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

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(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

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

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(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?

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