I know that my reasoning is incorrect, I just don't know where I went wrong. I did discuss this with my Maths teacher, and even she could not find what I did wrong.  Let us begin by assuming a function, $f(x)$ that is continuous and has an antiderivative in the interval $[0, 2\pi]$. Let $A$ be the area under the curve for $f(x)$ in the interval $[0, 2\pi]$  $A = \displaystyle \int_{0}^{2\pi}{f(x)\space\mathrm{d}x}$  Now there must exist a function, $g(x)$ such that:    $f(x) = g(x)\cdot \cos(x)$  Substituting the value of $f(x)$:  $A = \displaystyle\int_{0}^{2\pi}{g(x)\cdot \cos(x)\space\mathrm{d}x}$  Using t substitution: Let $t = \sin(x)$ Then: $\mathrm{d}t = \cos(x)\space\mathrm{d}x$ And:  $x = \arcsin(t)$  Changing the limits: $t = \sin(x)$ $0$ becomes $\sin(0) = 0$ $2\pi$ becomes $\sin(2\pi) = 0$  Substituting in the definite integral:  $A = \displaystyle \int_{0}^{0}{g(\arcsin(t))\space\mathrm{d}t}$  But Definite Integral where the lower and upper bounds are the same is $0$. So:  $A = 0$, which is not possible.  Thanks for the help.
122986
Quick question, might end up having a simple answer, but I have here a "proof" that any integral from 0 to $2\pi$ is zero, as follows:  $$\int^{2\pi}_0f(x)dx$$  Now using u-substitution, let $u = \sin x$, so $dx = \frac{du}{\cos x}$, so:  $$\int^{0}_{0}\frac{f(x)du}{\cos x}$$  (The bounds were evaluated to zero due to the u-substituion).  However, this is zero, since the integral goes from zero to zero.  Any help as to why this is wrong would be appreciated.
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?
118717
I've been reading through Zorich's "Analysis I" book recently, and I came across this nice little exercise.  Let $f: [0,1]\to \mathbb R$ be a continuous function such that $f(0)=f(1)$. Show that    for any $n\in \mathbb N$ there exists a horizontal closed interval of length $\frac 1n$ with endpoints on the graph of this function; if the number $\ell$ is not of the form $\frac 1n$ there exists a function of this form on whose graph one cannot inscribe a horizontal chord of length $\ell$.   The first part can be proven like this: Consider $g: [0,(n-1)/n] \to \mathbb R$ given by $g(x) = f(x) - f(x+1/n)$. Then   $$\sum_{k=0}^{n-1} g(k/n) = 0$$  and therefore either all of these points are zero or there exists both a point where $g$ is positive and a point where $g$ is negative. By continuity, there must then also be a point where $g = 0$. So we are done.  Now, the second statement seemed rather counterintuitive, and I have given it some time now, but don't see a counterexample for $\ell &lt; 1/2$.  (For $\ell &gt; 1/2$ the function $f(x) = \sin(2\pi x)$ will do.)  Can anyone help me out?  Cheers,
21069
I am trying to understand the relationship between differentiation and integration.  Differentiation has been introduced to me by this diagram:    Which displays that the derivative of a point $x$ on a continuous function $f(x)$ is the gradient of a line which is a tangent to that particular point, as shown in the diagram.  This can be written as  $$\lim_{h \to 0} \left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)=f'\left(x\right)$$  and this gives the gradient of the point $(x, f(x))$.  Next examine this diagram:     The area under the continuous function from $f(x)$ from $a$ to $x$ is $A(x)$, shaded pink.  The area of $A(a)$ is $0$ and the area from $a$ to $b$ is $A(b)$.  Following this convention the area of $x$ to $x+h$ is  $$A\left(x+h\right)-A\left(x\right)$$  This section has with $h$ and the area is close to that of a rectangle so it can be said that   $$A\left(x+h\right)-A\left(x\right)\approx hf(x)$$  If you divide this through by $h$ you get an equation which shows that the derivative of $f(x)$ is $A'(x)$.    $$\lim_{h \to 0} \left(\frac{A\left(x+h\right)-A\left(x\right)}{h}\right)=f\left(x\right)$$  and hence   $$A'(x)=f(x)$$  This is where I have a problem, earlier it was stated that the derivative is the gradient of a line which is a tangent to a point on a continuous function.  Here however, we have an area, which encloses many points, which are not a line which is a tangent to a continuous function.  So I do not see how you can find the derivative of an area because it is not a point that you can find the gradient of a line which is a tangent to it.?  (Images taken from course texts provided by the Open University, Chapters C1 &amp; C2 of the course MST121).
