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

Let $f(x) = \int\limits_0^x {(t^2 + 2t + 2)dt}$ where $x$ is the set of real numbers satisfying the inequation $\log_{\sqrt{2}}(1 + \sqrt{6x - x^2 - 8}) \ge 0$. If the range of $f(x)$ is $[a, b]$,then $(a + b)$ is:

For each real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$,and let $\{x\} = x - [x]$. Then the smallest positive integer $M$ for which $\int_1^M \{x\}^{[x]} dx > 1$ is

Let $[t]$ denote the greatest integer less than or equal to $t.$ Then,the value of the integral $\int\limits_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x$ is equal to

The value of the limit $\lim _{n \rightarrow \infty} \int _{0}^{1} x^{10} \sin (n x) d x$ equals

The value of $\int_{0}^{1} 9x^8 dx + \int_{0}^{\pi/2} \cos x dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module