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Mathematics Question on applications of integrals

The area bounded by the y−axis,y=cosxy-axis,y=cosx and y=sinxy=sinx when 0≤x≤π20≤x≤\frac{π}{2}

A

2(2−1)2(\sqrt{2}-1)

B

2−1\sqrt{2}-1

C

2+1\sqrt{2}+1

D

2\sqrt{2}

Answer

2−1\sqrt{2}-1

Explanation

Solution

The correct answer is B:2−1\sqrt{2}-1
The given equations are
y=cosx...(1)y=cosx...(1)
And,y=sinx...(2)y=sinx...(2)
Area
Required area=Area(ABLA)+area(OBLO)
=∫121xdy+∫012xdy=∫^1_\frac{1}{\sqrt{2}}xdy+∫^\frac{1}{\sqrt{2}}_0xdy
=∫121cos−1y dy+∫012sin−1xdy=∫^1_{\frac{1}{\sqrt{2}}}cos^{-1}y\,dy+∫^{\frac{1}{\sqrt{2}}}_0sin^{-1}xdy
Integrating by parts,we obtain
=[ycos−1y−1−y2]121+[xsin−1x+1−x2]012=\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)−12cos−1(12)+1−12]+[12sin−1(12)+1−12−1]=[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]
=−π42+12+π42+12−1=\frac{-π}{4\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{π}{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1
=22−1=\frac{2}{\sqrt{2}}-1
=2−1units=\sqrt{2}-1units
Thus,the correct answer is B.
Put 2x=t⇒dx=dt22x=t⇒dx=\frac{dt}{2}
When x=32,t=3x=\frac{3}{2},t=3 and when x=12,t=1x=\frac{1}{2},t=1
=∫0122xdx+14∫13(3)2−(t)2dt=∫^{\frac{1}{2}}_02\sqrt{x}dx+\frac{1}{4}∫^3_1\sqrt{(3)^2-(t)^2}dt
=2[x3232]012+14[t29−t2+92sin−1(t3)]13=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[23(12)32]+14[329−(3)2+92sin−1(33)]−[129−(1)2+92sin−1(13)]=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]
=232+14[0+92sin−1(1)−128+92sin−1(13)]=\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})}]
=23+9π16−24−98sin−1(13)=\frac{\sqrt2}{3}+\frac{9π}{16}-\frac{\sqrt{2}}{4}-\frac{9}{8}sin^{-1}(\frac{1}{3})
=9π16−98sin−1(13)+212=\frac{9π}{16}-\frac{9}{8}sin^{-1}(\frac{1}{3})+\frac{\sqrt{2}}{12}
Therefore,the required area is[2×(9π16−98sin−1(13)+212)]=9π8−94sin−1(13)+132units[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