Question

question
let f be the function defined by $f(x)=\int_{-2}^{\sqrt{x}} \frac{1}{(2t + 2)^6} dt$. find an expression for $f(x)$.

Explanation:

Step1: Recall Leibniz Rule

For $f(x)=\int_{a}^{u(x)} g(t)dt$, $f'(x)=g(u(x))\cdot u'(x)$

Step2: Identify $g(t)$ and $u(x)$

$g(t)=\frac{1}{(2t+2)^6}$, $u(x)=\sqrt{x}=x^{\frac{1}{2}}$

Step3: Compute $u'(x)$

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

Step4: Substitute $t=u(x)$ into $g(t)$

$g(u(x))=\frac{1}{(2\sqrt{x}+2)^6}=\frac{1}{[2(\sqrt{x}+1)]^6}=\frac{1}{64(\sqrt{x}+1)^6}$

Step5: Multiply $g(u(x))$ and $u'(x)$

$f'(x)=\frac{1}{64(\sqrt{x}+1)^6}\cdot\frac{1}{2\sqrt{x}}$

Answer:

$\frac{1}{128\sqrt{x}(\sqrt{x}+1)^6}$