MereTeacher

JEE • NEET • FOUNDATION

JEE Previous Year Question

$Let [t] denote the greatest integer less than or equal to t . Then the value of \int_{1}^{2} ∣2 x - [3 x] ∣ d x is$

Topic: JEE PYQ

Answer: 1. Full step-by-step solution below.

Correct answer: 1

Solution

(1) \sum | 2 x - [3 x] | dx = \sum | 3x - x | dx = \sum (x - 3x) dx = \sum xdx - \sum 3x dx = \frac{( 4 - 1 )}{2} - 3 \sum 3 xdx = \frac{3}{2} - \frac{1}{2} = 1

More Definite Integration PYQs

$Let g (x) = \int_{0}^{x} f (t) d t, where f is continuous function in [0,3] such that \frac{1}{3} \leq f (t) \leq 1 for all\dots$
$If \int_{0}^{K} \frac{dx}{2 + 18 x ^{2}} = \frac{π}{24}, then the value of K is$
$\int_{0}^{4} \frac{1}{1 + x} dx =$
$Let a differentiable function f satisfy f (x) + \int_{3}^{x} \frac{f ( t )}{t} d t = x + 1, x \geq 3 . Then 12 f (8) is equal to$
$The value of the integral \int_{- 1}^{2} log_{e} (x + x^{2} + 1) dx is$