Definite Integration
Definite Integral — Finding Parameter
nta_pyq_2023_jan
Grade 12
Question:
Let $\alpha>0$. If $\displaystyle\int_0^\alpha\dfrac{x}{\sqrt{x+\alpha}-\sqrt{x}}\,dx=\dfrac{16+20\sqrt{2}}{15}$, then $\alpha$ is equal to:
2
4
$\sqrt{2}$
$2\sqrt{2}$
Step-by-Step Solution
Key Concept: Rationalize: $\frac{x}{\sqrt{x+\alpha}-\sqrt{x}}=\frac{x(\sqrt{x+\alpha}+\sqrt{x})}{\alpha}$. $I=\frac{1}{\alpha}\int_0^\alpha x(\sqrt{x+\alpha}+\sqrt{x})dx$.
$\alpha=2$.
Correct Answer: 1