Indefinite Integration 3 Question 2
2. Let $f:[0,2] \rightarrow R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$.
Let $F(x)=\int _0^{x^{2}} f(\sqrt{t}) d t$, for $x \in[0,2]$. If $F^{\prime}(x)=f^{\prime}(x), \forall$ $x \in(0,2)$, then $F(2)$ equals
(a) $e^{2}-1$
(b) $e^{4}-1$
(c) $e-1$
(d) $e^{4}$
(2014 Adv)
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Answer:
Correct Answer: 2. (b)
Solution:
- PLAN Newton-Leibnitz’s formula
$$ \frac{d}{d x} \int _{\varphi(x)}^{\psi(x)} f(t) d t=f{\psi(x)} \quad \frac{d}{d x} \Psi(x)-f{\varphi(x)} \quad \frac{d}{d x} \varphi(x) $$
Given,
$$ F(x)=\int _0^{x^{2}} f(\sqrt{t}) d t $$
$$ \begin{aligned} & \therefore \quad F^{\prime}(x)=2 x f(x) \\ & \text { Also, } \quad F^{\prime}(x)=f^{\prime}(x) \\ & \Rightarrow \quad 2 x f(x)=f^{\prime}(x) \\ & \Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}=2 x \\ & \Rightarrow \quad \int \frac{f^{\prime}(x)}{f(x)} d x=\int 2 x d x \Rightarrow \operatorname{In} f(x)=x^{2}+c \\ & \Rightarrow \quad f(x)=e^{x^{2}+c} \Rightarrow f(x)=K e^{x^{2}} \quad\left[K=e^{c}\right] \\ & \text { Now, } \quad f(0)=1 \\ & \therefore \quad 1=K \end{aligned} $$
Hence,
$$ f(x)=e^{x^{2}} $$
$$ F(2)=\int _0^{4} e^{t} d t=\left[e^{t}\right] _0^{4}=e^{4}-1 $$
