Definite Integral Problem

Sep 2008
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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 2009
959
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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 \(\displaystyle \int (5-2t)^{4} \ dt = \int u^{4}\Big(-\frac{du}{2}\Big) = -\frac{u^{5}}{10} + C = -\frac{(5-2t)^{5}}{10} + C\)
 
Sep 2008
222
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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 :(
 
Apr 2010
384
153
Canada
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

\(\displaystyle \int f(x)dx \) 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 2009
959
362
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?
 
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Sep 2008
222
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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!
 
Apr 2010
384
153
Canada
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!
\(\displaystyle (5-2t)^4 \) 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

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

...
 
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chiph588@

MHF Hall of Honor
Sep 2008
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Champaign, Illinois
\(\displaystyle (5-2t)^4 = 2^4(t-\tfrac52)^4 \)

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