question\ngiven $f(x)$ below, find $f(x)$.\n$f(x)=int_{0}^{sqrt{x}}\frac{t}{t + 8}dt$\nprovide your answer…

Explanation:

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

Let $u = \sqrt{x}$, then $F(x)=\int_{0}^{u}\frac{t}{t + 8}dt$. By the fundamental theorem of calculus and chain - rule, if $F(x)=\int_{a}^{u(x)}f(t)dt$, then $F'(x)=f(u(x))\cdot u'(x)$.

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

We have $f(t)=\frac{t}{t + 8}$, substituting $t = u=\sqrt{x}$ into $f(t)$, we get $f(u(x))=\frac{\sqrt{x}}{\sqrt{x}+8}$.

Step3: Find $u'(x)$

Since $u = \sqrt{x}=x^{\frac{1}{2}}$, then $u'(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$.

Step4: Calculate $F'(x)$

$F'(x)=f(u(x))\cdot u'(x)=\frac{\sqrt{x}}{\sqrt{x}+8}\cdot\frac{1}{2\sqrt{x}}=\frac{1}{2(\sqrt{x}+8)}$.

Answer:

$\frac{1}{2(\sqrt{x}+8)}$