Definite Integration Ques 2
- For any real number $x$, let $[x]$ denotes the largest integer less than or equal to $x$. Let $f$ be a real valued function defined on the interval $[-10,10]$ by
$ f(x)=\left\{\begin{array}{cc} x-[x[, & \text { if } f(x) \text { is odd } \\ 1+[x[-x, & \text { if } f(x) \text { is even } \end{array}\right. $
Then, the value of $\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos $ $\pi x $ $d x$ is
(2010)
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Answer:
Correct Answer: 2.$(4)$
Solution: 6. Given, $f(x)=\left\{\begin{array}{cc}x-[x], & \text { if }[x] \text { is odd. } \\ 1+[x]-x, & \text { if }[x] \text { is even. }\end{array}\right.$
$f(x)$ and $\cos \pi x$ both are periodic with period 2 and both are even.
$ \therefore \quad \int_{-10}^{10} f(x) \cos \pi x d x=2 \int_0^{10} f(x) \cos \pi x d x $
$ =10 \int_0^2 f(x) \cos \pi x d x $
Now, $\int_0^1 f(x) \cos \pi x d x$
$ =\int_0^1(1-x) \cos \pi x d x=-\int_0^1 u \cos \pi u d u $
and $\int_1^2 f(x) \cos \pi x d x=\int_1^2(x-1) \cos \pi x d x$
$ =-\int_0^1 u \cos \pi u d u $
$ \therefore \quad \int_{-10}^{10} f(x) \cos \pi x d x=-20 \int_0^1 u \cos \pi u d u=\frac{40}{\pi^2} $
$ \Rightarrow \frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x d x=4 $
BETA
