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NDA Math Area Under the Curves

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Consider the function

f(x)=\left|x-1\right|+x^2

where

x\in\mathbb{R}

[NDA I 2016]

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Question : 17 of 27

Marks: +1, -0

What is the area of the region bounded by x-axis, the curve

y=f(x)

and the two ordinates

x=1

x=\frac{3}{2}

$\frac{5}{12}$ square units
$\frac{7}{12}$ square units
$\frac{2}{3}$ square units
$\frac{11}{12}$ square units

Solution:

Area required for given region is

A_{2}=\int\limits_{1}^{\frac{3}{2}}f(x)dx

=\int\limits_{1}^{\frac{3}{2}}[x^{2}+x-1]dx

=\left[\;\frac{1}{3}x^{3}+\;\frac{1}{2}x^{2}-x\right]_{1}^{\frac{3}{2}}

A_{2}=\left(\;\frac{1}{3}\left(\;\frac{27}{8}\right)+\;\frac{1}{2}\left(\;\frac{9}{4}\right)-\;\frac{3}{2}\right)-\left(\;\frac{1}{3}(1)+\;\frac{1}{2}(1)-1\right)

A_{2}=\;\frac{11}{12}

square unit

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