# Definite Integral Problem

#### fabxx

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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#### Random 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 $$\displaystyle \int (5-2t)^{4} \ dt = \int u^{4}\Big(-\frac{du}{2}\Big) = -\frac{u^{5}}{10} + C = -\frac{(5-2t)^{5}}{10} + C$$

#### 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 #### AllanCuz

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!

#### Random 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?

• fabxx

#### 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!

#### AllanCuz

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

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

...

• fabxx

#### chiph588@

MHF Hall of Honor
$$\displaystyle (5-2t)^4 = 2^4(t-\tfrac52)^4$$

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

• fabxx