Question

If \(I=\int_{0}^{4}  \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4-x}}dx\), then 8\(I\) is:

• A. 8
• B. 6
• C. 16
• D. 4

Answer 1

The Correct Option is C

To solve the integral \(I=\int_{0}^{4}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{4-x}}dx\), we apply the property of definite integrals: \(\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx\). Let \(f(x)=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{4-x}}\), and define \(g(x)=\frac{\sqrt{4-x}}{\sqrt{4-x}+\sqrt{x}}\). Observe that:
\(f(x)+g(x)=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{4-x}}+\frac{\sqrt{4-x}}{\sqrt{4-x}+\sqrt{x}}=1\).

Now, we have \(I=\int_{0}^{4}f(x)dx=\int_{0}^{4}g(x)dx\). Thus:
\(2I=\int_{0}^{4}[f(x)+g(x)]dx=\int_{0}^{4}1dx=4\). Solving for \(I\), we get:
\(I=2\).

Finally, we calculate \(8I\):
\(8I=8 \cdot 2=16\). Thus, the value of \(8I\) is 16.