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0

Since $|x^n f(x)| \le |f(x)|$ for $x \in [0, 1]$ and $x^n f(x) \to 0$ as $n \to \infty$, by dominated convergence, $$\lim_{n \to \infty} \int_{0}^1 x^n f(x) dx = 0$$


0

Find the (real) value of $a$ such that the curves \begin{eqnarray} y &=& a^x \\ y &=& \log_a (x) \end{eqnarray} intersect exactly once. Find also the $x$ and $y$ values where they intersect. Note that's the logarithm of base $a$ in the second curve. I think this is a pretty tough problem. It doesn't involve advanced calculus, but you need to ...


0

Since $f$ continuous on a compact so is $|f|$. So $|f(x)|$ has a maximum value $M$ in $[0,1]$, hence $$ \left | \int_0^1 x^n f(x) dx \right | \le \int_0^1 x^n |f(x) |dx \le M \int_0^1 x^n dx = \frac{M}{n+1} $$ take the limit and you are done.


2

There are basically two approaches to the problem. See mookid's or lcv's for one. Here's another: Can you evaluate this limit in the case that $f$ is a polynomial? How does this help you find the limit for an arbitrary continuous function $f$? (Hint: use the Weierstrass theorem). This is a good method to keep in mind when you have no expectation of what ...


2

Hint: as $f$ is continuous it is bounded. details: $$ \left|\int f(x) x^n dx\right| \le \int\left| f(x) \right| x^n dx \le M \int x^n dx \\ = \frac M{n+1} \to 0 $$


2

$$\int\frac x{\sqrt{x^2+k^2}}dx=\frac12\int\frac{(x^2+k^2)'}{\sqrt{x^2+k^2}}dx=\frac122\sqrt{x^2+k^2}+C=\sqrt{x^2+k^2}+C$$


0

Euler's substitutions solve all integrals of the form $\int R(x,\sqrt{ax^2+bx+c})$, where $R(x,y)$ is a rational function. In your case $R(x,y)=\frac{x}{y}$ and $a=1$, $b=0$, and $c=k^2$.


0

So we let $\inf f, \inf g, \inf (f+g)$ and $I$ be the interval. We have that $\inf f + \inf g \leq f(t) + g(t) = (f+g)(t)$ $\forall t \in I$. Now, suppose $\exists \epsilon > 0$ such that $\forall t>0$, $f(t) + g(t) - \inf(f+g) \geq \epsilon$. Then we have that $\inf(f+g) + \epsilon \leq f(t) + g(t)\; \forall t>0$. But this is a contradiction ...


2

Hint: Let $u=x^{2}+k^{2}$. Then $du=2xdx$ since $k$ is a constant.


1

The argument looks fine, although the notation is somewhat sloppy: for instance I would write the first step as $$\int_0^{\infty} f(x)\cos(nx)\,dx = \cos(nx)\int_0^x f(y)\,dy\ \biggr\vert_{x=0}^{\infty} + n\int_0^{\infty} f(x)\sin(nx)\,dx.$$ The claim becomes false if $f(x)$ is locally integrable but not $L^1$, i.e. if $\lim_{x\to \infty} \int_0^x ...


1

Yes. We could do this another way, too. $$\begin{align} \frac{d}{dt}e^{it}&=\frac{d}{dt}\left(\cos{t}+i\sin{t}\right)\\ &=-\sin{t}+i\cos{t}\\ &=i\left(\cos{t}+i\sin{t}\right)\\ &=ie^{it} \end{align}$$ Be careful, because you technically need a more rigorous definition of complex exponentials and derivatives for complex numbers. However, it ...


2

Try taking the indefinite integral of both sides instead: \begin{align*} \frac{ds}{s} &= \mu \, dt \\ \int \frac{1}{s} \, ds &= \int \mu \, dt \\ \ln|s| &= \mu t + C &\text{for some constant $C \in \mathbb R$}\\ e^{\ln|s|} &= e^{\mu t + C}\\ |s| &= e^{\mu t} \cdot e^C \\ s &= \underbrace{(\pm e^C)}_{A}e^{\mu t} \\ s(t) &= ...


0

Let $(f+g)(a) = m_i$, $f(a') = m_i'$, and $g(a'') = m_i''$. Then either $a = a' = a''$, in which case we have: $$ f(a) + g(a) = (f + g)(a)$$ $$ f(a) + g(a) = f(a) + g(a)$$ (Valid step? $(f+g)(a) = f(a) + g(a)$) which means, in this case: $$ m_i'' + m_i' = m_i$$ Then we have the case where one or more of $a$, $a'$ or $a''$ differ. In this case: $$f(a') ...


1

Continuing your calculation: \begin{align}k-\lfloor kx\rfloor-\lfloor k-kx\rfloor=k-\lfloor kx\rfloor-k-\lfloor-kx\rfloor=-\lfloor-kx\rfloor-\lfloor kx\rfloor=\lceil kx\rceil-\lfloor kx\rfloor\end{align} So it's $0$ if $kx\in\mathbb Z$, else it's $1$.


3

One way of viewing the hyperbolic functions is that they are "abbreviations" for combinations of exponential functions, so whatever can be done using hyperbolic functions can equally be done using exponentials - albeit in a less concise manner. Similarly, inverse hyperbolics are really just log functions in various combinations. What this means is that if ...


1

Hint: do it explicitly for the function $f(z) = 1$.


0

$f_k(x) = \{kx\}$. Then let $f_k(x) = a$. Note that $f_k(1-x) = \{k(1-x)\} = \{-kx\} = \{1-kx\}$ since $\{k\} = 0$ so the result follows.


