Let f : R rightarrow R be a twice differentia | vedclass

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

Question

(B) Given $\int_0^2 x F^{\prime}(x) dx = 6$. Using integration by parts: $[x F(x)]_0^2 - \int_0^2 F(x) dx = 6$ $2 F(2) - 0 - \int_0^2 F(x) dx = 6$. Si

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) Given $\int_0^2 x F^{\prime}(x) dx = 6$. Using integration by parts: $[x F(x)]_0^2 - \int_0^2 F(x) dx = 6$ $2 F(2) - 0 - \int_0^2 F(x) dx = 6$. Since $F(x) = x f(x)$,$F(2) = 2 f(2) = 2(1) = 2$. So,$2(2) - \int_0^2 F(x) dx = 6 \implies \int_0^2 F(x) dx = 4 - 6 = -2$. Now,consider $\int_0^2 x^2 F^{\prime \prime}(x) dx = 40$. Using integration by parts: $[x^2 F^{\prime}(x)]_0^2 - \int_0^2 2x F^{\prime}(x) dx = 40$ $4 F^{\prime}(2) - 2 \int_0^2 x F^{\prime}(x) dx = 40$. Substituting the given value $\int_0^2 x F^{\prime}(x) dx = 6$: $4 F^{\prime}(2) - 2(6) = 40$ $4 F^{\prime}(2) - 12 = 40 4 F^{\prime}(2) = 52 F^{\prime}(2) = 13$. Finally,$F^{\prime}(2) + \int_0^2 F(x) dx = 13 + (-2) = 11$.

Let $f : R \rightarrow R$ be a twice differentiable function such that $f(2)=1$. If $F(x) = x f(x)$ for all $x \in R$,$\int_0^2 x F^{\prime}(x) dx = 6$ and $\int_0^2 x^2 F^{\prime \prime}(x) dx = 40$,then $F^{\prime}(2) + \int_0^2 F(x) dx$ is equal to:

Similar Questions

If $f(x)$ is a function satisfying $f^{\prime}(x)=f(x)$ with $f(0)=1$ and $g(x)$ is a function that satisfies $f(x)+g(x)=x^2$. Then the value of the integral $\int_0^1 f(x) g(x) d x$ is

Let $f(x)$ be a function satisfying $f'(x) = f(x)$ with $f(0) = 1$ and $g(x)$ be a function satisfying $f(x) + g(x) = x^2$. The value of the integral $\int_{0}^{1} f(x)g(x) \, dx$ is

Value of the $\int_0^9 \sqrt{x} \,dx + \int_0^{\pi/2} (\cos x + \sin x) \,dx$ is:

If $I_n = \int_{0}^{1} \frac{dx}{(1 + x^2)^n}$; $n \in N$,then which of the following statements hold good?

Let $f(\theta) = \sin \theta + \int_{-\pi / 2}^{\pi / 2} (\sin \theta + t \cos \theta) f(t) dt$. Then the value of $\left| \int_{0}^{\pi / 2} f(\theta) d\theta \right|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module