Differentiable functions (Fundamental theorem of calculus)

Question

Determine a differentiable function $g'(x)$, when

a) $f: [a, b] \to \mathbb R$ is continuous and $g: [a, b]\to \mathbb R$, $g(x) = \int_x^b f(t) dt$

b) $g: \mathbb R \to \mathbb R$, $g(x) = \int_0^{\sin x} t^2e^t dt$

$f(x)$ is continuous, so then $g'(x) = f(x)$?

for b, $g(x) = t^2 e^t$ since that is also continuous everywhere?

Answer 1

Surely you know that $\int_x^b f(t)dt= -\int_b^x f(t)dt$. Apply the fundamental theorem of Calculus to that.

For (b), first all, g(x), a function of x, cannot be written as a formula in t. Now, use the chain rule: Let u= sin(x) so that $g(x)= \int_0^{sin(x)} t^2e^t dt= \int_0^u t^2e^t dt$. So $g'(x)= \frac{\int_0^u t^2e^t dt}{du}\frac{du}{dt}$