Definite Integral Problem

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May 8th 2010, 08:24 PM
fabxx

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Definite 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 PM
Random Variable

simple substitution

$\int (5-2t)^{4} \ dt$

let $u = 5-2t$

then $du = -2 \ dt$

which means that $dt = -\frac{du}{2}$

so $\int (5-2t)^{4} \ dt = \int u^{4}\Big(-\frac{du}{2}\Big) = -\frac{u^{5}}{10} + C = -\frac{(5-2t)^{5}}{10} + C$
May 8th 2010, 08:48 PM
fabxx

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 PM
AllanCuz

Quote:

Originally Posted by fabxx

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 :(

You do know what the chain rule is? What we're looking for when we take

$\int f(x)dx$ is some function $g(x)$ that when we differentiate it, it becomes $f(x)$ . In other words, $g`(x) = f(x)$

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 PM
Random Variable

Because when you make the substitution, you switch from integrating with respect to $t$ to integrating with respect to $u$ . To do that you need to know how to write $dt$ in terms of $du$ . Have you covered integration by substitution in class?
May 8th 2010, 09:02 PM
fabxx

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 PM
AllanCuz

Quote:

Originally Posted by fabxx

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!

$(5-2t)^4$ this is what we want to get back to...

If we differentiate

$\frac { (5-2t)^5}{5}$ we get $-2(5-2t)^4$ 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

$\frac { (5-2t)^5}{5} * - \frac{1}{2}$

...
May 8th 2010, 10:03 PM
chiph588@

$(5-2t)^4 = 2^4(t-\tfrac52)^4$

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