \Big( { e^{2x} \over 7 } + { 3 \over e^{3x} } \Big... ...int \Big( { e^{5x} e^{2x} \over 7 } + { 3 e^{5x} \over e^{3x} } \Big) \,dx } $

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

$ = \displaystyle{ \int \Big( { e^{5x+2x} \over 7 } + 3 e^{5x-3x} \Big) \,dx } $

$ = \displaystyle{ \int \Big( { e^{7x} \over 7 } + 3 e^{2x} \Big) \,dx } $

(Use the properties of integrals.)

$ = \displaystyle{ (1/7) \int e^{7x} \,dx } + \displaystyle{ 3 \int e^{2x} \,dx }$

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

$ = \displaystyle{ (1/7) { e^{7x} \over 7 } } + \displaystyle{ 3 { e^{2x} \over 2 } } + C $

$ = \displaystyle{ (1/49) e^{7x} } + \displaystyle{ (3/2) e^{2x} } + C $ .

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Duane Kouba
1999-05-19