Fundamental Theorm of Calculus

**sfgiants13** · October 6th 2008, 11:00 AM

I understand it for the most part and have got the previous problems, but this one is throwing me off quite a bit. The question is to find the derivative using FTC part 1.

It's the integral of 1-3x to 1 of u^3/(1+u^2) du

**Moo** · October 6th 2008, 11:19 AM

Hello,

Originally Posted by sfgiants13

I understand it for the most part and have got the previous problems, but this one is throwing me off quite a bit. The question is to find the derivative using FTC part 1.

It's the integral of 1-3x to 1 of u^3/(1+u^2) du

You have $\int_{1-3x}^1 \frac{u^3}{1+u^2} ~du$

What does FTC part 1 says ?

$F(x)=\int_a^x f(u) ~du ~ \implies ~ F'(x)=f(x)$

So now, here is what I propose to you : arrange the original integral in order to get a smilar form to what the FTC gives.

$F(x)=\int_{1-3x}^1 \frac{u^3}{1+u^2} ~du=-\int_1^{1-3x} \frac{u^3}{1+u^2} ~du$ (because in the FTC, t is on the upper boundary - and there's a - sign because you inverted the order of boundaries)

Now, what if you let $t=1-3x$ ? You will be able to let t in the upper boundary of the integral. And $x=\frac{1-t}{3}$

Hence $F \left(\tfrac{1-t}{3}\right)=-\int_1^t \frac{u^3}{1+u^2} ~du$

Apply FTC :

$\left(F \left(\tfrac{1-t}{3}\right)\right)'=-\left[\frac{t^3}{1+t^2}\right]$

Apply chain rule in the LHS :
$\boxed{-\left[\frac{t^3}{1+t^2}\right]}=\left(F \left(\tfrac{1-t}{3}\right)\right)'=\boxed{\frac{-1}{3} F' \left(\tfrac{1-t}{3}\right)}$

Substitute back $t=1-3x$ :

$-\left[\frac{(1-3x)^3}{1+(1-3x)^2}\right]=\frac{-1}{3} \cdot F'(x)$

$\boxed{F'(x)=3 \cdot \frac{(1-3x)^3}{1+(1-3x)^2}}$

Does this look clear enough ?

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