Question

The area bounded by the \(y-axis,y=cosx\) and \(y=sinx\) when \(0≤x≤\frac{π}{2}\)

• A. \(2(\sqrt{2}-1)\)
• B. \(\sqrt{2}-1\)
• C. \(\sqrt{2}+1\)
• D. \(\sqrt{2}\)

Answer 1

The Correct Option is B

The correct answer is B:\(\sqrt{2}-1\)
The given equations are
\(y=cosx...(1)\)
And,\(y=sinx...(2)\)

Required area=Area(ABLA)+area(OBLO)
\(=∫^1_\frac{1}{\sqrt{2}}xdy+∫^\frac{1}{\sqrt{2}}_0xdy\)
\(=∫^1_{\frac{1}{\sqrt{2}}}cos^{-1}y\,dy+∫^{\frac{1}{\sqrt{2}}}_0sin^{-1}xdy\)
Integrating by parts,we obtain
\(=\bigg[ycos^{-1}y-\sqrt{1-y^2}\bigg]^1_{\frac{1}{\sqrt{2}}}+[xsin^{-1}x+\sqrt{1-x^2}]^{\frac{1}{\sqrt{2}}}_0\)
\(=[cos^{-1}(1)-\frac{1}{\sqrt{2}}cos^{-1}(\frac{1}{\sqrt{2}})+\sqrt{1-\frac{1}{2}}]+[\frac{1}{\sqrt{2}}sin^{-1}(\frac{1}{\sqrt{2}})+\sqrt{1-\frac{1}{2}}-1]\)
\(=\frac{-π}{4\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{π}{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1\)
\(=\frac{2}{\sqrt{2}}-1\)
\(=\sqrt{2}-1units\)
Thus,the correct answer is B.
Put \(2x=t⇒dx=\frac{dt}{2}\)
When \(x=\frac{3}{2},t=3\) and when \(x=\frac{1}{2},t=1\)
\(=∫^{\frac{1}{2}}_02\sqrt{x}dx+\frac{1}{4}∫^3_1\sqrt{(3)^2-(t)^2}dt\)
\(=2\bigg[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\bigg]^{\frac{1}{2}}_0+\frac{1}{4}\bigg[\frac{t}{2}\sqrt{9-t^2}+\frac{9}{2}sin^{-1}(\frac{t}{3})\bigg]^3_1\)
\(=2\bigg[\frac{2}{3}(\frac{1}{2})^{\frac{3}{2}}\bigg]+\frac{1}{4}\bigg[{\frac{3}{2}\sqrt{9-(3)^2}+\frac{9}{2}sin^{-1}(\frac{3}{3})}\bigg]-\bigg[\frac{1}{2}\sqrt{9-(1)^2}+\frac{9}{2}sin^{-1}(\frac{1}{3})\bigg]\)
\(=\frac{2}{3\sqrt{2}}+\frac{1}{4}[{0+\frac{9}{2}sin^{-1}(1)}-{\frac{1}{2}\sqrt{8}+\frac{9}{2}sin^{-1}(\frac{1}{3})}]\)
\(=\frac{\sqrt2}{3}+\frac{9π}{16}-\frac{\sqrt{2}}{4}-\frac{9}{8}sin^{-1}(\frac{1}{3})\)
\(=\frac{9π}{16}-\frac{9}{8}sin^{-1}(\frac{1}{3})+\frac{\sqrt{2}}{12}\)
Therefore,the required area is\([2\times(\frac{9π}{16}-\frac{9}{8}sin^{-1}(\frac{1}{3})+\frac{\sqrt{2}}{12})]=\frac{9π}{8}-\frac{9}{4}sin^{-1}(\frac{1}{3})+\frac{1}{3\sqrt{2}}units\)