SATHEE: Indefinite Integration 3 Question 2

Indefinite Integration 3 Question 2

2. Let $f : [0, 2] \to R$ 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 $F^{'} (x) = f^{'} (x), \forall$ $x \in (0, 2)$ , then $F (2)$ equals

(a) $e^{2} - 1$

(b) $e^{4} - 1$

(d) $e^{4}$

(2014 Adv)

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Answer:

Correct Answer: 2. (b)

Solution:

PLAN Newton-Leibnitz’s formula

$\frac{d}{d x} \int_{φ (x)}^{ψ (x)} f (t) d t = f ψ (x) \frac{d}{d x} Ψ (x) - f φ (x) \frac{d}{d x} φ (x)$

Given,

$F (x) = \int_{0}^{x^{2}} f (\sqrt{t}) d t$

$\begin{aligned} ∴ F^{'} (x) = 2 x f (x) \\ Also, F^{'} (x) = f^{'} (x) \\ \Rightarrow 2 x f (x) = f^{'} (x) \\ \Rightarrow \frac{f^{'} (x)}{f (x)} = 2 x \\ \Rightarrow \int \frac{f^{'} (x)}{f (x)} d x = \int 2 x d x \Rightarrow In f (x) = x^{2} + c \\ \Rightarrow f (x) = e^{x^{2} + c} \Rightarrow f (x) = K e^{x^{2}} [K = e^{c}] \\ Now, f (0) = 1 \\ ∴ 1 = K \end{aligned}$

Hence,

$f (x) = e^{x^{2}}$

$F (2) = \int_{0}^{4} e^{t} d t = {[e^{t}]}_{0}^{4} = e^{4} - 1$

Indefinite Integration 3 Question 2

2. Let $f : [0, 2] \to R$ be a function which is continuous on $[0, 2]$ and is differentiable on $(0, 2)$ with $f (0) = 1$ .

Answer:

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Indefinite Integration 3 Question 2

2. Let f:[0,2]→R be a function which is continuous on [0,2] and is differentiable on (0,2) with f(0)=1.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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2. Let $f : [0, 2] \to R$ be a function which is continuous on $[0, 2]$ and is differentiable on $(0, 2)$ with $f (0) = 1$ .