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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Area Under Curves

NDA 2016 · Mathematics · Area Under Curves

Consider the function f(x)=|x-1|+x^2 , where x ∈ R . What is the...

Consider the function f(x)=∣x−1∣+x2\mathrm{f}(\mathrm{x})=|\mathrm{x}-1|+\mathrm{x}^2, where x∈R\mathrm{x} \in \mathrm{R}. What is the area of the region bounded by x -axis, the curve y=f(x)y=f(x) and the two ordinates x=12x=\frac{1}{2} and x=1x=1 ?

Options

  1. A512\frac{5}{12} square unit
  2. B56\frac{5}{6} square unit
  3. C76\frac{7}{6} square units
  4. D2 square units

Correct answer

A. 512\frac{5}{12} square unit

Step-by-step solution

f(x)=f(x)=\begin{cases}x^2+x-1 & x \geq 1 \\ x^2-x+1 & x < 1\end{cases}∀x∈R\quad \forall x \in R Hence area required for given region is A1=∫1/21f(x)dx=∫1/21(x2−x+1)dx=[13x3−x22+x]1/21=(13×1−12+1)−(13×18−12×14+12)A1=512 square unit. \begin{aligned} A_1 & =\int_{1/2}^1 f(x) d x \\ & =\int_{1/2}^1\left(x^2-x+1\right) d x \\ & =\left[\frac{1}{3} x^3-\frac{x^2}{2}+x\right]_{1/2}^1 \\ & =\left(\frac{1}{3} \times 1-\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} \text{ square unit. } \end{aligned}
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