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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Definite Integration

NDA 2016 · Mathematics · Definite Integration

Consider the functions f(x)=x g(x) and g(x)=[(1)/(x)] Where [·] is...

Consider the functions f(x)=xg(x) and g(x)=[1x]Where[⋅]f(x)=x g(x) \text{ and } g(x)=\left[\frac{1}{x}\right] Where [\cdot] is the greatest integer function. What is \int_{13\frac{1}{3}}1f(x)dx^1 f(x) d x equal to ?

Options

  1. A3772\frac{37}{72}
  2. B23\frac{2}{3}
  3. C1772\frac{17}{72}
  4. D37144\frac{37}{144}

Correct answer

A. 3772\frac{37}{72}

Step-by-step solution

∫1/31f(x)dx=∫1/31/2f(x)dx+∫1/21f(x)dx...(1)f(x)=xg(x)...(1)\begin{aligned} & \int_{1/3}^1 f(x) d x=\int_{1/3}^{1 / 2} f(x) d x+\int_{1/2}^1 f(x) d x...(1)\\ & \mathrm{f}(\mathrm{x})=\mathrm{xg}(\mathrm{x}) ...(1) \end{aligned}g\left(\frac{1}{3}\right)=3 \quad g\left(\frac{1}{2}\right)=2 \quad g(1)=1ThevalueofThe value ofg(x)invalue(12,13)in value \left(\frac{1}{2}, \frac{1}{3}\right) will be 2 and in range \left(\frac{1}{2}, 1\right)itwillbe1from(1)it will be 1 from (1)∫1/31f(x)dx=∫1/31/2xg(x)dx+∫1/21xg(x)dx=∫1/31/2x⋅2dx+∫1/21x×1dx=[x2]1/31/2+[x22]1/21=[14−19]+12[1−14]\begin{aligned} & \int_{1/3}^1 f(x) d x=\int_{1/3}^{1 / 2} x g(x) d x+\int_{1/2}^1 x g(x) d x \\ & =\int_{1/3}^{1 / 2} x \cdot 2 d x+\int_{1/2}^1 x \times 1 d x \\ & =\left[x^2\right]_{1/3}^{1 / 2}+\left[\frac{x^2}{2}\right]_{1/2}^1 \\ & =\left[\frac{1}{4}-\frac{1}{9}\right]+\frac{1}{2}\left[1-\frac{1}{4}\right] \end{aligned}=536+38=3772=\frac{5}{36}+\frac{3}{8}=\frac{37}{72}
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