Question Bank [2022] — Maths (commerce) STD 12 Commerce / Arts — Question
Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Question Bank [2022]3 Marks
Question
Evaluate $\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} d x$
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Answer
$\text { Let } I =\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} d x$
$=\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} \times \frac{\sqrt{1+x}-\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} d x$
$=\int_0^1 \frac{\sqrt{1+x}-\sqrt{x}}{(\sqrt{1+x})^2-(\sqrt{x})^2} d x$
$=\int_0^1 \frac{\sqrt{1+x}-\sqrt{x}}{1+x-x} d x$
$=\int_0^1\left[(1+x)^{\frac{1}{2}}-x^{\frac{1}{2}}\right]^{ d x}$
$=\int_0^1(1+x)^{\frac{1}{2}} d x-\int_0^1 x^{\frac{1}{2}} d x$
$=\left[\frac{(1+x)^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^1-\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^1$
$=\frac{2}{3}\left[(2)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]-\frac{2}{3}\left[(1)^{\frac{3}{2}}-0\right]$
$=\frac{2}{3}(2 \sqrt{2}-1)-\frac{2}{3}(1)$
$=\frac{4 \sqrt{2}}{3}-\frac{2}{3}-\frac{2}{3}$
$\therefore I =\frac{4}{3}(\sqrt{2}-1)$
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