51587
I will be TA this semester for the second course on Calculus, which contains the definite integral.  I have thought this since the time I took this course, so how do I convince my students that for a definite integral   $$\int_a^b f(x)\ dx=\int_a^b f(z)\ dz=\int_a^b f(☺)\ d☺$$  i.e. The choice of variable of integration is irrelevant?   I still do not have an answer to this question, so I would really hope someone would guide me along, or share your thoughts. (through comments of course)  NEW EDIT: I've found a relevant example from before, that will probably confuse most new students. And also give new insights to this question.  Example: If $f$ is continuous, prove that   $$\int_0^{\pi/2}f(\cos x)\ dx = \int_0^{\pi/2}f(\sin x)\ dx$$  And so I start proving...  Note that $\cos x=\sin(\frac{\pi}{2} -x)$ and that $f$ is continuous, the integral is well-defined and  $$\int_0^{\pi/2}f(\cos x)\ dx=\int_0^{\pi/2}f(\sin(\frac{\pi}{2}-x))\ dx $$  Applying the substitution $u=\frac{\pi}{2} -x$, we obtain $dx =-du$ and hence  $$\int_0^{\pi/2}f(\sin(\frac{\pi}{2}-x))\ dx=-\int_{\pi/2}^{0}f(\sin u)\ du=\int_0^{\pi/2}f(\sin u)\ du\color{red}{=\int_0^{\pi/2}f(\sin x)\ dx}$$  Where the red part is the replacement of the dummy variable. So now, students, or even some of my peers will ask: $u$ is now dependent on $x$, what now? Why is the replacement still valid?  For me, I guess I will still answer according to the best answer here (by Harald), but I would love to hear more comments about this.
25438
I know the integral is possible with a simple fourier series expansion of $-\log(\sin(x))$  But I am interested in complex analysis, so I want to try this.  $$I = \int_{0}^{\pi} \log(\sin(x)) dx$$  The substitution $x = \arcsin(t)$ first comes to mind.   But that substitution isnt valid as,   The upper and lower bounds would both be $0$ because $\sin(\pi) = \sin(0) = 0$  What is a workaround using inverse trig or some other way?  Inverse sine is good, because that gives us a denominator from which we will be able to find poles.  $$x = arctan(t)$$  Is also good, but it would hard to do.  Idea?
117220
$S := \{x \in \Bbb R^3: ||x||_2 = 1 \} \subset (\Bbb R^3, || \cdot ||_2 )$ and $T:S^2 \to (\Bbb R, |\cdot |)$ is  a continuous function.  I've already shown that $$T_{\mathrm{max}} := \mathrm{sup}\{ T(x) : x \in S^2\}  \: \: \mathrm{ and }\: \: T_{\mathrm{min}}:= \mathrm{inf}\{ T(x) : x \in S^2\}$$ are reached by $T$ with the argument that $S^2$ is closed and bounded and therefore after the Heine-Borel-Theorem compact, and therefore since T is continuous it assumes its max and min in $S^2$.   Now i have answer the following:     1) Is there a point $x_0 \in S^2$ with $T(x_0)=T(-x_0)$ ?      2) Is there a value $T \in(T_{\mathrm{min}}, T_{\mathrm{max}})$ which is only assumed at exactly one point of $S^2$ ?   I wanted to show that $T$ is injective to get that 1) is false and 2) is true but i didn't get anywhere. Any ideas? Thanks in advance!
