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!

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- May 8th 2010, 08:24 PMfabxxDefinite Integral Problem
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!

- May 8th 2010, 08:46 PMRandom Variable
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 $\displaystyle \int (5-2t)^{4} \ dt = \int u^{4}\Big(-\frac{du}{2}\Big) = -\frac{u^{5}}{10} + C = -\frac{(5-2t)^{5}}{10} + C$ - May 8th 2010, 08:48 PMfabxx
I flipped through my textbook and it didn't say that it had to multiply (-1/2). How come it needs to multiply that? I'm still confused :(

- May 8th 2010, 08:53 PMAllanCuz
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! - May 8th 2010, 08:58 PMRandom Variable
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?

- May 8th 2010, 09:02 PMfabxx
I understand that if I don't get (-1/2) it won't be the original problem and i do understand the chain rule.

if its the chain rule

should the anti differentiation of (-2t) = -2 [(t^2)/2] ?

Thanks! - May 8th 2010, 09:05 PMAllanCuz
$\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} $

... - May 8th 2010, 10:03 PMchiph588@
$\displaystyle (5-2t)^4 = 2^4(t-\tfrac52)^4 $

So $\displaystyle \int (5-2t)^4dt = 2^4\int (t-\tfrac52)^4dt = \frac{2^4}{5}(t-\tfrac52)^5+C = \frac{2^5}{2\cdot5}(t-\tfrac52)^5+C =$ $\displaystyle \left(-\frac12\right)\cdot \frac{(5-2t)^5}{5}+C $