Confusion with derivate of integral

Question

I am asked to find the derivative in two ways first by evaluating the integral and then to differentiate the result the second way is by evaluating the integral directly. I can't solve any of the problems. For the first part how does one evaluate an integral of the form $$\int_a^{f(x)}f(t)dt$$ And for the second part how does one use the FTC with the form $$\frac{d}{dx}\int_a^{f(x)}f(t)dt$$

Answer 1

$f(x)$ is a real number, for $\displaystyle\int_{a}^{f(x)}f(t)dt$, just imagine you are doing some sort like $\displaystyle\int_{0}^{1}f(t)dt$.

On the other hand, $\dfrac{d}{dx}\displaystyle\int_{a}^{f(x)}f(t)dt=\left(\dfrac{d}{du}\displaystyle\int_{a}^{u}f(t)dt\right)\left(\dfrac{du}{dx}\right)=f(u)\dfrac{du}{dx}=f(f(x))\dfrac{df}{dx}=f(f(x))f'(x)$, where $u=f(x)$.

Answer 2

Remember the second fundamental theorem of calculus. Let $F'(x) = f(x)$, then we know:

$$\int_{h(x)}^{g(x)} f(t)\,dt = F(g(x)) - F(h(x))$$

Then differentiating both sides we get:

$$\begin{align} {d\over dx} \int_{h(x)}^{g(x)} f(t)\,dt &= {d\over dx} (F(g(x)) - F(h(x))) \\ &= F'(g(x))g'(x) - F'(h(x))h'(x)\\ &= f(g(x))g'(x) - f(h(x))h'(x)\\ \end{align}$$

In this case, the lower bound is a constant so the latter term is zero. In general, the fundamental theorem of calculus states:

$${d\over dx} \int_a^{g(x)} f(t)\,dt = f(x)g'(x)$$