Let f: R rightarrow R be a function defined b | vedclass

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

Question

(A) Given $f(x) = \begin{cases} [x], & x \leq 2 \ 0, & x > 2 \end{cases}$.We need to evaluate $I = \int_{-1}^2 \frac{x f(x^2)}{2+f(x+1)} dx$.For $x \

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given $f(x) = \begin{cases} [x], & x \leq 2 \ 0, & x > 2 \end{cases}$.We need to evaluate $I = \int_{-1}^2 \frac{x f(x^2)}{2+f(x+1)} dx$.For $x \in [-1, 2]$,$x^2 \in [0, 4]$.If $x^2 \leq 2$,$f(x^2) = [x^2]$. If $x^2 > 2$,$f(x^2) = 0$.$x^2 \leq 2 \implies x \in [-\sqrt{2}, \sqrt{2}]$.Since the integral is from $-1$ to $2$,we split it:For $x \in [-1, \sqrt{2}]$,$f(x^2) = [x^2]$. For $x \in (\sqrt{2}, 2]$,$f(x^2) = 0$.Also,$f(x+1) = [x+1]$ for $x+1 \leq 2 \implies x \leq 1$,and $f(x+1) = 0$ for $x > 1$.Thus,$I = \int_{-1}^1 \frac{x [x^2]}{2+[x+1]} dx + \int_1^{\sqrt{2}} \frac{x [x^2]}{2+0} dx + \int_{\sqrt{2}}^2 \frac{x \cdot 0}{2+0} dx$.For $x \in [-1, 0)$,$[x^2] = 0$,so the first part is $\int_{-1}^0 0 dx = 0$.For $x \in [0, 1)$,$[x^2] = 0$,so $\int_0^1 0 dx = 0$.For $x \in [1, \sqrt{2}]$,$[x^2] = 1$,so $\int_1^{\sqrt{2}} \frac{x \cdot 1}{2} dx = \frac{1}{2} [\frac{x^2}{2}]_1^{\sqrt{2}} = \frac{1}{4} (2-1) = \frac{1}{4}$.Thus $I = \frac{1}{4}$.Then $4I - 1 = 4(\frac{1}{4}) - 1 = 0$.

Let $f: R \rightarrow R$ be a function defined by $f(x)=\begin{cases} [x], & x \leq 2 \\ 0, & x>2 \end{cases}$,where $[x]$ is the greatest integer less than or equal to $x$. If $I=\int_{-1}^2 \frac{x f(x^2)}{2+f(x+1)} dx$,then the value of $(4I-1)$ is

Similar Questions

If $x$ satisfies the equation $\left( \int_{0}^{1} \frac{dt}{t^2 + 2t \cos \alpha + 1} \right) x^2 - \left( \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt \right) x - 2 = 0$ for $0 < \alpha < \pi$,then the value of $x$ is

The value of $\int\limits_0^{\frac{\pi }{2}} {\sin 8x \cot x \, dx} + \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\ln \left( {\frac{{1 - \sin x}}{{1 + \sin x}}} \right)dx}$ is equal to

$\int_{0}^{2} ( |2x^2 - 3x| + [x - \frac{1}{2}] ) dx$,where $[ \cdot ]$ is the greatest integer function,is equal to:

$ 6\int_{0}^{\pi}|(\sin 3x+\sin 2x+\sin x)| dx $ is equal to ....

Vedclass Test Series

Exam Paper Generator

Online Exam Module