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:

Always remember to update your integration limits when using the substitution method for definite integrals.

Answer 1

The Correct Option is C

Step 1: Concept
Use the method of substitution to simplify the radical expression.

Step 2: Meaning
Let $t = x^2 + 4$. Then $dt = 2x dx$, or $x dx = \frac{1}{2} dt$.

Step 3: Analysis
Change the limits: when $x=0, t=4$; when $x=1, t=5$. The integral becomes $\int_{4}^{5} \frac{1}{2} \sqrt{t} dt = \frac{1}{2} [\frac{t^{3/2}}{3/2}]_{4}^{5} = \frac{1}{3} [t^{3/2}]_{4}^{5}$.

Step 4: Conclusion
Evaluating at the limits: $\frac{1}{3} [5^{3/2} - 4^{3/2}] = \frac{1}{3} [5\sqrt{5} - 8]$.
Final Answer: (C)