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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Choose the correct answer in the the area bounded by the \(y\) - axis, \(y = \cos\: x\) and \(y = \sin\: x\) when \(0 \leq x \leq \frac{\large \pi}{2}\) is

\[ (A) 2 (\sqrt{2-1}) \qquad (B) \sqrt 2 - 1 \qquad (C) \sqrt 2 + 1 \qquad (D) \sqrt 2\]

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  • Suppose we are given two curves represented by Y=f(x);y=g(x) where $f(x)\geq g(x)$ in [a,b] the points of intersection of the two curves are given by x=a and x=b by taking common values of y from the equation of the curves.
  • Hence the required area is given by
  • $f(x)=\int_a^b[f(x)-g(x)]dx$.
  • Integration by parts in $\int udv=uv-\int vdu.$
Step 1:
Here the given curves are $y=\cos x$ and $y=\sin x.$
To find the point of intersection let us solve the two equations.
cos x=sin x.This is possible only if $x=\frac{\pi}{4}$.If $x=\frac{\pi}{4},y=\frac{1}{\sqrt 2}.$
The point of intersection is $\big(\large\frac{\pi}{4},\frac{1}{\sqrt{2}}\big)$
we can now find the required area by integrating with respect to y.
Hence $A=\int_0^{\frac{1}{\sqrt 2}}x_1dy+\int_{\frac{1}{\sqrt 2}}^1x_2dy$
where $x_2=\cos^{-1}y;and;x_1=\sin^{-1}y.$
$A=\int_0^{\frac{1}{\sqrt 2}}\sin^{-1}ydy+\int_{\frac{1}{\sqrt 2}}^1\cos^{-1}ydy$
But this can be done only by integration of parts,because the function is of the form $\int udv$
$\int u dv=uv-\int vdu$.
Step 2:
Let $u=\sin^{-1}y$.hence $\large\frac{du}{dy}=\frac{1}{\sqrt {1-y^2}}$
and dy=dv $\Rightarrow y=v$
Now substituting for u and v we get,
$\bigg[y\sin^{-1}y\bigg]_0^{\large\frac{1}{\sqrt 2}}-\int_0^{\large\frac{1}{\sqrt 2}}\frac{1}{\sqrt 1}-y^2.y$
Again on integrating by method of substitution;
we get,
$\bigg[y\sin^{-1} y\bigg]_0^{\large\frac{1}{\sqrt 2}}+\big[\sqrt 1-y^2\big]_0^{\large\frac{1}{\sqrt 2}}\big]$
Step 3:
On applying limits we get,
$\big(\large\frac{1}{\sqrt 2}$$2\sin^{-1}(\large\frac{1}{\sqrt 2}+$$\sqrt 1-(\large\frac{1}{\sqrt 2})^2-$$1\big)$
similarly if we integrate $x_2$ by the same method we get,
$\int_0^1\frac{1}{\sqrt 2}\cos^{-1}y dy=y\cos^{-1}y-\sqrt {1-y^2})\frac{1}{\sqrt 2}$
Applying limits we get,
$0-\frac{\pi}{4\sqrt 2}+\frac{1}{\sqrt 2}$
Step 4:
Now combining both areas we get
$A=\large\frac{\pi}{4\sqrt 2}+\frac{1}{\sqrt 2}-1+\frac{1}{\sqrt 2}-\frac{\pi}{4\sqrt 2}$.
$A=\frac{2}{\sqrt 2}-1=\sqrt 2-1$sq.units.
Hence the required area is $\sqrt 2-1$sq.units
Hence the correct answer is B.
answered Dec 21, 2013 by yamini.v

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