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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} $

1 Answer

  • 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.
$5x=0\Rightarrow x=0$ and $y=5$
$5y=10\Rightarrow y=2$ and $x=3$
$(3)\times 4\Rightarrow 4x-16y=-20$
$15y=15\Rightarrow y=1$ 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 $y_3$
On integrating we get,
$A=\begin{bmatrix}\large\frac{4x^2}{2}\normalsize dx+5x\end{bmatrix}_{-1}^0+\begin{bmatrix}5x-\large\frac{x^2}{2}\end{bmatrix}_0^3-\large\frac{1}{4}\begin{bmatrix}\large\frac{x^2}{2}+\normalsize 5x\end{bmatrix}_{-1}^3$
Step 4:
On applying limits,
On simplifying we get,
answered May 6, 2013 by sreemathi.v
edited May 7, 2013 by sreemathi.v

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