Question

\[ \int_0^1 x\sqrt{x^2+4}\,dx= \]

• A. \(\frac{1}{3}[5\sqrt{5}-4]\)
• B. \(\frac{1}{2}[5\sqrt{5}-8]\)
• C. \(\frac{1}{3}[5\sqrt{5}-8]\)
• D. \(\frac{1}{3}[5\sqrt{5}+4]\)

Hint:

For integrals containing \(x\sqrt{x^2+a}\), use \(u=x^2+a\).

Answer 1

The Correct Option is C

Concept: Use substitution when the derivative of the inside expression is present outside.

Step 1: Given: \[ I=\int_0^1 x\sqrt{x^2+4}\,dx \]

Step 2: Put: \[ u=x^2+4 \] Then: \[ \frac{du}{dx}=2x \] \[ du=2x\,dx \] \[ x\,dx=\frac{du}{2} \]

Step 3: Change limits.
When \(x=0\): \[ u=0^2+4=4 \] When \(x=1\): \[ u=1^2+4=5 \]

Step 4: Substitute in the integral. \[ I=\int_4^5 \sqrt{u}\cdot\frac{du}{2} \] \[ I=\frac{1}{2}\int_4^5 u^{\frac{1}{2}}\,du \]

Step 5: Integrate. \[ I=\frac{1}{2}\cdot\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\Bigg|_4^5 \] \[ I=\frac{1}{3}u^{\frac{3}{2}}\Bigg|_4^5 \] \[ I=\frac{1}{3}\left[5^{\frac{3}{2}}-4^{\frac{3}{2}}\right] \] \[ I=\frac{1}{3}\left[5\sqrt{5}-8\right] \] Therefore, \[ \boxed{\frac{1}{3}\left[5\sqrt{5}-8\right]} \]