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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Using integration find the area of region bounded by the triangle whose vertices are $(-1, 0), (1, 3) $ and $(3, 2)$.

$\begin{array}{1 1}4 \\ 2\\ 6\\ 8 \end{array} $

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  • Equation of a line when two points are given is\[\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}.\]
Given the vertices of the triangle which is formed by the intersection of three lines is A(-1,0),B(1,3) and C(3,2).
Let us now find the equation of the line segments AB,BC and AC of the triangle.
Equation of line segment AB is\[\frac{y-0}{3-0}=\frac{x+1}{1+1}\Rightarrow \frac{y}{3}=\frac{x+1}{2}\]
Area of the region bounded by this line and the x-axis is $y_1$.
on integrating we get $y_1=\frac{3}{2}[\frac{x^2}{2}+x]_{-1}^1.$
On applying limits we get,
$y_1=\frac{3}{2}[\frac{1}{2}+1-\frac{1}{2}+1]=3$ sq.units.
Equation of the line segment BC is \[\frac{y-3}{2-3}=\frac{x-1}{3-1}\]
$\Rightarrow y-3=\frac{(-1)(x-1)}{2} \Rightarrow y=\frac{-(x-1)}{2}+3$
Area of the region bounded by this line and x-axis is $y_2$.
Area of $y_2=\int_1^3y^2dx=\int_1^3\frac{1}{2}(-x+7)dx$
on integrating we get,
Area of $y_2=\frac{1}{2}[\frac{x^2}{2}+7x]_1^3$
on applying limits we get,
$\;\;\;\;\;\quad\;\;\;\;\;\;\;=5$ sq.units.
Equation of line AC is\[\frac{y-0}{2-0}=\frac{x+1}{3+1}\Rightarrow y=\frac{1}{2}(x+1)\]
The area bounded by this line and the x-axis is $y_3$.
Area of $y_3=\int_{-1}^3\frac{1}{2}(x+1)dx$
On integrating we get,
Area of $y_3=\frac{1}{2}[\frac{x^2}{2}+x]_{-1}^3$
on applying limits we get,
Area of $y_3=\frac{1}{2}[\frac{9}{2}+3-\frac{1}{2}+1]=4\;units.$
Now area of the required triangle is\[y_1+y_2-y_3.\]
$\Rightarrow A=3+5-4=4$ sq.units.
Hence the required area is 4 sq. units.


answered Jan 25, 2013 by sreemathi.v
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<div class="clay6-toolbox"><b>Toolbox:</b><ul><li class="clay6-basic" id="pr00">
If we are given two curves represented by y=f(x),y=g(x) where $f(x)\geq g(x)$ in [a,b],the point of intersection of two curves are given by x=a and x=b by taking common values of y from the equation of the two curves.</li></ul></div><div class="clay6-step-odd"><div class="clay6-advanced" id="pr10">
The curve $x^2=4y$ is the equation of the parabola with vertex at the origin and axis along y-axis and open upwards.</div><div class="clay6-basic" id="pr11">
Let $R_1$ be the region lying inside the parabola $x^2=4y\Rightarrow y=\frac{x^2}{4}$</div><div class="clay6-basic" id="pr12">
We have $4x^2+4y^2=9$</div><div class="clay6-basic" id="pr13">
$x^2+y^2=\frac{9}{4}\Rightarrow x^2+y^2=\frac{9}{4}\Rightarrow y^2=\frac{9}{4}-x^2$</div><div class="clay6-basic" id="pr14">
$y=\sqrt {\frac{9}{4}-x^2}$</div><div class="clay6-advanced" id="pr15">
Clearly this represents the equation of a circle with the centre at the origin and radius $\frac{3}{2}$.</div><div class="clay6-advanced" id="pr16">
Let $R_2$ be the region lying inside the circle.</div><div class="clay6-basic" id="pr17">
Hence the area of the required region is bounded by the parabola $x^2=4y$ and the circle $x^2+y^2=\frac{9}{4}.$</div><div class="clay6-basic" id="pr18">
The area of the required region is the shaded portion shown in the fig.</div><div class="clay6-image" id="pr19">http://clay6.com/mpaimg/1_test17.png</div></div><div class="clay6-step-even"><div class="clay6-advanced" id="pr20">
To find the point of intersection let us substitute $x^2=4y$ in the equation of the circle.</div><div class="clay6-basic" id="pr21">
$4y+y^2=\frac{9}{4}\Rightarrow 4y^2+16y-9=0$</div><div class="clay6-basic" id="pr22">
On factorising we get,</div><div class="clay6-basic" id="pr23">
(2y+9)(2y-1)=0</div><div class="clay6-basic" id="pr24">
$\Rightarrow x=\frac{-9}{2}$ or $y=\frac{1}{2}$.</div><div class="clay6-basic" id="pr25">
If $y=\frac{1}{2},x=\pm\sqrt 2$ and if $y=\frac{-9}{2},$x is imaginary.</div><div class="clay6-basic" id="pr26">
Hence the points of intersection are ($\sqrt 2,\frac{1}{2})(\sqrt 2,\frac{-1}{2}).$</div><div class="clay6-basic" id="pr27">
Clearly the curves are symmetrical about the x-axis.</div><div class="clay6-advanced" id="pr28">
Hence the required area is</div><div class="clay6-advanced" id="pr29">
$A=\int_0^\frac{1}{\sqrt 2}(R_2-R_1)dx$</div><div class="clay6-basic" id="pr210">
$\;\;\;=\int_0^\frac{1}{\sqrt 2}\frac{\sqrt {9-4x^2}}{2}-\int_0^\frac{1}{\sqrt 2}\frac{x^2}{4}dx$.</div><div class="clay6-basic" id="pr211">
$A=\frac{1}{2}\int_0^\frac{1}{sqrt 2}\sqrt {9-4x^2}-\frac{1}{4}\int_0^\sqrt 2x^2dx.$</div></div><div class="clay6-step-odd"><div class="clay6-basic" id="pr30">
On integrating we get,</div><div class="clay6-basic" id="pr31">
$\;\;\;\;=\frac{1}{2}.\frac{1}{2}\begin{bmatrix}\frac{2x}{2}\sqrt{9-4x^2}+\frac{9}{2}\sin^{-1}\frac{2x}{3}\end{bmatrix}_0^\sqrt 2-\frac{1}{4}\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_0^\sqrt 2$</div><div class="clay6-basic" id="pr32">
On applying the limits we get,</div><div class="clay6-basic" id="pr33">
$\;\;\;=\frac{1}{4}\begin{bmatrix}\sqrt 2\sqrt{9-8}+\frac{9}{2}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)\end{bmatrix}-\frac{1}{12}(\sqrt 2)^3$</div><div class="clay6-advanced" id="pr34">
$\;\;\;\;=\frac{\sqrt 2}{4}+\frac{9}{8}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)-\frac{\sqrt 2}{6}$.</div><div class="clay6-basic" id="pr35">
$\;\;\;=\frac{\sqrt 2}{12}+\frac{9}{8}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)$.</div><div class="clay6-advanced" id="pr36">
But since the curves are symmetrical about x-axis the required area is</div><div class="clay6-basic" id="pr37">
$A=2\begin{bmatrix}\frac{\sqrt 2}{12}+\frac{9}{8}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)\end{bmatrix}$.</div><div class="clay6-basic" id="pr38">
$\;\;\;=\frac{\sqrt 2}{6}+\frac{9}{4}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)$.</div><div class="clay6-basic" id="pr39">
Hence the required area is $\frac{\sqrt 2}{6}+\frac{9}{4}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)$.</div></div>
answered Dec 21, 2013 by yamini.v

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