SATHEE: Definite Integration Question 29

Definite Integration Question 29

Question 29

If for a real number $y, [y]$ is the greatest integer less than or equal to $y$ , then the value of the integral $\int_{π / 2}^{3 π / 2} [2 \sin x] d x$ is

$(1999, 2 M)$ (a) $- π$ (b) 0 (c) $- \frac{π}{2}$ (d) $\frac{π}{2}$

Show Answer

Solution:

The graph of $y = 2 \sin x$ for $π / 2 \leq x \leq 3 π / 2$ is given in figure. From the graph, it is clear that

$[2 \sin x] = \begin{array}{ccc} 2, & if & x = π / 2 1, & if & π / 2 < x \leq 5 π / 6 0, & if & 5 π / 6 < x \leq π - 1, & if & π < x \leq 7 π / 6 - 2, & if & 7 π / 6 < x \leq 3 π / 2 \end{array}$

Therefore, $\int_{π / 2}^{3 π / 2} [2 \sin x] d x$

$= \int_{π / 2}^{5 π / 6} d x + \int_{5 π / 6}^{π} 0 d x + \int_{π}^{7 π / 6} (- 1) d x + \int_{7 π / 6}^{3 π / 2} (- 2) d x$

$=[x]{\pi / 2}^{5 \pi / 6}+[-x]{\pi}^{7 \pi / 6}+[-2 x]_{7 \pi / 6}^{3 \pi / 2}$

$= \frac{5 π}{6} - \frac{π}{2} + - \frac{7 π}{6} + π + \frac{- 2 \cdot 3 π}{2} + \frac{2 \cdot 7 π}{6}$

$= π \frac{5}{6} - \frac{1}{2} + π 1 - \frac{7}{6} + π \frac{7}{3} - 3$

$= π \frac{5 - 3}{6} + π - \frac{1}{6} + π \frac{7 - 9}{3} = - π / 2$

Definite Integration Question 29

Question 29

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Definite Integration Question 29

Question 29

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu