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
Hello,
You have $\displaystyle \int_{1-3x}^1 \frac{u^3}{1+u^2} ~du$
What does FTC part 1 says ?
$\displaystyle F(x)=\int_a^x f(u) ~du ~ \implies ~ F'(x)=f(x)$
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.
$\displaystyle F(x)=\int_{1-3x}^1 \frac{u^3}{1+u^2} ~du=-\int_1^{1-3x} \frac{u^3}{1+u^2} ~du$ (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 $\displaystyle t=1-3x$ ? You will be able to let t in the upper boundary of the integral. And $\displaystyle x=\frac{1-t}{3}$
Hence $\displaystyle F \left(\tfrac{1-t}{3}\right)=-\int_1^t \frac{u^3}{1+u^2} ~du$
Apply FTC :
$\displaystyle \left(F \left(\tfrac{1-t}{3}\right)\right)'=-\left[\frac{t^3}{1+t^2}\right]$
Apply chain rule in the LHS :
$\displaystyle \boxed{-\left[\frac{t^3}{1+t^2}\right]}=\left(F \left(\tfrac{1-t}{3}\right)\right)'=\boxed{\frac{-1}{3} F' \left(\tfrac{1-t}{3}\right)}$
Substitute back $\displaystyle t=1-3x$ :
$\displaystyle -\left[\frac{(1-3x)^3}{1+(1-3x)^2}\right]=\frac{-1}{3} \cdot F'(x)$
$\displaystyle \boxed{F'(x)=3 \cdot \frac{(1-3x)^3}{1+(1-3x)^2}}$
Does this look clear enough ?