question_answer

                       

Find the area bounded by the curve \[y=\cos x\]between x = 0 and \[x=2\,\pi .\]

OR

Find the area bounded by the curve \[{{y}^{2}}=4{{a}^{2}}(x-1)\] and the lines \[x=1\] and \[y=4a.\]

Answer:

The graph of \[y=\cos x\] between 0 to \[2\pi \] is the curve shown in given figure.

\[\therefore \] Required area = Area of region OABO

+ Area of region BCDB

+ Area of region DEFD

\[=\int_{0}^{\pi /2}{\cos x\,dx+\left| \int_{\pi /2}^{3\pi /2}{\cos x\,dx} \right|}+\int_{3\pi /2}^{2\pi }{\cos \,xdx}\]

[\[\therefore \] in region BCDB, the curve is below the X-axis, where value come out to be negative, so we take the absolute value]

\[=[\sin x]_{0}^{\pi /2}+\left| [\sin x]_{\pi /2}^{3\pi /2} \right|+[\sin x]_{3\pi /2}^{2\pi }\]

\[=\left( \sin \frac{\pi }{2}-sin0 \right)+\left| \sin \frac{3\pi }{2}-sin\frac{\pi }{2} \right|+\left( \sin 2\pi -sin\frac{3\pi }{2} \right)\]      \[=(1-0)\,+|-1-1|+\,[0-(-1)]\]

\[=1+2+1\]

= 4 sq units

Therefore, the required area is 4 sq units.

OR

The equation of the given curve is \[{{y}^{2}}=4{{a}^{2}}(x-1)\]or, \[{{(y-0)}^{2}}=4{{a}^{2}}(x-1).\]

Clearly, this equation represents a parabola with vertex at (1, 0) as shown in figure. The region enclosed by \[{{y}^{2}}=4{{a}^{2}}(x-1),\] x = 1 and y = 4a is the area of shaded portion in figure.

When we slice the area of the shaded portion in horizontal strips, we observe that each strip has left end on the line x = 1 and the right end on the parabola \[{{y}^{2}}=4{{a}^{2}}(x-1).\] So, the approximating rectangle shown in figure has, length \[=x-1,\] width = dy and area \[=(x-1)\,dy.\] Since, the approximating rectangle can move from y = 0 to y = 4a. So, required area A is given by

\[A=\int_{0}^{4a}{(x-1)}\,dy\]

\[A=\int_{0}^{4a}{\frac{{{y}^{2}}}{4{{a}^{2}}}}\,dy\]

[\[\because P(x,\,\,y)\] line on \[{{y}^{2}}=4{{a}^{2}}(x-1)\] \[\therefore \,\,x-1={{y}^{2}}/4{{a}^{2}}\] ]

\[A=\frac{1}{4{{a}^{2}}}\left[ \frac{{{y}^{3}}}{3} \right]_{0}^{4a}=\frac{1}{4a}\left( \frac{64{{a}^{3}}}{3} \right)=\frac{16a}{3}\] sq units