Question

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

Explanation:

Step1: Substitute \( u = 1 + x \)

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

Rewrite \(\frac{1}{\sqrt{u}}\) as \( u^{-\frac{1}{2}} \). So the integral is \(\int_{1}^{9} u^{-\frac{1}{2}} du\).

Step3: Integrate using power rule

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

Step4: Evaluate the definite integral

Evaluate \(2\sqrt{u}\) from \(1\) to \(9\): \(2\sqrt{9}-2\sqrt{1}=2\times3 - 2\times1=6 - 2 = 4\).

Answer:

\(4\)