SATHEE: Definite Integration Question 31

Definite Integration Question 31

Question 31

Let $f (x) = x - [x]$ , for every real number $x$ , where $[x]$ is the integral part of $x$ . Then, $\int_{- 1}^{1} f (x) d x$ is

(1998, 2M) (a) 1 (b) 2 (c) 0 (d) $- \frac{1}{2}$

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Solution:

Let $\int_{- 1}^{1} f (x) d x = \int_{- 1}^{1} (x - [x]) d x = \int_{- 1}^{1} x d x - \int_{- 1}^{1} [x] d x$ $= 0 - \int_{- 1}^{1} [x] d x [∵ x$ is an odd function $]$

$- 1, if - 1 \leq x < 0$

But

$[x] = 0$ , if $0 \leq x < 1$

$1, if x = 1$

$∴ \int_{- 1}^{1} [x] d x = \int_{- 1}^{0} [x] d x + \int_{0}^{1} [x] d x$ $= \int_{- 1}^{0} (- 1) d x + \int_{0}^{1} 0 d x$ $=-[x]{-1}^{0}+0=-1 ; \quad \therefore \quad \int{-1}^{1} f(x) d x=1$

Definite Integration Question 31

Question 31

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Definite Integration Question 31

Question 31

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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