Question

question 10 - 1 point given (f(x)) below, find (f(x)). (f(x)=int_{sin(x)}^{2}(t^{2}+10)dt) provide your answer below: (f(x)=)

Explanation:

Step1: Apply the fundamental theorem of calculus and chain - rule

If $F(x)=\int_{a}^{u(x)}f(t)dt$, then $F^{\prime}(x)=f(u(x))\cdot u^{\prime}(x)$. Here, $a = 2$, $u(x)=\sin(x)$ and $f(t)=t^{2}+10$.

Step2: Find $f(u(x))$

Substitute $t = u(x)=\sin(x)$ into $f(t)$. So $f(u(x))=\sin^{2}(x)+10$.

Step3: Find $u^{\prime}(x)$

Since $u(x)=\sin(x)$, then $u^{\prime}(x)=\cos(x)$.

Step4: Calculate $F^{\prime}(x)$

By the formula $F^{\prime}(x)=f(u(x))\cdot u^{\prime}(x)$, we have $F^{\prime}(x)=(\sin^{2}(x)+10)\cos(x)$.

Answer:

$F^{\prime}(x)=(\sin^{2}(x)+10)\cos(x)$