Question

$\int_{0}^{8} \frac{dx}{sqrt{1 + x}}$

Explanation:

Step1: Use substitution

Let $u = 1 + x$, then $du=dx$. When $x = 0$, $u = 1$; when $x = 8$, $u=9$. The integral becomes $\int_{1}^{9}\frac{du}{\sqrt{u}}$.

Step2: Rewrite the integrand

We know that $\frac{1}{\sqrt{u}}=u^{-\frac{1}{2}}$.

Step3: Apply the power - rule for integration

The power - rule for integration is $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n
eq - 1$). For $n=-\frac{1}{2}$, we have $\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C = 2u^{\frac{1}{2}}+C$.

Step4: Evaluate the definite integral

$\int_{1}^{9}u^{-\frac{1}{2}}du=2u^{\frac{1}{2}}\big|_{1}^{9}=2\sqrt{u}\big|_{1}^{9}=2\sqrt{9}-2\sqrt{1}$.

Step5: Calculate the result

$2\times3 - 2\times1=6 - 2 = 4$.

Answer:

$4$