Let $0 &lt;a &lt;b\ $ let $f &gt;0 $  be continuous and strictly increasing  function on $ [a,b]$. Prove that $$\int_{a}^b f \  + \int_{f(a)}^{f(b)} \ f^{-1}=bf(b)-af(a)$$  How to approach this problem . Any Hint? I am suppose to do it without using antiderivatives.
3297
Suppose $f$ is a continuous, strictly increasing function defined on a closed interval $[a,b]$ such that $f^{-1}$ is the inverse function of $f$. Prove that,   $$\int_{a}^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)$$   A high school student or a Calculus first year student will simply, possibly, apply change of variable technique, then integration by parts and he/she will arrive at the answer without giving much thought into the process. A smarter student would probably compare the integrals with areas and conclude that the equality is immediate.  However, I am an undergrad student of Analysis and I would want to solve the problem "carefully". That is, I wouldn't want to forget my definitions, and the conditions of each technique. For example, while applying change of variables technique, I cannot apply it blindly; I must be prudent enough to realize that the criterion to apply it includes continuous differentiability of a function. Simply with $f$ continuous, I cannot apply change of variables technique.  Is there any method to solve this problem rigorously? One may apply the techniques of integration (by parts, change of variables, etc.) only after proper justification.  The reason I am not adding any work of mine is simply that I could not proceed even one line since I am not given $f$ is differentiable. However, this seems to hold for non-differentiable functions also.  I would really want some help. Pictorial proofs and/or area arguments are invalid.
34866
$f:[0,1]\to \mathbb R$ , be continuous function then prove that    $$\int_0^1f^2(x)dx\geq \biggl(\int_0^1|f(x)| \biggr) ^2$$   I tried this for $x^2$  For that above is true   But I checked following proof Which is complete opposite to above.   Please help me to find that where is I am making wrong ?  Any help will be appreciated
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.
89568
Let $f$ be a continuous and integrable function over $[a,b]$. Prove or disprove that $$\int_a^b |f(x)|\ \mathrm{d}x\geq \left | \int_a^b f(x)\ \mathrm{d}x\right| $$
39103
Suppose $f : [a,b] \to \mathbb{R}$ is continuous and $g \in \mathcal{R}[a,b]$ with $g(x) \ge 0$ for all $x \in [a,b]$. Show that there exists a $c \in [a,b]$ such that $$\int_a^b f(x)g(x) \, dx = f(c) \int_a^b g(x) \, dx.$$   My attempt:  My class proved a theorem which states that $f$ attains its minimum and maximum ($m$ and $M$, respectively), since $f$ is continuous on a closed interval like $[a,b]$. So we would have $m \le f(x) \le M$. Multiplying both sides by $g$ and integrating, we get $$m \int_a^b g(x) \, dx \le \int_a^b f(x) g(x) \, dx \le M \int_a^b g(x) \, dx.$$  I am stuck after this. Where can I go from here?
43481
PROBLEM. Let $f:[a,b]\to\mathbb R$ be a continuous function, such that $f'(x)\ge 0$, for all $x\in [a,b]\setminus A$, where $A\subset [a,b]$ is a countable set. Show that $f$ is increasing.  Attention. In this problem, we DO NOT assume that $f$ is differentiable in the whole $[a,b]$.  Notes. (1) If we assume that $f$ is differentiable in the whole interval, then we can easily show that $f'(x)\ge 0$, everywhere. For otherwise, if $f'(x_0)=c&lt;0$, for some $x_0\in [a,b]$, then by virtue of , $(c,0)\subset f'([a,b])$, and hence, $f'(x)&lt;0$, for uncountably many $x$'s.   (2) The conclusion of the problem does not hold if we replace the assumption $A$ is countable with $A$ is a set of measure zero. Take for example the , with a negative sign in front.  (3) If the hypothesis $f'(x)\ge 0$, is replaced by $f'(x)=0$, then the conclusion becomes f is constant.
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.
69307
Let $f:[0,\infty) \to \mathbb R$ be a twice differentiable function, such that  $\,f''(x) \ge f(x)$, for all $x\in\ [0,\infty)$, and $ f(0)=f'(0)=1$. Can we deduce that $f$ is increasing?   I feel like it is, but I cannot see it. I can only show that to get it increasing it is enough to show that $f$ is non negative .
44777
Prove that if $f:(0,\infty)→\mathbb{R}$ satisfying $f(xy)=f(x)+f(y)$, and if $f$ is continuous at $x=1$, then $f$ is continuous for $x&gt;0$.   I let $x=1$ and I find that $f(x)=f(x)+f(1)$ which implies that $f(1)=0$. So, $\lim_{x\to1}f(x)=0$, but how can I use this to prove continuity of $f$ for every $x \in \mathbb R$?   Any help would appreciated. Thanks
55375
Can Some one tell me what this method is called and how it works With a detailed proof   $$\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$$  I've been using this a lot in definite integration but haven't seemed to have realized why it is true.  But whatever it is it always seems to work.  Basically a proof of how it is always true.
81832
Please help me to solve this question:  Suppose $f:[a,b] \to \Bbb R$ satisfies:  $f''(x)+f(x)&gt;0$ and $f(x)&gt;0$ for all $x\in(a ,b)$; $f(a)=f(b)=0$.  Prove that $b-a&gt;\pi$.  Thanks in advance.
50389
Suppose $f$ is continuous, $f(x) \ge 0$ on $[a,b]$ and $\int_a^b f(x) dx=0$. Show that $f(x)=0$ on $[a,b]$.   I know that if $f$ in continuous at $x_0$ and $f(x_0) &gt; 0$, then there exists $\delta &gt; 0$ s.t. $f(x) \ge \frac{1}{2} f(x_0)$ for $|x-x_0| &lt; \delta$ and I think it should help me but I'm stuck at this point.
33713
$f(x)$ is positive and continuous function on $\mathbb{R}$ and, moreover, $\int_{-\infty}^{+\infty}f(x)dx=1$. $\alpha\in(0;1)$ and $[a;b]$ is the interval having a minimum length such that the $\int_{a}^{b}f(x)dx=\alpha$. Prove $f(a)=f(b)$.  With mean value theorem easy to show, that for every $\alpha\in(0;1)$, exist such $[a;b]$, so  $\int_{a}^{b}f(x)dx=\alpha$. The statement looks like Rolle's theorem, but I have no idea. I will be grateful for help.
27701
If $f: [0,1]\rightarrow \mathbb{R}$ is continuous function positive, so $$\int_{0}^{1} \frac{f(x)}{f(x)+f(1-x)}dx=\frac{1}{2}$$???  all examples that I tested have worked.
119987
Let $f:(a,b)\to \mathbb{R}$ be a strictly increasing function. Does the limit $\lim_{x\to a^+}f(x)$ necessarily exist and is a real number or $-\infty$?  If so, is it true that $\ell=\lim_{x\to a}f(x)\le f(x) \ \ \forall x\in (a,b)$? Please provide proofs.
107154
if $f:[a,b]\longrightarrow [a,b]$ is continuous and $\alpha , \beta \in \mathbf{R}^+$ then  there is some $c\in [a,b]$, $$f(c)=\dfrac {\alpha f(a)+\beta f(b)}{\alpha+\beta}.$$ Should I use Intermediate Value Theorem?
66475
How to prove that a strictly increasing function $f:[a, b]\rightarrow \mathbb{R}$ which has the intermediate value property is continuous on $[a, b]$.