Q92 / 180 · KCET 2009 · Mathematics

$\int_{1}^{3} \frac{\sqrt{4-\mathrm{x}}}{\sqrt{\mathrm{x}}+\sqrt{4-\mathrm{x}}} \mathrm{dx}$ is equal to

  • Correct answer
    A
    1
  • B
    3
  • C
    2
  • D
    0

Solution

Let $I=\int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} d x \quad \text{...(i)}$ $$ \Rightarrow \quad I=\int_{1}^{3} \frac{\sqrt{4-(4-x)}}{\sqrt{4-x}+\sqrt{4-(4-x)}} d x $$ $$ \begin{gathered} {\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]} \\ \Rightarrow \quad I=\int_{1}^{3} \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} d x \quad \text{...(ii)} \end{gathered} $$ On adding Eqs. (i) and (ii), we get $$ \begin{aligned} 2 \mathrm{I} &=\int_{1}^{3} 1 \mathrm{dx}=[\mathrm{x}]_{1}^{3} \\ \Rightarrow \quad \mathrm{I} &=\frac{2}{2}=1 \end{aligned} $$

Hence, the correct answer is option: 'A'.

Relevant Questions