g(x)=f(-x), where $g:[-b,-a] \rightarrow R$ and it is given that $f$ is integrable on $[a,b]$  Need to prove that $\int_{-b}^{-a}f(-x)dx$ is integrable if $\int_{a}^{b}f(x)dx$ is.  Is there a way to do this using the properties of integrals?
124149
$f:[a,b]\rightarrow R$ that is integrable on [a,b]  So we need to prove:  $$\int_{-b}^{-a}f(-x)dx=\int_{a}^{b}f(x)dx$$  1.) So we'll use a property of definite integrals: (homogeny I think it's called?)  $$\int_{-b}^{-a}f(-x)dx=-1\int_{-b}^{-a}f(x)dx$$  2.) Great, now using the fundamental theorem of calculus:  $$-1\int_{-b}^{-a}f(x)dx=(-1)^2\int_{-a}^{-b}f(x)dx=\int_{-a}^{-b}f(x)dx$$  This is where I'm stuck. For some reason I think it might be smarter to skip step 2, to leave it asL  $$-1\int_{-b}^{-a}f(x)dx$$   because graphically, we've "flipped" the graph about the x-axis, but we're still calculating the same area. Proving that using properties seems to have stumped me.  I prefer hints over solutions, thanks.
1645
Can you please help me to understand this proof:     Consider $g(x)=f(x)-x$. $f(a)\ge a$ so $g(a)=f(a)-a\ge 0$. $f(b)\le b$ so $g(b)=f(b)-b\le 0$. By the Intermediate Value Theorem, since $g$ is continuous and $0\in[g(b),g(a)]$ there exists $c\in[a,b]$ such that $g(c)=f(c)-c=0$, so $f(c)=c$ for some $c\in[a,b]$."   My question is why $f(a)\ge a$ and $f(b)\le b$??  Thanks a lot!
39612
How to show $f=g$ almost everywhere  Let two real-valued measurable functions $f$ and $g$ be such that for any measurable set $E$ the integrals $\int_E f\,d\mu$  and  $\int_E g\,d\mu$ coincide.  Could you please help.  It seems obvious. we could define the integrals as charges. can we do sth from here. but I do not know how to approach.
89568
Let $f$ be a continuous and integrable function over $[a,b]$. Prove or disprove that $$\int_a^b |f(x)|\ \mathrm{d}x\geq \left | \int_a^b f(x)\ \mathrm{d}x\right| $$
77431
If $f$ is bounded by $M$ on $[a, b]$ and if the restriction of $f$ to every interval $[c, b]$ where $c$ in $(a, b)$ is Riemann integrable, then $f$ is Riemann integrable and that $\int _c^b f \to \int_a^b f$ as $c \to a+$.  I am stuck with this problem.
75449
Just would like to see if I'm right.  Proof: Let $G$ be generated by $x$ (Let $G=\langle x \rangle$ for some $x$ in $G$.) We wish to show that there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$. First, we will show that for all $x^n$ in $G$, $f(x^n)=f^n(x)$. We proceed by induction.   Base case: $n=1$. $f(x*e_G)=f(x)f(e_G)=f(x)$. Now, assume that $f(x^k)=f^k(x)$ for some arbitrary positive integer k. We show that $f(x^{k+1})=f^{k+1}(x)$. Observe that $f(x^{k+1})=f(x^kx)=f(x^k)f(x)=f^k(x)f(x)=f^{k+1}(x).$   Next we will show that for all $x^{-n}$ in $G$, $f(x^{-n})=f^{-n}(x)$. We proceed by induction. Base case: $n=-1$. $f(x^{-1}*e_G)=f(x^{-1})f(e_G)=f(x)$. Now, assume that $f(x^{-k})=f^{-k}(x)$ for some arbitrary positive integer k. We show that $f(x^{-k-1})=f^{-k-1}(x)$. Observe that $f(x^{-k-1})=f(x^{-k}x)=f(x^{-k})f(x)=f^{-k}(x)f(x)=f^{-k-1}(x).$ Because $f$ is onto, $Im(f)=G'$ and therefore $Im(f)=(...,f^{-2}(x),f^{-1}(x),e_G,f(x),f^2(x),...)=G'$.  Thus, there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$ and the proof is complete by the principle of mathematical induction. Q.E.D.  I have a feeling the proof shouldn't be this long. Any suggestions for a shorter, cleaner proof?
