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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area bounded by the lines $y=4x+5,y=5-x$ and $4y=x+5.$

$\begin{array}{1 1} \frac{13}{1}\;sq.units. \\ 26\;sq.units. \\30\;sq.units. \\ \frac{15}{2}\;sq.units\end{array} $

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Toolbox:
  • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
  • $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c$.
Step 1:
Given :$y=4x+5,y=5-x$ and $4y=x+5.$
First let us sketch the graph of the given lines.
Consider $y=4x+5$
Clearly the lines meets the $x$-axis at $(-\large\frac{5}{4}$,$0)$ and the $y$-axis at $(0,5)$.Join these two points,we get the graph of the above line.Similarly the graphs of the other two lines can be obtained.
Step 2:
Now let us find the points of intersection of these lines.
$4x-y=-5$-------(1)
$x+y=5$-------(2)
_______________(+)
$5x=0\Rightarrow x=0$ and $y=5$
$x+y=5$
$x-4y=-5$-------(3)
________________(-)
$5y=10\Rightarrow y=2$ and $x=3$
$\qquad\qquad\;4x-y=-5$
$(3)\times 4\Rightarrow 4x-16y=-20$
________________________
$15y=15\Rightarrow y=1$ and $x=-1$
Hence the points of intersection are $(-1,1),(0,5)$ and $(3,2)$
Step 3:
Now the required area is the shaded portion shown in the fig.
Area of the shaded portion is $A=\int_{-1}^0y_1dx+\int_0^3y_2dx-\int_{-1}^3y_3dx.$
On substituting for $y_1,y_2$ and $y_3$
$A=\int_{-1}^0(4x+5)dx+\int_0^3(5-x)dx-\int_{-1}^3\large\frac{x+5}{4}$$dx.$
On integrating we get,
$A=\begin{bmatrix}\large\frac{4x^2}{2}\normalsize dx+5x\end{bmatrix}_{-1}^0+\begin{bmatrix}5x-\large\frac{x^2}{2}\end{bmatrix}_0^3-\large\frac{1}{4}\begin{bmatrix}\large\frac{x^2}{2}+\normalsize 5x\end{bmatrix}_{-1}^3$
$\;\;=2(x^2)_{-1}^0+5(x)_{-1}^0+5(x)_0^3-\large\frac{1}{2}$$(x^2)_0^3-\large\frac{1}{8}$${(x^2)}_{-1}^3-\large\frac{5}{4}$$(x)_{-1}^3$
Step 4:
On applying limits,
$\;\;=2[0-1]+5[0+1]+5[3-0]-\large\frac{1}{2}$$[9-0]-\large\frac{1}{8}$$[9-1]-\large\frac{5}{4}$$[3+1]$
On simplifying we get,
$\;\;=-2+5+15-\large\frac{9}{2}$$-1-5$
$\;\;=12-\large\frac{9}{2}=\large\frac{15}{2}$sq.units.
answered May 6, 2013 by sreemathi.v
edited May 7, 2013 by sreemathi.v
 

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