Let f(x) = x - [x] for every real number x , where [x] is the greatest integer less than or equal to x . Then the value of _-1^1 f(x) dx is: - Mathematics MCQ Mathematics Mock Test - 10 2026 | ExamTest.live

mediumMCQMathematics Mock Test - 102026Mathematics

11 attempts0% success rate1 mark

Let f (x) = x - [x] for every real number x, where [x] is the greatest integer less than or equal to x . Then the value of \int_{- 1}^{1} f (x) d x is:

Solution & Step-by-step Explanation

I = \int_{- 1}^{1} (x - [x]) d x = \int_{- 1}^{0} (x - [x]) d x + \int_{0}^{1} (x - [x]) d x .In the interval [- 1, 0), [x] = - 1 .In the interval [0, 1), [x] = 0 . I = \int_{- 1}^{0} (x - (- 1)) d x + \int_{0}^{1} (x - 0) d x I = \int_{- 1}^{0} (x + 1) d x + \int_{0}^{1} x d x I = [\frac{x ^{2}}{2} + x]_{- 1}^{0} + [\frac{x ^{2}}{2}]_{0}^{1} I = [0 - (\frac{1}{2} - 1)] + [\frac{1}{2} - 0] I = \frac{1}{2} + \frac{1}{2} = 1 .

Try it yourself before checking the explanation above.

Let f (x) = x - [x] for every real number x, where [x] is the greatest integer less than or equal to x . Then the value of \int_{- 1}^{1} f (x) d x is:

0

-1

\frac{1}{2}

1

Practice with timed mock tests and track your performance across Mathematics.