Definite Integration Ques 97
- Given a function $f(x)$ such that it is integrable over every interval on the real line and $f(t+x)=f(x)$, for every $x$ and a real $t$, then show that the integral $\int _a^{a+t} f(x) d x$ is independent of $a$.
$(1984,4$ M)
Integer Answer Type Question
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Solution:
- Let $\varphi(a)=\int _a^{a+t} f(x) d x$
On differentiating w.r.t. $a$, we get
$\varphi^{\prime}(a)=f(a+t) \cdot 1-f(a) \cdot 1=0$ [given, $f(x+t)=f(x)$ ]
$\therefore \varphi(a)$ is constant.
$\Rightarrow \quad \int _a^{a+t} f(x) d x$ is independent of $a$.
$f(x)$ 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 $\int _1^{2} f(x) \cos \pi x d x=\int _1^{2}(x-1) \cos \pi x d x$
$$ =-\int _0^{1} u \cos \pi u d u $$
$\therefore \quad \int _{-10}^{10} f(x) \cos \pi x d x=-20 \int _0^{1} u \cos \pi u d u=\frac{40}{\pi^{2}}$
$\Rightarrow \quad \frac{\pi^{2}}{10} \int _{-10}^{10} f(x) \cos \pi x d x=4$
BETA