21010
What is the intuition/idea behind integration of a function with respect to another function? Say $$\int f(x)d(g(x)) \;\;\;\;\;?$$ or may be a more particular example $$\int x^2d(x^3)$$  My concern is not at the level of problem solving. To solve we could simply substitute $u=x^3$ and then $x^2=u^{2/3}$. My concern is rather about what meaning physical/geometrical does this impart?  ADDED if you ask what kind of meaning I seek, integration of a function w.r.t a variable gives the area under the curve and above x-axis.
98623
Let $f:(a,b)\rightarrow \mathbb{R}$.   $f$ satisfied the following property:  If $\forall x_{1},x_{0},x_{2}\in(a,b)$ and $x_{1}&lt;x_{0}&lt;x_{2};$then$\frac{f(x_{0})-f(x_{1})}{x_{0}-x_{1}}\geq \frac{f(x_{2})-f(x_{0})}{x_{2}-x_{0}}.$   My question : whether the function $f\in C((a,b))?$    We can get  $\displaystyle\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}\geq \frac{f(x_{2})-f(x_{0})}{x_{2}-x_{0}}. $Let $\displaystyle g(x)=\frac{f(x)-f(x_{0})}{x-x_{0}},$ if we can prove $g(x)$ is bound and  decreasing (not strictly ) in $(x_{0}-\delta ,x_{0});$ then we have $f^{'}_{-}(x_{0}) $ exists .In a similar way ,$f^{'}_{+}(x_{0}) $ exists.  $f^{'}_{-}(x_{0}) $ exists $\Rightarrow\displaystyle\lim_{x\rightarrow x_{0}-}f(x)=f(x_{0});$ $f^{'}_{+}(x_{0}) $ exists $\Rightarrow\displaystyle\lim_{x\rightarrow x_{0}+}f(x)=f(x_{0}).$  Obviously, $f$ is coutinuous at $x=x_{0}.$Further, $f\in C((a,b))!$     But  I failed to prove  $g(x)$ is bound and  decreasing (not strictly ) in $(x_{0}-\delta ,x_{0}).$Sometimes  I doubt the conculsion that  $f\in C((a,b)).$  Either make a counterexample to deny it ,or prove the conculsion is correct?
20882
If $f$ is a continuous function with $\displaystyle \int_{0}^{x}f(t)dt\rightarrow \infty$ as $|x|\rightarrow \infty\;,$ Then show that every line $y=mx$ Intersect the curve $\displaystyle y^2+\int_{0}^{x}f(t)dt=2$  $\bf{My\; Try::}$ Put $y=mx$ in $$\displaystyle y^2+\int_{0}^{x}f(t)dt=2\;,$$ We get $$\displaystyle m^2x^2+y^2+\int_{0}^{x}f(t)dt=2.$$ We can write it as $$\displaystyle 2\int_{0}^{x}2m^2tdt+\int_{0}^{x}f(t)dt=2$$ So $$\displaystyle \int_{0}^{x}\left[2m^2t+f(t)\right]dt-2 =0$$ Now Let $$g(x) = \displaystyle \int_{0}^{x}\left[2m^2t+f(t)\right]dt-2.\;,$$ When $x\rightarrow \infty\;,$ Then $g(x)\rightarrow \infty.$ (Where $g(x)$ is an Continuous function.) Now I did not Understand How can i solve that, Help me Thanks
111217
Find $$\int \limits_0^1 \int \limits_x^1 \arctan \bigg(\frac yx \bigg)dx \, \, dy$$  So obviously using cylindrical is the way to go to give $\theta r$ inside the integral (after considering the jacobian term).  But how can the limits of $x$ be $(x,1)$. I am guessing it is meant to say $dy \, \, dx$ right so those limits are for $y$? Please tell me if you disagree.   But even if this is the case, it is not possible to compute the integral.  After sketching out $x&lt;y&lt;1$ and $0&lt;x&lt;1$, I realised that the region is the triangle bounded by origin, $(0,1)$ and $(1,0)$. So the limits of $r$ and $\theta$ are $(0,1/\cos\theta)$ and $(\pi/4 , \pi /2)$. So we have $$\int \limits_{\pi/4}^{\pi/2} \int \limits_0^{1/\cos\theta} r\theta \, \, \, dr \, \, d\theta$$ which gives $$\int \limits_{\pi/4}^{\pi/2} \frac 12 \theta \sec^2 \theta \, \, \,  d\theta$$  The integral of $\sec^2 \theta $ is $ln |\cos \theta |$ and one of the limits is $\pi/2$ which is undefined. Please help.
