SATHEE: Definite Integration Question 34

Definite Integration Question 34

Question 34

The value of $\int_{0}^{2 π} [2 \sin x] d x$ , where [.] represents the greatest integral functions, is (a) $- \frac{5 π}{3}$ (b) $- π$ (c) $\frac{5 π}{3}$ (d) $- 2 π$

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Solution:

It is a question of greatest integer function. We have, subdivide the interval $π$ to $2 π$ as under keeping in view that we have to evaluate $[2 \sin x]$

We know that, $\sin \frac{π}{6} = \frac{1}{2}$

$\begin{array}{lll} ∴ & \sin π + \frac{π}{6} & = \sin \frac{7 π}{6} = - \frac{1}{2} \Rightarrow & \sin \frac{11 π}{6} = \sin 2 π - \frac{π}{6} = - \sin \frac{π}{6} = - \frac{1}{2} \Rightarrow & \sin \frac{9 π}{6} = \sin \frac{3 π}{6} = - 1 \end{array}$

Hence, we divide the interval $π$ to $2 π$ as

$\begin{aligned} π, \frac{7 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}, \frac{11 π}{6}, 2 π & \sin x = 0, - \frac{1}{2}, - 1, - \frac{1}{2}, - \frac{1}{2}, 0 & \Rightarrow 2 \sin x = (0, - 1), (- 2, - 1), (- 1, 0) & \Rightarrow [2 \sin x] = - 1 & = \int_{π}^{7 π / 6} [2 \sin x] d x + \int_{7 π / 6}^{11 π / 6} [2 \sin x] d x & + \int_{11 π / 6}^{2 π} [2 \sin x] d x & = \int_{π}^{7 π / 6} (- 1) d x + \int_{7 π / 6}^{11 π / 6} (- 2) d x + \int_{11 π / 6}^{2 π} (- 1) d x & = - \frac{π}{6} - 2 \frac{4 π}{6} - \frac{π}{6} = - \frac{10 π}{6} = - \frac{5 π}{3} \end{aligned}$

Definite Integration Question 34

Question 34

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Definite Integration Question 34

Question 34

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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