Draw a rough sketch of \[{{y}^{2}}=x+1\] and \[{{y}^{2}}=-\,x+1\] and determine the area enclosed by the two curves. |
OR |
Draw a rough sketch of \[y=\sin 2x\] and determine the area enclosed by the two curves, X-axis and lines \[x=\frac{\pi }{4}\] and \[\frac{3\pi }{4}.\] |
Answer:
Given curves are \[{{y}^{2}}=x+1\] ...(i) and \[{{y}^{2}}=-\,x+1\] ?(ii) On solving Eqs. (i) and (ii), we get \[x+1=-x+1\] \[\Rightarrow \] \[2x=1-1\] \[\Rightarrow \] \[2x=0\] \[\Rightarrow \] x = 0 On putting x = 0 in Eq. (i), we get \[{{y}^{2}}=1\] \[\Rightarrow \] \[y=\pm \,1\] \[\therefore \] Two given parabolas intersect each other at and \[B(0,\,\,-\,1)\] as shown in the figure. \[\therefore \] Required area = 2 [Area of the region OCAO + Area of the region ODAO] \[=2\left[ \int_{-\,1}^{0}{\sqrt{x+1}}dx+\int_{0}^{1}{\sqrt{1-x}dx} \right]\] \[=2\left\{ \left[ \frac{{{(x+1)}^{^{\frac{3}{2}}}}}{\frac{3}{2}} \right]_{-\,1}^{0}+\left[ (-\,1)\frac{{{(1-x)}^{\frac{3}{2}}}}{\frac{3}{2}} \right]_{0}^{1} \right\}\] \[=2\times \frac{2}{3}\{[{{(x+1)}^{3/2}}]_{-1}^{0}-[{{(1-x)}^{3/2}}]_{0}^{1}\}\] \[=\frac{4}{3}[\{{{(0+1)}^{3/2}}-{{(-1+1)}^{3/2}}\}-\{{{(1-1)}^{3/2}}-{{(1-0)}^{3/2}}\}]\]\[=\frac{4}{3}[1\,+1]=\frac{8}{3}\text{sq}\,\,\text{units}\] OR Given equation of curve is\[y=\sin 2x\] and the equation of lines are \[x=\frac{\pi }{4}\] and \[x=\frac{3\pi }{4}.\] To draw the graph of y = sin2x , et us consider the following values of sin2x at various points.
X 0 \[\pi /4\] \[\pi /2\] \[3\pi /4\] \[\pi \] \[5\pi /4\] 2x 0 \[\pi /2\] \[\pi \] \[3\pi /2\] \[2\pi \] \[5\pi /2\] \[\mathbf{y=sin 2x}\] 0 1 0 \[-1\] 0 1 Now, let us draw the graph of y = sin2x as shown below Clearly, the required area is the shaded area, which is given by \[\int_{\pi /4}^{\pi /2}{y\,dx}+\int_{\pi /2}^{3\pi /4}{(-\,y)}\,dx\] \[=\int_{\pi /4}^{\pi /2}{(\sin \,2x)dx}+\int_{\pi /2}^{3\pi /4}{(-\,\sin 2x)}\,dx\] \[=-\frac{1}{2}\left[ \cos 2x \right]_{\pi /4}^{\pi /2}+\frac{1}{2}\left[ \cos 2x \right]_{\pi /2}^{3\pi /4}\] \[=-\frac{1}{2}\left[ \cos \pi -cos\frac{\pi }{2} \right]+\frac{1}{2}\left[ \cos \frac{3\pi }{2}-\cos \pi \right]\]\[=-\frac{1}{2}\left[ -\,1-0 \right]+\frac{1}{2}[0-(-\,1)]=\frac{1}{2}+\frac{1}{2}=1\,\,\text{sq}\,\,\text{unit}\]
You need to login to perform this action.
You will be redirected in
3 sec