Let f:[-1, 2] rightarrow [0, infty) be a cont | vedclass

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

Question

(B) Given $f(x) = f(1-x)$ and $R_1 = \int_{-1}^2 x f(x) dx$. Using the property $\int_{a}^b g(x) dx = \int_{a}^b g(a+b-x) dx$,where $a = -1$ and $b =

Vedclass · Accepted Answer

$2 R_1$

Vedclass · Answer

(B) Given $f(x) = f(1-x)$ and $R_1 = \int_{-1}^2 x f(x) dx$. Using the property $\int_{a}^b g(x) dx = \int_{a}^b g(a+b-x) dx$,where $a = -1$ and $b = 2$,we have $a+b = 1$. Thus,$R_1 = \int_{-1}^2 (1-x) f(1-x) dx$. Since $f(1-x) = f(x)$,we get $R_1 = \int_{-1}^2 (1-x) f(x) dx$. $R_1 = \int_{-1}^2 f(x) dx - \int_{-1}^2 x f(x) dx$. $R_1 = \int_{-1}^2 f(x) dx - R_1$. $2 R_1 = \int_{-1}^2 f(x) dx$. Since $R_2$ is the area bounded by $y = f(x)$ from $x = -1$ to $x = 2$,$R_2 = \int_{-1}^2 f(x) dx$. Therefore,$R_2 = 2 R_1$.

Let $f:[-1, 2] \rightarrow [0, \infty)$ be a continuous function such that $f(x) = f(1-x)$ for all $x \in [-1, 2]$. Let $R_1 = \int_{-1}^2 x f(x) dx$ and $R_2$ be the area of the region bounded by $y = f(x)$,$x = -1$,$x = 2$,and the $X$-axis. Then $R_2$ is:

Similar Questions

Let $I_n = \int_0^{\pi / 2} x^n \cos x \, dx$,where $n$ is a non-negative integer. Then,$\sum_{n=2}^{\infty} \left( \frac{I_n}{n!} + \frac{I_{n-2}}{(n-2)!} \right)$ equals

Find the value of $\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$.

The value of $\int_{-\pi/4}^{\pi/4} \sin^{103} x \cdot \cos^{101} x \, dx$ is

The value of the definite integral $\int_{2}^{4} (x(3 - x)(4 + x)(6 - x)(10 - x) + \sin x) dx$ equals

If the value of the integral $\int_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x$ is $\frac{2}{\pi}$,then a value of $\alpha$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module