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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Sketch the region and find the area bounded between $y=\sqrt {5-x^2}$ and $y=|x-1|$ using integration.

$\begin{array}{1 1} \large\frac{3\pi+2}{4} \\\large\frac{3\pi-2}{4} \\ \large\frac{5\pi-2}{4} \\ \large\frac{5\pi+2}{4} \end{array} $

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Toolbox:
  • The area enclosed by a curve $y=f(x)$,the $x$-axis and the ordinate $x=a$ and $y=b$ is given by $\int_a^b ydx.$
  • $\int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big)$$+c.$
  • $sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}$
Step 1:
Given curves are $y=\sqrt {5-x^2}$...........(i) and $y=|x-1|$.....(ii)
Clearly the equation $y=\sqrt {5-x^2}$ represents a semi circle of radius $\sqrt 5$
and having centre $(0,0)$ which is above $x\:axis$.
Step 2:
$y=|x-1| $ can be written as $y=x-1$........(iii) when $x>1$ and
$y=1-x$...........(iv) when $x<1$
Clearly this represents a broken line broken at $(1,0)$
Step 3
Solving the equations (i) and (iii) we get the point of intersection $B$
Solving $y^2=5-x^2$ and $y=x-1$ we get
$\Rightarrow\:x^2+(x-1)^2=5$ $\Rightarrow\:x=2$ and $y=1$
$B(2,1)$
Similarly by solving the equations (i) and (iv) we get the point $A$
$\Rightarrow\:x^2+(1-x)^2=5\:\:\Rightarrow\:x=-1,\:\:y=2$
$A(-1,2)$
Step 4
From the figure the area of the shaded region is given by
$A=\int _{-1} ^{2}\sqrt {5-x^2}\:dx-\int _{-1} ^{1}(1-x)\:dx-\int _{1}^{2}(x-1)\:dx$
On integrating
$A=\begin{bmatrix}\large\frac{x}{2}\normalsize\sqrt{5-x^2}+\large\frac{5}{2}\normalsize\sin^{-1}\big(\large\frac{x}{\sqrt 5}\big)\end{bmatrix}_{-1}^{2}$-$\begin{bmatrix}x-\large\frac{x^2}{2}\end{bmatrix}_{-1}^1$-$\begin{bmatrix}\large\frac{x^2}{2}-x\end{bmatrix}_{1}^2$
On applying the limits
$1+\large\frac{5}{2}$$sin^{-1}(\large\frac{2}{\sqrt 5}$$)+1-\large\frac{5}{2}$$sin^{-1}(\large\frac{-1}{\sqrt 5}$$)-1+\large\frac{1}{2}$$-1-\large\frac{1}{2}$$-2+2+\large\frac{1}{2}$$-1$
$\Rightarrow\:A=\large\frac{5}{2}$$(sin^{-1}\large\frac{2}{\sqrt 5}$$+sin^{-1}\large\frac{1}{\sqrt 5})-\large\frac{1}{2}$
But $sin^{-1}\large\frac{1}{\sqrt 5}$$=cos^{-1}\large\frac{2}{\sqrt 5}$
$\Rightarrow\:A=\large\frac{5}{2}$$(sin^{-1}\large\frac{2}{\sqrt 5}$$+cos^{-1}\large\frac{2}{\sqrt 5})-\large\frac{1}{2}$
Also we know that $sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}$
$\Rightarrow\:\large\frac{5}{2}.\large\frac{\pi}{2}-\frac{1}{2}=\large\frac{5\pi-2}{4}$$sq.units.$
Note: $\large\frac{5\pi-2}{4}$$sq.units = 3.43 sq units.
answered Feb 10, 2014 by rvidyagovindarajan_1
edited Feb 14, 2014 by balaji.thirumalai
 
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