Let $f(x)$ be a real valued function whose first, second derivatives are continuous in $[0,2\pi]$. Also, $f''(x)\geq0$.  Show that    $\int_{0}^{2\pi}f(x)\cos{x} dx\geq 0$.  I tried using Integration by parts twice, by which I got   $I= \int_{0}^{2\pi}f(x)\cos{x} dx\geq 0$  $\implies f'(2\pi)-f'(0)\geq\int_{0}^{2\pi}f''(x)\cos{x}dx$  I know that this LHS should be $\geq0$, but how do I compare it with RHS?  Any hints, solutions or suggestions would be really helpful. Thank you!
59735
Let $f(x)$ be a continuous function, whose first and second derivatives are continuous on $[0,2π]$ and $f&#39;&#39;(x)≥0\:\:∀\:x∈[0,2π]$. Show that$$ \int _0^{2\pi} f(x) \cos x dx \ge 0$$
88196
Let $f$ be a continuous function on the unit interval $[0,1]$. Show that $$\lim_{n\to\infty}\int_{0}^{1}\cdots\int_0^1\int_{0}^{1}f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=f\left(\dfrac{1}{2}\right)$$  This problem is from Selected Problems in Real Analysis. But the author doesn't include a solution. Maybe there is a method for this sort of problem?  Maybe we can use this: $$\int_0^1\cdots\int_{0}^{1}\int_{0}^{1}\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2}?$$  I also found a similar problem in .
53743
Let $f$ be continuous function on $[a,b]\times [c,d]$. Prove that      Prove that $$\int_a^b \left( \int_c^d f(x,y)dy\right) dx=\int_c^d  \left( \int_a^b f(x,y)dx\right) dy$$   First of all, note that $ \int_c^d f(x,y)dy$ is continuous in $x$ and $ \int_a^b f(x,y)dx$ is continuous in $y$ so both integrals exist.  We have  $$\frac{d\left[\int_a^b \left( \int_c^t f(x,y)dy\right) dx \right]}{dt}=\int_a^b f(x,t) dx$$  (We used differentiation under the integral sign and the fundamental theorem of calculus)  Also  $$\frac{d\left[\int_c^t  \left( \int_a^b f(x,y)dx\right) dy \right]}{dt}=\int_a^b f(x,t) dx$$  (Here we used the fundamental theorem of calculus)  Thus $$\int_a^b \left( \int_c^t f(x,y)dy\right) dx-\int_c^t  \left( \int_a^b f(x,y)dx\right) dy$$  as a function of $t$ is constant on $(c,d)$.  It remains to show that this constant is $0$. How can I do this?
52466
Let $f:[0,1]\to\mathbb{R}$ be continuous and $f(0)=f(1)$. Show that for any integer $n\geqslant 2$, there is $\xi\in(0,1)$ s.t.$f(\xi)=f\left(\xi+\frac{1}{n}\right).$  I think this requires the intermediate value theorem somewhere.
9297
Let $f : [0, 2] → \mathbb{R}$ be a continuous, positive function and  $$\int_{0}^1 f(x)\,dx = \int_{1}^2 f(x)\,dx= 1$$ For each $x \in [0, 1]$, prove that there exists a unique function $g(x) \in [1, 2]$ such that $$\int_{x}^{g(x)} f(t)\,dt = 1$$ and $g \in C^1$. Well, I think in $g(x) = x +1$, but I'm not sure. Any hint is appreciated.  Thank you.
94975
Suppose $f$ is a continuous single-variable function, prove that:  $$2\int_a^b \int_a^x f(x)f(y) \, dy \, dx = \left[ \int_a^b f(x) \, dx \right]^2$$  This question was just on my Calculus III final exam and I couldn't figure it out, could someone provide some insight, please?
22884
Assuming $a&lt;b$, function f(x) is continous at [a,b], and we have  $\int_a^bf(x)dx=\int_a^bxf(x)dx=\int_a^bx^2f(x)dx=0$  Prove that $\exists \ x_1,x_2,x_3\text {(different from each other)}  \in(a,b)$ satisfying $f(x_1)=f(x_2)=f(x_3)=0$.  I have proved the existence of one zero point by differential mean value theorem, but have no idea about going on.  Thanks for any solution or hint.
120740
Let $f$ be $C^1$ in $[-\pi, \pi]$ and satisfies $\int_{-\pi}^\pi f(x)dx=0$, periodic boundary condition.  Then, prove that $\int_{-\pi}^\pi (f(x))^2dx\le \int_{-\pi}^\pi (f'(x))^2dx$.  I try to prove it by Parseval's equality(with $X_n=\sin nx$) and Schwartz inequality,   but then some constants come out. Also why condition $\int_{-\pi}^\pi f(x)dx=0$ needs?  Give some advice.
6428
(a) Let a function $g:I\to\mathbb{R}$ be continuous at $x_0\in I$. Prove that the function $f$ given by $f(x)=(x-x_0)g(x)$ is differentiable at $x_0$.  (b) Let a function $f:I\to\mathbb{R}$ be differentiable at $x_0\in I$ and $f(x_0)=0$. Prove that there exists a function $g:I\to \mathbb{R}$ that is continuous at $x_0 \in I$ such that $f(x)=(x-x_0)g(x)$.    Thought:  (a)  By $Fermat's$, since $g$ is continous, if $g'(x_0)=0$ then $g(x)$ is differentiable?  but how does it connect back with $f(x)$?  (b)  use the definition of every differentiable function is continuous?  Thus $f(x)$ is continuous and also $(x-x_0)$ so $g(x)$ too?  Thank you!!
