If F(x) = frac{1}{x^2} int_4^x (4t^2 - 2F'(t) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $F(x) = \frac{1}{x^2} \int_4^x (4t^2 - 2F'(t)) \, dt$. Multiplying by $x^2$,we get $x^2 F(x) = \int_4^x (4t^2 - 2F'(t)) \, dt$. Differentiat

Vedclass · Accepted Answer

$\frac{32}{9}$

Vedclass · Answer

(C) Given $F(x) = \frac{1}{x^2} \int_4^x (4t^2 - 2F'(t)) \, dt$. Multiplying by $x^2$,we get $x^2 F(x) = \int_4^x (4t^2 - 2F'(t)) \, dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $2x F(x) + x^2 F'(x) = 4x^2 - 2F'(x)$. At $x = 4$: $2(4) F(4) + (4)^2 F'(4) = 4(4)^2 - 2F'(4)$. From the original equation,$F(4) = \frac{1}{4^2} \int_4^4 (4t^2 - 2F'(t)) \, dt = 0$. Substituting $F(4) = 0$ into the differentiated equation: $8(0) + 16 F'(4) = 64 - 2F'(4)$. $16 F'(4) + 2 F'(4) = 64$. $18 F'(4) = 64$. $F'(4) = \frac{64}{18} = \frac{32}{9}$.

If $F(x) = \frac{1}{x^2} \int_4^x (4t^2 - 2F'(t)) \, dt$,then $F'(4)$ equals:

Similar Questions

If $f(x) = \int_a^x {t^3 e^t \, dt}$,then $\frac{d}{dx} f(x) = $

$\int_0^{\pi /2} \sin^{2m} x \, dx = $

If $I=\int_{-a}^a(x^4-2x^2)dx$,then $I$ is minimum at $a=$

$\int_0^\pi x \cdot \sin^5 x \cdot \cos^6 x \, dx =$

Let $F(x) = \int_x^{x^2+\frac{\pi}{6}} 2 \cos^2 t \, dt$ for all $x \in \mathbb{R}$ and $f: [0, \frac{1}{2}] \rightarrow [0, \infty)$ be a continuous function. For $a \in [0, \frac{1}{2}]$,if $F'(a) + 2$ is the area of the region bounded by $x=0, y=0, y=f(x)$ and $x=a$,then $f(0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module