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Application of Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Integrals1 Mark
MCQ
The area bounded by the $y-$ axis, $\text{y}=\cos\text{x}$ and $\text{y}=\sin\text{x}$ when $0\leq\text{x}\leq\frac{\pi}{2}$ is:
  • A
    $2\big(\sqrt{2}-1\big)$
  • ✓
    $\sqrt{2}-1$
  • C
    $\sqrt{2}+1$
  • D
    $\sqrt{2}$
✓

Answer

Correct option: B.
$\sqrt{2}-1$
Points of intersection is obtained by solving $y = \sin x$ and $y = \cos x$
$\therefore\sin\text{x} = \cos \text{x}$
$\Rightarrow \text{x}=\frac{\pi}{4}$
Thus the two functions intesect at $\text{x}=\frac{\pi}{4}$
$\Rightarrow \text{y} = \sin \frac{\pi}{4} =\frac{1}{\sqrt{2}}$
Hence $\text{A}\Big(\frac{\pi}{4},\frac{1}{\sqrt{2}}\Big)$ is the point of intersection.
$\therefore$ Area bound by the curves and the $y -$ axis when $0\leq\text{x}\leq\pi2$
$\text{A} = \int\limits^\frac{1}{\sqrt{2}}_0|\text{x}_1|\text{dy}+\int\limits^1_\frac{1}{\sqrt{2}}|\text{x}_2|\text{dy}$
$=\int\limits^\frac{1}{\sqrt{2}}_0\text{x}_1\text{ dy}+\int\limits^1_\frac{1}{\sqrt{2}}\text{x}_2\text{ dy}$
$= \int\limits^\frac{1}{\sqrt{2}}_0\sin^{-1}\text{y}\text{ dy}+\int\limits^1_\frac{1}{\sqrt{2}}\cos^{-1}\text{y}\text{ dy}$
$= \Big[\text{y}\sin^{-1}\text{y}+\sqrt{1-\text{y}^2}\Big]^\frac{1}{\sqrt{2}}_0+\Big[\text{y}\cos^{-1}\text{y}-\sqrt{1-\text{y}^2}\Big]^1_\frac{1}{\sqrt{2}}$
$= \Big[\frac{1}{\sqrt{2}}\sin^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{2}}-1\Big]$
$+\bigg[1\times\cos^{-1}-0-\frac{1}{\sqrt{2}}\cos^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{2}}\bigg]$
$= \frac{2}{\sqrt{2}}-1$
$=\big(\sqrt{2}-1\big)\text{sq. units}$

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