24299
Suppose $x \in (0,\frac{\pi}{2})$ and we want to prove $$x&lt;\frac{\sin x +\tan x}{2}$$I tried to prove it by taking $f(x)=\sin x+ \tan x -2x$ and show $f(x) &gt;0 ,when\space  x \in (0,\frac{\pi}{2})$ take f'$$f'=\cos x +1+\tan ^2 x-2\\=\tan^2 x-(1-\cos x)\\=\tan ^2 x-2sin^2(\frac x2)$$ I get stuck here ,because the last line need to be proved $\tan ^2 x&gt;2sin^2(\frac x2) ,when\space  x \in (0,\frac{\pi}{2})$ $\bf{Question}:$ Is there a geometrical method to prove the first inequality ? (or other idea)  Thanks in advance.    $\bf{Remark}: $I can see the function is increasing $when\space  x \in (0,\frac{\pi}{2})$ like below :
79662
I use tangent half-angle substitution to calculate this indefinite integral: $$ \int \frac{1}{2+\sin x}\,dx = \frac{2}{\sqrt{3}}\tan^{-1}\frac{2\tan \frac{x}{2}+1}{\sqrt{3}}+\text{constant}. $$  Wolfram Alpha also give the same answer. However, $\frac{2}{\sqrt{3}}\tan^{-1}\frac{2\tan \frac{x}{2}+1}{\sqrt{3}}$ is discontinuous on $(n+1)\pi$ where $n$ is any integer. Why is an anti-derivative of a continuous function discontinuous?
30048
Show that   $$\displaystyle1 - \frac{x^2}2\leq\cos x\quad\forall x \in \mathbb{R}$$   Let $f(x) = \cos x - 1 + \frac{x^2}2$; then we need to show that $f(x) \geq 0\quad\forall x \in \mathbb{R}$.  Since $$\lim_{x \to \pm\infty} f(x) = +\infty$$ and $$\operatorname{im}f'' = [0, 2]$$ (as $f''(x) = 1 -\cos x$), it remains to show that $x = 0$ is a minimum for $f(x)$, but I cannot do that because I would need to solve $f'(x) = x -\sin x = 0$, which I cannot. I think it would suffice to show that plugging in $x = 0$ we have that $f'(x) = 0$, and combining it with the fact that the function $f$ is convex, we would be done.    Is it enough? What's a better proof?  EDIT: It just occurred to me that there is a way to solve $f'(x) = 0$. Just observe that $f'(x)$ is an odd function and that $f''(x) \geq 0$ implies that $f'(x)$ is monotonically increasing. Since an odd function, if defined at $x = 0$, must be $0$ there, $x = 0$ is a solution to the equation and it's the only one.
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?
10597
The following integral can be obtained using the :   $$\int \frac{dx}{1+\cos^2 x} = \frac{\tan^{-1}(\frac{\tan x}{\sqrt{2}})}{\sqrt{2}}$$  Now assume we are performing this integration between $0$ and $2\pi$. Hence the result of the integration is zero.  On the other hand when looking at the integrand, $\displaystyle \frac{1}{1+\cos^2 x}$, we see that it is a periodic function that is never negative. The fact that it is never negative guarantees that the result of the integration will never be zero (i.e., intuitively there is a positive area under the curve).  What is going on here?
16406
I cant seem to work this out, can someone pls. help,  $$\int_{0}^{\frac{\pi }{2}}\frac{x dx}{\sin x +\cos x}$$  You are meant to use the property of integral that:   $$\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$$