Question:
If $[x]$ is the greatest integer $ \le x,$ then the value of the integral $\int\limits^{0.9}_{-0.9}\left(\left[x^{2}\right]+\left(\frac{2-x}{2+x}\right)\right)dx$ is
If $[x]$ is the greatest integer $ \le x,$ then the value of the integral $\int\limits^{0.9}_{-0.9}\left(\left[x^{2}\right]+\left(\frac{2-x}{2+x}\right)\right)dx$ is
Updated On: Jul 28, 2022
- 0.486
- 0.243
- 1.8
- 0
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The Correct Option is D
Solution and Explanation
$\int\limits^{0.9}_{-0.9}\left\{\left[x^{2}\right]+\left(\frac{2-x}{2+x}\right)\right\}dx$
$= \int \limits^{0.9}_{-0.9} \left[x^{2}\right]dx+\int \limits^{0.9}_{-0.9} log \left(\frac{2-x}{2+x}\right)dx$
$= 0 +\int \limits^{0.9}_{-0.9} log \left(\frac{2-x}{2+x}\right)dx$
Put $x = - x \Rightarrow f \left(x\right) = log \frac{2-x}{2+x}$
and $f\left(-x\right) = log \frac{2-x}{2+x}$
$= - log \frac{\left(2-x\right)}{2+x} = -f\left(x\right)$
So, it is an odd function, hence Required integral = 0.
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.