7. if $f(t)=int_{-pi}^{5t} sin^{4}(\theta^{3}-2)sqrt{400 - \theta^{2}}d\theta$, what is $f(t)$?

Explanation:

Step1: Apply the fundamental theorem of calculus

By the fundamental theorem of calculus, if $F(t)=\int_{a}^{t}g(\theta)d\theta$, then $F^\prime(t) = g(t)$. Here, $a = -\pi$ and $g(\theta)=\sin^{4}(\theta^{3}-2)\sqrt{400 - \theta^{2}}$.

Step2: Find the derivative of the integral

Since $f(t)=\int_{-\pi}^{5t}\sin^{4}(\theta^{3}-2)\sqrt{400 - \theta^{2}}d\theta$, let $u = 5t$. By the chain - rule and the fundamental theorem of calculus, $f^\prime(t)=\frac{d}{du}\left(\int_{-\pi}^{u}\sin^{4}(\theta^{3}-2)\sqrt{400 - \theta^{2}}d\theta\right)\cdot\frac{du}{dt}$. We know that $\frac{d}{du}\left(\int_{-\pi}^{u}\sin^{4}(\theta^{3}-2)\sqrt{400 - \theta^{2}}d\theta\right)=\sin^{4}(u^{3}-2)\sqrt{400 - u^{2}}$ and $\frac{du}{dt}=5$. Substituting $u = 5t$ back in, we get $f^\prime(t)=5\sin^{4}((5t)^{3}-2)\sqrt{400-(5t)^{2}}$.

Answer:

$5\sin^{4}(125t^{3}-2)\sqrt{400 - 25t^{2}}$