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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=1x=1 and x=32?x=\frac{3}{2} ?

Options

  1. A512\frac{5}{12} square unit
  2. B712\frac{7}{12} square unit
  3. C23\frac{2}{3} square unit
  4. D1112\frac{11}{12} square unit

Correct answer

D. 1112\frac{11}{12} square unit

Step-by-step solution

Area required for givenregionisgiven region isA2=∫13/2f(x)dx=∫13/2[x2+x−1]dx=[13x3+12x2−x]13/2A2=(13(278)+12(94)−32)−(13(1)+12(1)−1)A2=1112 square unit \begin{aligned} A_2 & =\int_1^{3 / 2} f(x) d x \\ & =\int_1^{3 / 2}\left[x^2+x-1\right] d x \\ & =\left[\frac{1}{3} x^3+\frac{1}{2} x^2-x\right]_1^{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} \text{ square unit } \end{aligned}$$
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