From:  Suppose $f$ is a real function on $(0, 1]$ and $f \in \mathscr{R}$ on $[c,1]$ for every $c&gt;0$. Define $\int_0^1 f(x)dx=\lim_{c\to 0} \int_c^1 f(x)dx$ if this limit exists (and is finite). If $f \in \mathscr{R}$ on $[0,1]$, show that this definition of the integral agrees with the old one.  Can I use the fundamental theorem of calculus to prove this problem? $$|\int_c^1 f(x)dx-\int_0^1 f(x)dx|=|\int_0^c f(x)dx|$$, and define $$F(c)=\int_0^c f(x)dx$$ is continuous so $$\lim_{c\to 0}|\int_0^c f(x)dx|=|\lim_{c\to 0}F(c)|=0$$
116225
Here is Prob. 7, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:      Suppose $f$ is a real function on $(0, 1]$ and $f \in \mathscr{R}$ on $[c, 1]$ for every $c &gt; 0$. Define    $$ \int_0^1 f(x) \ \mathrm{d} x = \lim_{c \to 0} \int_c^1 f(x) \ \mathrm{d} x $$    if this limit exists (and is finite).       (a) If $f \in \mathscr{R}$ on $[0, 1]$, show that this definition of the integral agrees with the old one.       (b) Construct a function $f$ such that the above limit exists, although it fails to exist with $\lvert f \rvert$ in place of $f$.    Here I'll only attempt Part (a):   My Attempt:   Here is the link to a post of mine here on Math SE where I've copied the definition of the Riemann and Riemann-Stieltjes integral that Rudin uses (i.e. Definitions 6.1 and 6.2 in Baby Rudin, 3rd edition):        As $f \in \mathscr{R}$ on $[0, 1]$, so $\int_0^1 f(x) \ \mathrm{d} x$ exists in $\mathbb{R}$.       According to the statement of the problem, we only need to show that    $$ \lim_{c \to 0+} \int_c^1 f(x) \ \mathrm{d} x = \int_0^1 f(x) \ \mathrm{d} x. \tag{0}$$       Let $\varepsilon &gt; 0$ be given. We need to find a real number $\delta&gt; 0$ such that    $$ \left\lvert \int_c^1 f(x) \ \mathrm{d} x \ - \ \int_0^1 f(x) \ \mathrm{d} x \right\rvert &lt; \varepsilon \tag{1} $$   for any real number $c$ such that $0 &lt; c &lt; \delta$.       Now let's choose a real number $\delta_0 \in (0, 1)$, and let us choose $c$ such that $0 &lt; c &lt; \delta_0$.       Then as $f \in \mathscr{R}$ on $[0, 1]$ and as $c \in (0, 1)$, so by Theorem 6.12 (c) in Baby Rudin $f \in \mathscr{R}$ on $[0, c]$ and on $[c, 1]$, and    $$ \int_0^c f(x) \ \mathrm{d} x \ + \ \int_c^1 f(x) \ \mathrm{d} x = \int_0^1 f(x) \ \mathrm{d} x. \tag{2} $$    Here is the link to my Math SE post on Theorem 6.12 (c) in Baby Rudin, 3rd edition:        In the light of (1) and (2), we can conclude that we now only need to show that there exists a real number $\delta &gt; 0$ such that    $$ \left\lvert \int_0^c f(x) \ \mathrm{d} x \right\rvert &lt; \varepsilon \tag{3} $$    for any real number $c$ such that $0 &lt; c &lt; \delta$, and we now also know that    $ 0 &lt; c &lt; \delta_0 &lt; 1$.       As $f \in \mathscr{R}$ on $[0, 1]$, so $f$ is also bounded on $[0, 1]$ and hence also on $[0, c]$. Let $M \colon= \sup \{ \ f(x) \ \colon \ 0 \leq x \leq c \ \}$.       Then by Theorem 6.12 (d) in Baby Rudin, we have    $$ \left\lvert \int_0^c f(x) \ \mathrm{d} x \right\rvert \leq M c. \tag{4} $$    Here is the link to my Math SE post on Theorem 6.12 (d) in Baby Rudin, 3rd edition:        So if we choose our $\delta$ such that $$0 &lt; \delta &lt; \min \left\{ \ \delta_0, \frac{\varepsilon}{M+1} \ \right\}, $$    then, for any real number $c$ such that $0 &lt; c &lt; \delta$, we have $0 &lt; c &lt; \delta_0$ so that $c \in (0, 1)$ and from (4) we also have    $$ \left\lvert \int_0^c f(x) \ \mathrm{d} x \right\rvert \leq M c \leq  \frac{M \varepsilon}{M+1} &lt; \varepsilon, $$    which by virtue of (3) implies that (1) holds.       Since $\varepsilon &gt; 0$ was arbitrary, therefore (0) holds as well, as required.    Is this proof correct and rigorous enough for Rudin? If not, then where is it lacking?   Is this proof the same as the proof asked for by Rudin?
