I am struggling with this question
a(t)= t/10e^t/10 so v(t) I think is the integral of this?
can somebody explain this to me using integration by parts
Do you mean $\displaystyle a(t) = \frac{t}{10}e^{\frac{t}{10}}$ ? v(t) would be the integral is v and a were velocity and acceleration respectively.
I would substitute $\displaystyle b = \frac{t}{10}$ but it doesn't really matter.
therefore $\displaystyle \frac{db}{dt} = 0.1$ and $\displaystyle dt = 10db$
$\displaystyle a(t) = be^b$ hence $\displaystyle v(t) = \int{be^b \cdot 10db}$
Using integration by parts letting $\displaystyle u = b$ and $\displaystyle \frac{dv}{db} = e^b$ so that
$\displaystyle \frac{du}{db} = 1$ and $\displaystyle v = e^b$
$\displaystyle I = uv - \int{v\frac{du}{db}} = be^b - \int{e^b} = e^b(b-1)$
remembering that $\displaystyle b = \frac{t}{10}$ we get
$\displaystyle e^{\frac{t}{10}}(1-\frac{t}{10}) + C$
Hello, Jaffa!
$\displaystyle \int \frac{t}{10}e^{\frac{t}{10}}dt $
We have: .$\displaystyle \frac{1}{10}\int t\, e^{\frac{t}{10}} dt$
I'll do it in conventional terms . . .
. . $\displaystyle \begin{array}{ccccccc}u &=&t & & dv &=& e^{\frac{t}{10}}dx \\ du &=& dx & & v &=& 10e^{\frac{t}{10}} \end{array}$
Then: .$\displaystyle \frac{1}{10}\bigg[t\cdot10e^{\frac{t}{10}} - \int10e^{\frac{t}{10}}dt\bigg] + C \;=\;\frac{1}{10}\bigg[10te^{\frac{t}{10}} - 100e^{\frac{t}{10}}\bigg] + C$
. . . $\displaystyle = \;e^{\frac{t}{10}}(t-10) + C$