106552
I'd really appreciate your help with the following problem:      Let $f$ be a integrable  function in $[a-1,b+1]$. I need to prove that:   $$\lim_{h\to 0} \int_{a}^{b} |f(x+h)-f(x)|dx=0 .$$   Basically I need to show that $\forall  \varepsilon  \exists  \delta .|h|&lt; \delta \to \left|\int_{a}^{b}|f(x+h)-f(x)|dx\right| &lt; \varepsilon $.  In addition, I know that because of $f$ being integrable I can write that $\sum \limits_{i=0}^{n-1}(\sup f-\inf f) \Lambda x_{i} &lt; \varepsilon$ for a division of $[a,b]$.  I tried to define $g=|f(x+h)-f(x)|$ and work with that, but it didn't lead me to any smart conclusion. Also I tried to put the limit inside of the integral and to use the derivative, but I'm not sure when I can do that.  Thank you for the help!
120329
Prove that if $f$ is Riemann integrable on $[a,b]$ and $0&lt;\rho \le f(x)$ for all $x \in [a,b]$ then $1/f$ is Riemann integrable on $[a,b]$.
30673
I am asked to prove the second mean value theorem:  Let $f$ and $g$ be defined on $[a,b]$ with $g$ continuous, $f\ge 0$, and $f$ integrable. Then there is a point $x_0 \in (a,b)$ such that $$ \int_{a}^{b}f(x)g(x)dx = g(x_0) \int_{a}^{b}f(x)dx. $$  My (incomplete) proof goes as follows:  Since $ [a,b] $ is compact and $ g $ is continuous, we know there exist $ x_1, x_2 \in [a,b] $ such that $ g(x_1) \le g(x) \le g(x_2) $ for any $ x \in [a,b] $. Thus, we have the following string of inequalities $$ f(x)g(x_1) \le f(x)g(x) \le f(x)g(x_2), \, \, \forall x \in [a,b], $$ $$ \Rightarrow g(x_1)\int_{a}^{b} f(x)dx \,\le \, \int_{a}^{b} f(x)g(x) dx \, \le  \, g(x_2)\int_{a}^{b}f(x)dx. $$  My problem now is to show the final equality. It seems to me the intermediate value theorem is somehow involved, but I cannot see how to use it. Furthermore, how do I make use of the $f\ge 0$ condition in the hypothesis?
69661
Let $f(x)$ be continuous on $[a,b]$ and suppose $\int_a^b f(x)g(x)dx = 0$ for every continuous function $g$ on $[a,b]$. Prove that $f(x)=0$ on $[a,b]$.   I understand that $f(x)$ must be zero otherwise the integral can't possibly be zero for all possible $g$. But, I am not sure how to approach this systematically. Any suggestions would be appreciated. Thanks!
93852
Let $f$ be a Lebesgue integrable function on $[0,2]$. If $\int_E fdx=0$ for all measurable set $E$, such that  $m(E)=\pi/2$. Is $f=0$ a.e. Prove or disprove  I could not figure out anything. Can a function be very oscillatory so that on every interval its integral is zero?  Any hint and approach are welcome.
105003
If $f \colon [a,b]\rightarrow \mathbb{R}$, then show that there exists $c \in [a,b]$ such that $c \in [a,b]$ such that $\int_a^c f = \int_c^b f$.  I think that I have to proceed with the Mean Value Theorem of Integrals...
