Let f:[0,1] rightarrow [0, infty) be a contin | vedclass

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

Question

(D) Given that $f(x) \geq 0$ and $\int_0^1 f(x) dx = 10$.For option $A$: Since $0 < e^{-x} \leq 1$ for $x \in [0, 1]$,we have $\int_0^1 e^{-x} f(x) dx

Vedclass · Accepted Answer

$\int_0^1 f(x)^2 dx \leq 100$

Vedclass · Answer

(D) Given that $f(x) \geq 0$ and $\int_0^1 f(x) dx = 10$.For option $A$: Since $0 For option $B$: Since $f(x) \geq 0$ and $(1+x)^2 > 0$,the integral $\int_0^1 -\frac{f(x)}{(1+x)^2} dx$ is $\leq 0$. Since $0 \leq 10$,this is true.For option $C$: Since $|\sin(100x)| \leq 1$,we have $|\int_0^1 \sin(100x) f(x) dx| \leq \int_0^1 |\sin(100x)| f(x) dx \leq \int_0^1 f(x) dx = 10$. Thus,$-10 \leq \int_0^1 \sin(100x) f(x) dx \leq 10$. This is true.For option $D$: Consider $f(x) = c$ (a constant). Then $\int_0^1 c dx = c = 10$. So $f(x) = 10$. Then $\int_0^1 f(x)^2 dx = \int_0^1 100 dx = 100$. However,if we take a function that is very large on a small interval and zero elsewhere,the integral of $f(x)^2$ can be arbitrarily large. For example,let $f_n(x) = (n+1)x^n$. Then $\int_0^1 f_n(x) dx = 1$,so let $f(x) = 10(n+1)x^n$. Then $\int_0^1 f(x)^2 dx = 100(n+1)^2 \int_0^1 x^{2n} dx = 100 \frac{(n+1)^2}{2n+1}$,which approaches $\infty$ as $n 	o \infty$. Thus,this statement is $NOT$ necessarily true.

Let $f:[0,1] \rightarrow [0, \infty)$ be a continuous function such that $\int_0^1 f(x) dx = 10$. Which of the following statements is $NOT$ necessarily true?

Similar Questions

Let the function $f :[0,2] \rightarrow R$ be defined as $f(x)=\begin{cases} e^{\min \{x^2, x-[x]\}}, & x \in[0,1) \\ e^{[x-\log_e x]}, & x \in[1,2] \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then the value of the integral $\int_0^2 x f(x) dx$ is

$I = \int_{\pi/4}^{\pi/3} \frac{8 \sin x - \sin 2x}{x} dx$. Then

If $f(x) = \int_{-1}^{x} |t| dt$,then for any $x \geq 0$,$f(x)$ is equal to

The approximate value of $\int_{1}^{9} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

$\int_{-2}^{1} [x+1] \, dx =$ (Where $[x]$ is the greatest integer function not greater than $x$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module