Definite Integration Ques 58
- Let $f(x)=\int _0^{x} g(t) d t$, where $g$ is a non-zero even function. If $f(x+5)=g(x)$, then $\int _0^{x} f(t) d t$ equals
(2019 Main, 8 April II)
(a) $5 \int _{x+5}^{5} g(t) d t$
(b) $\int _5^{x+5} g(t) d t$
(c) $2 \int _5^{x} g(t) d t$
(d) $\int _{x+5}^{5} g(t) d t$
Show Answer
Solution:
- Given, $f(x)=\int _0^{x} g(t) d t$
On replacing $x$ by $(-x)$, we get
$$ f(-x)=\int _0^{-x} g(t) d t $$
Now, put $t=-u$, so
$$ f(-x)=-\int _0^{x} g(-u) d u=-\int _0^{x} g(u) d u=-f(x) $$
$[\because g$ is an even function] $\Rightarrow \quad f(-x)=f(x) \Rightarrow f$ is an odd function.
Now, it is given that $f(x+5)=g(x)$
$$ \begin{aligned} & \therefore \quad f(5-x)=g(-x)=g(x)=f(x-5) \\ & \Rightarrow \quad f(5-x)=f(x+5) \end{aligned} $$
Put $t=u+5 \Rightarrow t-5=u \Rightarrow d t=d u$
$$ \begin{array}{ll} \therefore \quad & I=\int _{-5}^{x-5} f(u+5) d u=\int _{-5}^{x-5} g(u) d u \\ Put \quad u=-t \Rightarrow d u=-d t, \text { then we get } \\ & I=-\int _5^{5-x} g(-t) d t=\int _{5-x}^{5} g(t) d t \end{array} $$
$$ \begin{alignedat} & {\left[\because-\int _a^{b} f(x) d x=\int _b^{a} f(x) d x \text { and } g\right. \text { is an even function] }} \\ & \left.I=\int _{-x}^{5} f^{\prime}(t) d t \quad \text { [by Fundamental Theorem of Calculus } f^{\prime}(x)=g(x)\right] \\ & \quad=f(5)-f(5-x)=f(5)+f(5+x) \quad \text { [from Eq. (i)] } \\ & \quad=\int _{5+x}^{5} f^{\prime}(t) d t=\int _{5+x}^{5} g(t) d t \end{aligned} $$
BETA
