# Thread: Definite Integral Int by Parts

1. ## 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
$u = 7t - 42$

$du = 7 \, dt$

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

$v = -e^{7 - t}$

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

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

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

$I = -56e^4$

2. Hello, Macleef!

I would factor out the 7

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

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

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

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

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

3. Originally Posted by Soroban
Hello, Macleef!

I would factor out the 7

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

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

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...

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