21265
Suppose $f : [ 0, 1 ] → \mathbb{R}$ and $\lim\limits_{x \rightarrow c} f ( x )$ exists for all $c \in [ a , b ]$. Show that $f$ is Riemann integrable on $[ a , b ]$.  I can show this $f$ is bounded on $[a,b]$ since the limit exists at every point. But I'm not sure how to proceed. Any hints or help would be great.
88187
Let $$f\colon [0,1]\rightarrow [0, \infty )$$  be  continuous. Suppose that  $$\int_{0}^{x} f(t)\,\mathrm dt \ge {f(x)} \quad\text{for  all }x\in[0,1].$$  Then   $A.$ No  such  function  exists.  $B.$ There  are  infinitely  many  such  functions.  $C.$ There is  only  one  such  function  $D.$ There  are  exactly  two  functions.  Now  I  was  thinking,  since $\int_{0}^{x} f(t) dt$  is  the  area  under  the  graph  of  $f(t)$  from $0$  to  $x$ so it  can  be  easily  made  greater  than  the  value  of  $f(t)$  at  one  point, namely  $x$.  So  there   will  be  infinitely  many  such  functions  satisfying  this  condition.   Is  my  reasoning  correct or  not?  Thanks.
23132
Assume $f$ is a function over $\mathbb{R}$ satisfying $f(x+y) = f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Show that if $f$ is continuous, then $f(x) = cx$ for all $x \in \mathbb{R}.$   I find it hard to use the definition of continuity to derive anything from the equation. I think we could write the definition of the limit in terms to get $$\forall \epsilon,\exists \delta \quad 0&lt;|x-a|&lt;\delta \quad \implies \quad |f(x)-L| &lt; \epsilon.$$ I am not sure where to go from here.
60427
Prove that if the real-valued function $f$ on the interval $[a,b]$ is bounded and is continuous except at a finite number of points then $\int^1_0f(x)dx$ exists.   I know that I can break up the interval $[a,b]$ into subintervals where each subinterval has one discontinuity point. Then if $f$ is integrable on each subinterval then it is integrable on $[a,b]$. So if I let $c$ be my point of discontinuity in the subinterval $[a_1,b_1]\subset [a,b]$, where $c\in[a_1,b_1]$. Then how can I show this subinterval is integrable?
18829
I have a question:     Suppose $f$ is continuous and even on $[-a,a]$, $a&gt;0$ then prove that   $$\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$$   How can I do this? Don't know how to start.
21219
I need help to prove this in real analysis. I think it uses IMVT, but not sure how to do it.     Let $f(x)$ be a real valued continuous function on $[0,1]$. Show that   $$ \lim \limits_{n\to\infty}(n+1)\int_{0}^{1}x^nf(x)dx=f(1) $$
78643
Let $f(x)$ be a continuous function on $\mathbb{R}$, such that for any real number $x$ we have: $$\lim_{h\to 0}\dfrac{1}{h^3}\int_{-h}^{h}f(x+t)\cdot t\,dt=0.$$ Show that $f(x)$ is a constant function. Maybe we can use the following lemma? Lemma. If $g$ is a continuous function, then $$\lim_{h\to0}\frac{1}{2\,h}\int_{x-h}^{x+h}g(s)\,ds=g(x).$$ Proof. We may assume $h&gt;0$. $$ \left|g(x)-\frac{1}{2\,h}\int_{x-h}^{x+h}g(s)\,ds\right|=\frac{1}{2\,h}\left|\int_{x-h}^{x+h}(g(x)-g(s))\,ds\right|\le\frac{1}{2\,h}\int_{x-h}^{x+h}\left|g(x)-g(s)\right|\,ds.$$ Use that $g$ is continuous at $x$ to show that the last expression converges to $0$ as $h\to0$.