0

"Why isn't this a valid solution to the equation?": it absolutely is, setting $K_1=\frac{c}{a}$ and $K_2=0$ ! You just found a particular case of the general solution.


2

This is one of the simplest linear differential equation of the first order. $\frac{df}{dx}=f$ is rewritten $\frac{df}{f}=dx$. Then by integration, $\int\frac{df}{f}=\int dx$, and $\ln(f)=x+c$, or $f=Ce^x$.


1

Functions of the form $ce^x$ (where $c$ is a constant) are the only solutions to that differential equation by the Picard-Lindelöf theorem. ${}{}{}$


5

This limit has been floating around since the 1990s (if not longer), though I haven't seen it "in the wild" since 1996: If $a$ is real, evaluate $$ \lim_{x\to\infty} e^{e^{e^{[x + e^{-(a + x + e^{x} + e^{e^{x}})}]}}} - e^{e^{e^{x}}}. $$


4

Here's another one to see how you understand calculus. $f(x)\geqslant 0, \forall x \geqslant 0$ $f(x)\leqslant c\int_0^xf(t)dt, \forall x\geqslant 0 ,\exists c>0$ Prove that $f(x)$ is identically zero.


0

Try : $$ \int e^{-2x}\left[(e^x)^2 +(x^x)^2\right]\ln x dx$$


2

This will test all of your analytical and mathematical skills. $f(x)$ is a differentiable function and $g(x)$ is a double differentiable function such that $|f(x)|\leqslant 1$ and $f'(x)=g(x)$. If $$f(0)^2+g(0)^2=9$$ then prove that there exists some $c\in(-3,3)$ such that$ \ \ g(c) \cdot g''(c)<0$.


0

This was on this years intervarsity paper in Ireland. It's kinda fun. $\large{\int\limits_0^4 \frac{dx}{4+2^x}}$


0

If you haven't seen it before, then this should put your integration skills to the test: $$\int \! \sec^3(x) \, \mathrm{d}x$$


1

We can write \begin{align} \frac{x^3-11x^2+x+2}{x^4-2x^3} & = \frac{A_1}{x}+\frac{A_2}{x^2}+\frac{A_3}{x^3}+\frac{B}{x-2} \end{align} for $A_1$, $A_2$, $A_3$ and $B$ real numbers such that \begin{align} x^3-11x^2+x+2 & =\left[\frac{A_1}{x}+\frac{A_2}{x^2}+\frac{A_3}{x^3}+\frac{B}{x-2}\right]x^3(x-2) \\ & = A_1x^2(x-2)+A_2x(x-2)+A_3(x-2)+Bx^3 \\ ...


1

If the integral is evaluated over the real numbers, then you will want to use absolute values for the logarithmic terms: $$\frac{1}{2x^2} + \frac{1}{x} - 4 \log |2-x| + 5 \log |x| + C.$$ Also, if your web-based homework system is picky about distinguishing between $\log$ and $\ln$, you will want to use $\ln$. Mathematicians typically write $\log$ to mean ...


4

This will only be an illustration of Robjohn's excellent answer. There is large numerical instability while evaluating this integral (the numerical results returned by CAS will depend of the precision required and the method used : I got values ranging from $1.5$ to $5$). The problem is that the exponent will take the value $0$ for $\,z=\pi n\;$ as ...


8

When integrating $$ \int_0^\infty e^{-z^2\sin^2(z)}\,\mathrm{d}z $$ the problem points are near $z=k\pi$ where $\sin^2(z)$ vanishes. On $[(k-1/2)\pi,(k+1/2)\pi]$ $$ \begin{align} \int_{(k-1/2)\pi}^{(k+1/2)\pi} e^{-z^2\sin^2(z)}\,\mathrm{d}z &=\int_{-\pi/2}^{\pi/2}e^{-(z+k\pi)^2\sin^2(z)}\,\mathrm{d}z\tag{1}\\ ...


1

First we can see easily that $$\frac{d}{dx} (\cos^2 x+\sin^2 x)=\frac{d}{dx}1$$ and since $$\frac{d}{dx} =0$$ Then we can plug it back into the original function to get $$\frac{d}{dx} 1=0=\frac{d}{dx} (\cos^2 x+\sin^2 x)$$ But if you want to do this from scratch, we can also say that by the linearity of the derivative: $$\frac{d}{dx} (\cos^2 x+\sin^2 ...


3

$$ \begin{align} \frac{d}{dx} (\sin x)^2 & = 2\sin x\cdot\frac{d}{dx} \sin x = 2\sin x\cos x \\[15pt] \frac{d}{dx} (\cos x)^2 & = 2\cos x\cdot\frac{d}{dx}\cos x = 2\cos x\cdot(-\sin x) \end{align} $$ Now add those together.


1

Using lab bhattacharjee's result. The integral can be solven by using integration by parts. Let \begin{align} u=\arctan\frac{x}{2}\qquad\rightarrow\qquad du=\frac{2\,dx}{x^2+4}\qquad\text{and}\qquad dv=x\,dx\qquad\rightarrow\qquad v=\frac{1}{2}x^2 \end{align} Then \begin{align} 8\int_0^2x\arctan\frac {x}{2}\, ...


-2

...to prove that $(\sin^2(x)+\cos^2(x))=1$ one have(has) to prove that the derivative of $(\sin^2(x)+\cos^2(x))$ is 0. This is not true. It is the other way round. To prove that that the derivative of $(\sin^2(x)+\cos^2(x))$ is 0 we use the identity $f(x)=(\sin^2x+\cos^2x)=1$ (a constant) for every value of $x$. Since the derivative of a constant ...


0

We can indeed divide it up in that manner. