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Math Help - Definite Integral Int by Parts

  1. #1
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    Definite Integral Int by Parts

    Please check. I have problems solving for the second part of:

    \int_a^b u \, dv = uv \mid_a^b - \int_a^b v \, du

    Use integration by parts to solve the given integral.

    \int_1^9 (7t - 42)e^{7 - t} \, dt

    Let
    u = 7t - 42

    du = 7 \, dt

    dv = e^{7 - t} \, dt

    v = -e^{7 - t}

    I = [(7t - 42)(-e^{7-t}) \mid_1^9] - \int_1^9 (-7e^{7-t}) \, dt

    I = [(-7(9)e^{7 - 9}) - (-7(1)e^{7-1})] + 7 \int_1^9 e^{7-t} \, dt

    I = -56e^4 + [7(-e^{7-1})(-e^{7-1}) \mid_1^9]

    I = -56e^4
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  2. #2
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    Hello, Macleef!

    I would factor out the 7


    \int_1^9 (7t - 42)e^{7 - t} \, dt

    We have: . 7\!\int^9_1(t-6)\,e^{7-t}\,dt

    . . \begin{array}{ccccccc}<br />
u &=& t-6 & & dv &=& e^{7-t}dt \\ du &=& dt & & v &=& \text{-}e^{7-t} \end{array}

    We have: . 7\left[\text{-}(t-6)e^{7-t} + \int e^{7-t}\,dx\right]_1^9 \;=\;7\bigg[\text{-}(t-6)e^{7-t} - e^{7-t}\bigg]^9_1

    . . . . . . = \;\text{-}7e^{7-t}\bigg[(t-6) + 1\bigg]^9_1 \;=\;\text{-}7e^{7-t}(t - 5)\,\bigg]^9_1


    Then: . -7\bigg[e^{-2}(9-5) - e^6(1-5)\bigg] \;=\;-7\left[4e^{-2} + 4e^6\right]

    . . . . . = \;-28\left(e^{-2} + e^6\right) \;=\;\frac{\text{-}28(1 + e^8)}{e^2}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Macleef!

    I would factor out the 7



    We have: . 7\!\int^9_1(t-6)\,e^{7-t}\,dt

    . . \begin{array}{ccccccc}<br />
u &=& t-6 & & dv &=& e^{7-t}dt \\ du &=& dt & & v &=& \text{-}e^{7-t} \end{array}

    We have: . 7\left[\text{-}(t-6)e^{7-t} + \int e^{7-t}\,dx\right]_1^9 \;=\;7\bigg[\text{-}(t-6)e^{7-t} - e^{7-t}\bigg]^9_1

    . . . . . . = \;\text{-}7e^{7-t}\bigg[(t-6) + 1\bigg]^9_1 \;=\;\text{-}7e^{7-t}(t - 5)\,\bigg]^9_1


    Then: . -7\bigg[e^{-2}(9-5) - e^6(1-5)\bigg] \;=\;-7\left[4e^{-2} + 4e^6\right]

    . . . . . = \;-28\left(e^{-2} + e^6\right) \;=\;\frac{\text{-}28(1 + e^8)}{e^2}

    why is the last step  \frac{\text{-}28(1 + e^8)}{e^2} ? I thought (e^{-2} + e^6) would be... e^4...regardless...

    . = \;-28\left(e^{-2} + e^6\right) \;=\;\frac{\text{-}28(1 + e^8)}{e^2}
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