SATHEE: Definite Integration Question 54

Definite Integration Question 54

Question 54

Let $f : [1, \infty] \to [2, \infty]$ be differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) = d t = 3 x f (x) - x^{3}, \forall x \geq 1$ then the value of $f (2)$ is

(2011)

Show Answer

Solution:

Given, $f (1) = \frac{1}{3}$ and $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}, \forall x \geq 1$

Using Newton-Leibnitz formula.

Differentiating both sides

$\begin{aligned} \Rightarrow 6 f (x) \cdot 1 - 0 = 3 f (x) + 3 x f^{'} (x) - 3 x^{2} & \Rightarrow 3 x f^{'} (x) - 3 f (x) = 3 x^{2} \Rightarrow f^{'} (x) - \frac{1}{x} f (x) = x & \Rightarrow \frac{x f^{'} (x) - f^{'} (x)}{x^{2}} = 1 \Rightarrow \frac{d}{d x} \frac{x}{x} = 1 \end{aligned}$

On integrating both sides, we get

$\begin{aligned} \Rightarrow \frac{f (x)}{x} & = x + c \frac{1}{3} & = 1 + c \Rightarrow c = \frac{2}{3} and f (x) = x^{2} - \frac{2}{3} x ∴ f (2) & = 4 - \frac{4}{3} = \frac{8}{3} \end{aligned}$

NOTE Here, $f (1) = 2$ , does not satisfy given function.

$∴ f (1) = \frac{1}{3}$

For that $f (x) = x^{2} - \frac{2}{3} x$ and $f (2) = 4 - \frac{4}{3} = \frac{8}{3}$

Definite Integration Question 54

Question 54

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Definite Integration Question 54

Question 54

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu