Attached is a problem I hand wrote. I don't get why on the first step, it has to be multiplied by -(1/2). is that part of a rule? Thanks!
simple substitution
$\displaystyle \int (5-2t)^{4} \ dt $
let $\displaystyle u = 5-2t$
then $\displaystyle du = -2 \ dt$
which means that $\displaystyle dt = -\frac{du}{2} $
so $\displaystyle \int (5-2t)^{4} \ dt = \int u^{4}\Big(-\frac{du}{2}\Big) = -\frac{u^{5}}{10} + C = -\frac{(5-2t)^{5}}{10} + C$
You do know what the chain rule is? What we're looking for when we take
$\displaystyle \int f(x)dx $ is some function $\displaystyle g(x) $ that when we differentiate it, it becomes $\displaystyle f(x) $. In other words, $\displaystyle g`(x) = f(x) $
So to see why you need the -1/2 differentiate the answer and see what you get. You will notice that without the -1/2 you DO NOT obtain what was inside the integral!
Because when you make the substitution, you switch from integrating with respect to $\displaystyle t$ to integrating with respect to $\displaystyle u$. To do that you need to know how to write $\displaystyle dt$ in terms of $\displaystyle du$. Have you covered integration by substitution in class?
$\displaystyle (5-2t)^4 $ this is what we want to get back to...
If we differentiate
$\displaystyle \frac { (5-2t)^5}{5} $ we get $\displaystyle -2(5-2t)^4 $ which is not what we originally had. Note that if we divide by -2 we end up with the same result, thus the anti derivative is
$\displaystyle \frac { (5-2t)^5}{5} * - \frac{1}{2} $
...