Let f - -0- 2- - be a function which is continuous on -0- 2- and is differentiable on 0- 2 with f0 - 1- Let Fx - -0x2 f -t dt for x - -0- 2- If F- x - f-x for all x - 0- 2- then F2 equals-

F'(x) = f(x).2x = f'(x) ⇒f'(x)f(x) = 2x ⇒ f(x) = ke^x^2 Given f(0) = 1 ⇒ f(x) = e^x^2 So F(x) = nbsp;∫_0^x^2 e^x dx = e^x^2 1 So F(2) = e^4 1 ...

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Standard XII

Mathematics

Theorems on Integration

Let f : [0, 2...

Question

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

e^{4}

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e^{2} - 1

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e^{4} - 1

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e - 1

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Solution

The correct option is C $e^{4} - 1$
$F^{'} (x) = f (x) .2 x = f^{'} (x)$
$\Rightarrow \frac{f^{'} (x)}{f (x)} = 2 x$
$\Rightarrow f (x) = k e^{x^{2}}$
Given $f (0) = 1 \Rightarrow f (x) = e^{x^{2}}$
So $F (x) = \int_{0}^{x^{2}} e^{x} d x = e^{x^{2}} - 1$
So $F (2) = e^{4} - 1$

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Let f:[0,2]→ be a function which is continuous on [0,2] and is differentiable on (0,2) with f(0)=1. Let F(x)=∫x20f(√t)dt for x ∈ [0,2]. If F′(x)=f′(x) for all x ∈ (0,2), then F(2) equals.

The correct option is C e4−1F′(x)=f(x).2x=f′(x) ⇒f′(x)f(x)=2x ⇒f(x)=kex2 Given f(0)=1⇒f(x)=ex2 So F(x)=∫x20exdx=ex2−1 So F(2)=e4−1

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

The correct option is C $e^{4} - 1$
$F^{'} (x) = f (x) .2 x = f^{'} (x)$
$\Rightarrow \frac{f^{'} (x)}{f (x)} = 2 x$
$\Rightarrow f (x) = k e^{x^{2}}$
Given $f (0) = 1 \Rightarrow f (x) = e^{x^{2}}$
So $F (x) = \int_{0}^{x^{2}} e^{x} d x = e^{x^{2}} - 1$
So $F (2) = e^{4} - 1$