Question

use part i of the fundamental theorem of calculus to find the derivative of the function $y = \int_{-4}^{4x} (t^{2}+4)e^{t - 2}dt$.

Explanation:

Step1: Recall the FTC - Part I

Let $F(t)$ be an antiderivative of $(t^{2}+4)e^{t - 2}$, so $\int_{-4}^{4x}(t^{2}+4)e^{t - 2}dt=F(4x)-F(-4)$.

Step2: Differentiate using chain - rule

By the chain - rule, $\frac{d}{dx}(F(4x)-F(-4)) = F'(4x)\cdot4$. Since $F'(t)=(t^{2}+4)e^{t - 2}$, then $\frac{dy}{dx}=4((4x)^{2}+4)e^{4x - 2}$.

Answer:

$4((4x)^{2}+4)e^{4x - 2}$