Question

provide your answer below: f(x)=int_{1}^{e^{x}}(t^{3}+7)dt f(x)= square

Explanation:

Step1: Apply the fundamental theorem of calculus

If $F(x)=\int_{a}^{g(x)}f(t)dt$, then $F^{\prime}(x)=f(g(x))\cdot g^{\prime}(x)$. Here $a = 1$, $g(x)=e^{x}$, and $f(t)=t^{3}+7$.

Step2: Find $f(g(x))$ and $g^{\prime}(x)$

First, $f(g(x))=(e^{x})^{3}+7=e^{3x}+7$. Second, $g^{\prime}(x)=e^{x}$.

Step3: Calculate $F^{\prime}(x)$

$F^{\prime}(x)=(e^{3x}+7)\cdot e^{x}=e^{4x}+7e^{x}$.

Answer:

$e^{4x}+7e^{x}$