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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 : 16 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=\frac{1}{2}

x=1

$\frac{15}{2}$ square units
$\frac{5}{6}$ square units
$\frac{7}{6}$ square units
2 square units

Solution:

f(x)=\left\{\begin{array}{llll} x^2+x-1, & x \ge 1, & \forall, & x \in \mathbb{R} \\ x^2-x+1, & x < 1 & & \end{array}\right.

Hence area required for given region is

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

=\int\limits_{\frac{1}{2}}^{1}(x^{2}-x+1)\,dx

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

=\left(\;\;\frac{1}{3}\times1-\;\;\frac{1}{2}+1\right)-\left(\;\;\frac{1}{3}\times\;\;\frac{1}{8}-\;\;\frac{1}{2}\times\;\;\frac{1}{4}+\;\;\frac{1}{2}\right)

A_1=\frac{5}{12}

square units

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