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Question

Mathematics Question on integral

If [x][x] is the greatest integer ≤x, \le x, then the value of the integral ∫−0.90.9([x2]+(2−x2+x))dx\int\limits^{0.9}_{-0.9}\left(\left[x^{2}\right]+\left(\frac{2-x}{2+x}\right)\right)dx is

A

0.486

B

0.243

C

1.8

D

0

Answer

0

Explanation

Solution

\int\limits^{0.9}_{-0.9}\left\\{\left[x^{2}\right]+\left(\frac{2-x}{2+x}\right)\right\\}dx =∫−0.90.9[x2]dx+∫−0.90.9log(2−x2+x)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+∫−0.90.9log(2−x2+x)dx= 0 +\int \limits^{0.9}_{-0.9} log \left(\frac{2-x}{2+x}\right)dx Put x=−x⇒f(x)=log2−x2+xx = - x \Rightarrow f \left(x\right) = log \frac{2-x}{2+x} and f(−x)=log2−x2+xf\left(-x\right) = log \frac{2-x}{2+x} =−log(2−x)2+x=−f(x)= - log \frac{\left(2-x\right)}{2+x} = -f\left(x\right) So, it is an odd function, hence Required integral = 0.