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UPSC NDA Math Model Paper 5

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Question : 10 of 120

Marks: +1, -0

Find the area between the lines

y=|x-2|

y=5-|x+1|

10
8
6
4

Solution:

Shaded area has to be calculated

Curve 1:

y=|x-2|

\Rightarrow y=2-x

for

x<2

\Rightarrow y=x-2

for

x\geq 2

Curve 2:

y=5-|x+1|

\Rightarrow y=5+x+1=x+6

for

x<-1

\Rightarrow y=5-x-1=4-x

for

x\geq -1

Area enclosed

(A)=\int\limits_{x_{1}}^{x_{2}}(y_{1}-y_{2})\,dx

\Rightarrow A=\int\limits_{-1}^{2}(5-|x+1|)-(|x-2|)\,dx

\Rightarrow A=\left|\int\limits_{-2}^{-1}x+6-(2-x)\,dx\right|

+ \left|\int\limits_{-1}^{2}4-x-(2-x)\,dx\right|

+ \left|\int\limits_{2}^{3}4-x-(x-2)\,dx\right|

\Rightarrow A=\left|\int\limits_{-2}^{-1}(2x+4)\,dx\right|+\left|\int\limits_{-1}^{2}2\,dx\right|

+ \left|\int\limits_{2}^{3}(6-2x)\,dx\right|

\Rightarrow A=\left|[x^{2}+4x]_{-2}^{-1}\right|+\left|[2x]_{-1}^{2}\right|

+ \left|[6x-x^{2}]_{2}^{3}\right|

\Rightarrow A=\left|[-3+4(1)]\right|+\left|[2(3)]\right|

+ \left|[6(1)-5]\right|

\Rightarrow A=1+6+1=8

sq. units

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