73949
Let $f: \mathbb R \rightarrow \mathbb R $ be a continuous function such that $\int_{0}^{\infty} \,f(x) dx$ exists.  Then Prove that incase   (i) $f$ is a non negative function, then $\lim_{x\rightarrow \infty} f(x) $ must exist and is $0$.  (ii) $f$ is a positive differentiable function , $\lim_{x\rightarrow \infty} f'(x) $ must exist and is $0$  $Attempt$: For the first part, i don't have a rigorous proof except for the fact that the given condition can be visualised geometrically. Since, the definite integral is actually calculating the area beneath the non negative function, the only way the given limit can exist when limit of f(x) itself tends to 0 at infinity.  Please give me a direction so that i can make this proof rigorous enough.  For the second part, i took an example. We know that ( leaving out the finite integration parts from $0$ to $1$ ..) $\int_{1}^{\infty} e^{-x^2} dx \leq \int_{1}^{\infty} e^{-x} dx$ and the latter converges.  But the derivative of $e^{-x^2} = (-2x)e^{-x^2}$ whose integration does not exist when x $\in~[1,\infty)$ as it's a monotonic function after a finite $x$.  Any help in providing rigor to the above proof will be very helpful  Thanks
109196
Some days ago I answered a question that asked to find  $$\mathop {\lim }\limits_{x \to 0} {x^\alpha }\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^{\alpha  + 1}}}}dt} $$  given that $f$ is continuous in $[0,1]$  I proceeded as follows:  $$\eqalign{   &amp; t = x\cdot u  \cr    &amp; dt = x\cdot du \cr} $$  So this is produces:  $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du} $$  I then thought: "Well, if $f$ is continuous in the closed interval, then it is also uniformly continuous, so I can assume   $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  = f\left( 0 \right)\int\limits_1^\infty  {\frac{{du}}{{{u^{\alpha  + 1}}}}}  = \frac{1}{\alpha }f\left( 0 \right)$$  This turned out to be true. However, I wasn't very comfortable with such "move". So now I'm thinking, one can put  $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du} $$  And then   $$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du} } \right| &lt; \int\limits_1^{\frac{1}{x}} {\frac{{\left| {f\left( {xu} \right) - f\left( 0 \right)} \right|}}{{{u^{\alpha  + 1}}}}du}  &lt; \epsilon \frac{{1 - {x^\alpha }}}{\alpha }$$  However, this is still insufficient since I need to adress the behaviour of the upper limit too. Can someone show me how to adress both behaviours simultaneously?    Would this work?  Let $P$ be the statement that $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  = \frac{{f\left( 0 \right)}}{\alpha }$$  Then $P$ is true if and only if  $$ \forall \epsilon  &gt; 0\exists \delta  &gt; 0$$  Such that if $$\left| x \right| &lt; \delta $$ then $$ \left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  - \frac{{f\left( 0 \right)}}{\alpha }} \right| &lt; \epsilon $$  But then   $$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| &lt; $$  $$\left| {\int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| \leqslant $$  $$\varepsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } &lt; $$  And since  $$\frac{{1 - {\delta ^\alpha }}}{\alpha } &lt; \frac{1}{\alpha }$$ $$\epsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } &lt; \frac{\epsilon }{\alpha } &lt; \epsilon $$
11285
Assume that $f$ be continuous on $\mathbb{R}$, $f'(x)$ exists for all $x\neq 0$, and $\lim_{x\rightarrow 0} f'(x)=1$. We need to show $f'(0)$ exist and is equal to $1$.  $f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}$, $\lim_{x\rightarrow 0}f'(x)=1\Rightarrow\lim_{x\rightarrow 0}\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=1$...am I going in the right direction? Please help.
36487
I am trying to convince my friend that the integral of $0$ is $C$, where $C$ is an arbitrary constant. He can't seem to grasp this concept. Can you guys help me out here? He keeps saying it is $0$.
84181
If $f$ is continuous on $[a,b]$, $f(x)≥0$ on $[a,b]$ and $$\int_{a}^{b} f(x) =0$$   then prove that $f(x)=0$ for all $x \in [a,b]$.  I tried with Riemann's definite integral definition but couldn't proceed
50217
Suppose $f$ is Riemann integrable on $[0,1]$ and continuous at 0. Show that $$\lim\limits_{a \to 0^+} \int^1_0 ax^{a−1}f(x)dx=f(0).$$   Can anyone share some insight on this one? I have no idea where to begin. Maybe a substitution?
59694
Let $g:[0,1] \rightarrow \mathbb R$ be a continuous function and assume that    $$ \int_{0}^{1} g(x) \phi'(x) dx = 0 $$   for all continuously differentiable functions $\phi: [0,1] \rightarrow \mathbb R$, where $\phi (0) = \phi(1) = 0$. Show that $g(x) = c$ for all $x \in [0,1]$ for some constant $c \in \mathbb R$.    My thoughts are applying integration by parts with $G(t) = \int_{0}^{t} g(x) dx $, since now we get  $$ 0 = \int_{0}^{1} G(t) \phi'(t) dt  + \int_{0}^{1} g(t) \phi (t) dt $$  which doesn't really go far. The second thought was applying MVT for definite integrals, which doesn't go far either as the conditions are hardly satisfied.   Any hints?
116767
Let $f: [0,1] \to \mathbb{R}$ be a continuous function such that $$\int_0^1 f (t) dt =1 .$$ Show that there exists a point $c \in (0,1)$ such that $ f(c)=3c^2$.  I tried using mean value theorem of integrals on $f (x)$, so $$\int_0^1 f (t) dt = f (c) =1,$$ but it should be $3 c^2$. What is going wrong?
24011
We are asked to prove the following integral;  $\int_0^x(\int_0^tf(u) du) dt=\int_0^xf(u)(x-u)du$   I know that we have to use integration by parts which when we differentiate a function will help us use the fundamental theorem of calculus. I just don't know when and were to apply this.  Thanks in advance!