The value of integral, \int\limits_3^6 \sqrt x \over \sqrt 9 - x + \sqrt x dx is

Definite Integration PYQ — JEE Main 2005 · Shift 0 · Q84 with Solution

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Correct answer: B

We need to evaluate $I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}\,dx.$
A standard trick for such integrals is to use the substitution $x\mapsto 9-x,$ because the denominator contains both $\sqrt{x}$ and $\sqrt{9-x}$ .

Let $I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}\,dx.$ Now put $x=9-t \quad\Rightarrow\quad dx=-dt.$ When $x=3$ , $t=6$ ; when $x=6$ , $t=3$ . So

=\int_3^6 \frac{\sqrt{9-t}}{\sqrt{t}+\sqrt{9-t}}\,dt.$$ Renaming $t$ back to $x$, $$I=\int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}\,dx.$$ 3. Add the two expressions for $I$: $$I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}\,dx,$$ $$I=\int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}\,dx.$$ Therefore, $$2I=\int_3^6 \left(\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}+\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}\right)dx.$$ Since the denominators are the same, $$2I=\int_3^6 \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}\,dx =\int_3^6 1\,dx.$$ Hence, $$2I=6-3=3.$$ So, $$I=\frac{3}{2}.$$ 4. Evaluate the options: - A: $\frac12$ ❌ - B: $\frac32$ ✅ - C: $2$ ❌ - D: $1$ ❌ Therefore, the correct answer is $$\boxed{\frac{3}{2}}.$$

View written solutionFree

Correct answer: B

We need to evaluate $I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}\,dx.$
A standard trick for such integrals is to use the substitution $x\mapsto 9-x,$ because the denominator contains both $\sqrt{x}$ and $\sqrt{9-x}$ .

Let $I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}\,dx.$ Now put $x=9-t \quad\Rightarrow\quad dx=-dt.$ When $x=3$ , $t=6$ ; when $x=6$ , $t=3$ . So

=\int_3^6 \frac{\sqrt{9-t}}{\sqrt{t}+\sqrt{9-t}}\,dt.$$ Renaming $t$ back to $x$, $$I=\int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}\,dx.$$ 3. Add the two expressions for $I$: $$I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}\,dx,$$ $$I=\int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}\,dx.$$ Therefore, $$2I=\int_3^6 \left(\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}+\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}\right)dx.$$ Since the denominators are the same, $$2I=\int_3^6 \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}\,dx =\int_3^6 1\,dx.$$ Hence, $$2I=6-3=3.$$ So, $$I=\frac{3}{2}.$$ 4. Evaluate the options: - A: $\frac12$ ❌ - B: $\frac32$ ✅ - C: $2$ ❌ - D: $1$ ❌ Therefore, the correct answer is $$\boxed{\frac{3}{2}}.$$