113752
Let $f(x)$ be a strictly positive continuous function in $[0,1]$ such that $f^2(t) \leq 1 + 2\int_0^t f(s)\, \mathrm{d}s$. Prove that $f(t)\leq t+1$.    Obviously, if $f(x)$ can be differentiated, it is true. But I can't deal with the case that it is not differentiated.
99631
I have tried the following: $$ f(x) = \sin x-\frac{2x}{\pi} \\ f'(x)= \cos x-\frac{2}{\pi} \\ f''(x) = -\sin x \leq 0 $$ But this doesn't seem to be heading in the right direction as it would appear $f'(x)$ is decreasing and has a zero within the domain, so I'm not sure how this might be used to prove the inequality.
11660
I am given $f,f' \in L^1(\mathbb{R})$, and f is absolute continuous, I want to show that:  $$\lim_{|x|\rightarrow \infty} f(x)= 0$$  Not sure how to show this, I know that $f(x)=\int_0^x f'(t) \, dt+f(0)$, and I can assume without loss of generality that $f(0)=0$, any help?  Thanks in advance.
45139
Suppose for real numbers $a&lt;b$ one has a function with continuous derivative $f:[a,b] \to \mathbb{R}$ such that $f(a) = 0$ and there exists a real number $C$ with $$|f'(x)| \leq C|f(x)|$$ for all $x \in [a,b]$. Show that $f(x) = 0$ for all $x \in [a,b]$.  Well, since $f(a) = 0$, we have that $|f'(a)| \leq C|0|$, so $|f'(a)| = 0$. Since $f$ has a continuous derivative, we also know that $f$ is continuous. Since $f$ is continuous on a compact interval, $f$ obtains a maximum, say at $\xi \in [a,b]$. So, $|f'(x)| \leq C|f(\xi)|$. Since the derivative is bounded, we obtain that $f$ is Lipschitz, so $f$ is also uniformly continuous. Suppose to the contrary that $f(\xi) &gt; 0$? The above is pretty much everything I could figure out about $f$, so I'm not sure what to try next. This one is also from an old qual and possibly uses methods from beyond our course. I think maybe I should try to show that $|f'(x)| = 0$ for all $x$, but I don't know how.
102094
let $$f(x)=\int_{x}^{x+1}\sin{(e^t)}dt$$  show that: $$e^x|f(x)|\le 2$$  My idea: let $$e^t=u$$ then $$|f(x)|=|\int_{e^x}^{e^{x+1}}\dfrac{1}{u}d\cos{u}|$$
1902
let $f:[0,1]\longrightarrow R$ be a differentiable function with continuous derivative such that $f(1)=0$,show that:     $$4\int_{0}^{1}x^2|f'(x)|^2dx\ge\int_{0}^{1}|f(x)|^2+\left(\int_{0}^{1}|f(x)|dx\right)^2$$   I think it can be use Cauchy-schwarz inequality     $$\int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(x)dx\ge\int_{a}^{b}(f(x)g(x))^2dx$$
70712
Suppose that $f$ is continuous on $[0, 2\pi]$ and that $f(0) = f(2\pi)$.  Prove that there exists a $c$ in $[0, 2\pi]$ for which $f(c) = f(c+\pi)$.  Here is what i did: if $f(c) = f(c+\pi)$, then $f(c) = f(c-\pi)$.  $f(c+\pi) = f(c-\pi)$ for which $\lvert{\sin(x)}\rvert$ is the function that comes to my mind, and I'm stuck here. I must use the mean value theorem and maybe Rolle's theorem since it's a question from that section, but $\lvert{\sin(x)}\rvert$ is not differentiable. Yet the question itself doesn't mention anything about differentiability of the function. Could anyone solve this, please?
119685
Let $f(x)$  be a twice-differentiable function on $(a,b)$,show that   there exsit $\xi\in(a,b)$ ,such $$\int_{a}^{b}f(x)dx=\dfrac{1}{2}(b-a)[f(a)+f(b)]-\dfrac{1}{12}(b-a)^3f''(\xi)$$  if this problem condition is Amuss that $f(x)$ is a three-differentiable function,  $$f(x)=f(a)+f'(a)(x-a)+\frac{f''(\xi_{1})\cdot (x-a)^2}{2}$$ $$f(x)=f(b)+f'(b)(x-b)+\frac{f''(\xi_{2})\cdot (x-b)^2}{2}$$ so $(1)+(2)$ $$\Longrightarrow 2f(x)=f(a)+f(b)+f'(a)(x-a)+f'(b)(x-b)+\dfrac{1}{2}[f''(\xi_{1})+f''(\xi_{2})][(x-a)^2+(x-b)^2]$$ since $f(x)$ is a three-differentiable function,so $f''(x)$ is  continuous,so there exsit $\xi\in(\xi_{1},\xi_{2})$,such  $$\dfrac{1}{2}[f''(\xi_{1})+f''(\xi_{2})]=f''(\xi)$$ $$\int_{a}^{b}f(x)dx=\dfrac{1}{2}(b-a)[f(a)+f(b)]-\dfrac{1}{12}(b-a)^3f''(\xi)$$ But if $f(x)$ have twice-differentiable,this methods is not usefull  Then I use this methods can't prove it.