38829
Regarding real numbers, the following appears to be true, or at least true with some modifications. Could you help me for the proof? $$\int_0^af(x)dx+\int_{f(0)}^{f(a)}f^{-1}(x)dx=af(a)$$
110121
Suppose $f$ is Lebesgue integrable on $[a,b]$ and $F(x) = \int^x_a f(t) dt$, $x \in [a,b]$. Show that $F$ has bounded variation, and the total variation $T^b_a(F)$ satisfies $$ T^b_a(F) = \int^b_a |f(t)|dt. $$  Now, here is what I have so far. Show $F$ is of BV is simple: let $a=x_0 &lt; x_1 &lt; \dots &lt; x_k=b$ be any partition/subdivision of $[a,b]$. Then  $$ \sum^k_{i=1} |F(x_i)-F(x_{i-1})| = \sum^k_{i=1} \left|\int^{x_i}_{x_{i-1}} f(t) dt\right| \leq \sum^k_{i=1} \int^{x_i}_{x_{i-1}} |f(t)| dt = \int^b_a |f(t)| dt. $$ Thus $T^b_a(F) \leq \int^b_a|f(t)| dt &lt; \infty$. Therefore $F$ is of bounded variation.   On the other hand, I also know that $F$ is an indefinite integral iff $F$ is absolutely continuous. This would have also implied $F$ is BV since AC $\Rightarrow$ BV. I also know that if $F$ is BV, then $F'$ exists a.e.(almost everywhere). I also have an old homework question I did which showed $F'=f$ a.e. with the assumptions from the start. Thus all I need is that $\int|F'| \leq T^b_a(F)$.   I think this works: for every $x \in [a,b]$, we know that $F(x) = P^x_a(F) - N^x_a(F) + F(a)$; P and N are the positive and negative variations respectively. Then $\dfrac{d}{dx} [F(x)] = \dfrac{d}{dx}[P^x_a(F)] - \dfrac{d}{dx}[N^x_a(F)]$ a.e. Thus $|F'| \leq \dfrac{d}{dx}[P^x_a(F)] + \dfrac{d}{dx}[N^x_a(F)] = \dfrac{d}{dx}[T^x_a(F)]$ a.e. Finally  $$ \int^b_a |F'| \leq \int^b_a \dfrac{d}{dx} T^x_a(F) \leq T^b_a(F). $$ Assuming I can show anything I have stated and didn't show proof of, this looks like it works to me. Any comments would be greatly appreciated.
37731
I'm not entirely sure what I want to ask here, so please bear with me!  I think the explanation they give us in school for how finding the antiderivative by substitution works is: $$\int f(g(t))g&#39;(t)dt=(Fg)(t)=F(g(t))=F(u)=\int f(u)du$$  But I never really understood the equality $F(u)=\int f(u)du$. Why can we behave as if the identity $u=g(t)$ doesn't exist, compute the integral $\int f(u)du$, and then 'substitute' $u$ in the result yielding the correct answer? In a sense, it feels like the variable $u$ switches from being 'meaningful' in $F(u)$ (where it stands for $g(t)$) and being insignificant in $\int f(u)du$, since as I perceive it the "variable name" here is meaningless and we could say $ \int f(u)du = \int f(k)dk = \int f(x)dx = ... $ all the same. Another way to ask the same thing: why $(Fg)(t)=\int f(u)du$ and not, say, $(Fg)(t)=\int f(x)dx$? Where am I getting confused?  Maybe someone can explain this better than my high school teacher? Is there a 'formal' explanation for this? Thank you a lot!
30480
For $f \in C[a,b], g\ge0$ (or $g\le0$) on $[a,b]$, then there exists $c\in(a,b)$ $\int _a^b f(x)g(x)dx = f(c)\int_a^bg(x)dx$.   It is MVT for integral.  Why $c \in (a,b)$ ? How about $c\in[a,b]$? Is it wrong?  I am wondering this because, in proving progress,   by IVT, there exist $m,M$ s.t $m\le f(x)\le M$   so $m\int_a^bg(x)dx$ $\le$ $\int_a^b f(x)g(x)dx$ $\le$ $M\int_a^bg(x)dx$.  when $\int_a^bg(x)dx&gt;0$ then $m\le$ $\frac{\int_a^b f(x)g(x)dx} {\int_a^bg(x)dx}$   $\le$  $M$ and   $\frac{\int_a^b f(x)g(x)dx} {\int_a^bg(x)dx}$ becomes $f(c)$, so it seems $c\in [a,b]$ might work.  So, I want to know why $c\in(a,b)$, or $c\in[a,b] $ would be okay.
62793
I have two questions that i'm curious about.   If $f$ is differentiable real function on its domain, then $f'$ is Riemann integrable. If $g$ is a real function with intermediate value property, then $g$ is Riemann integrable.   Thank you in advance.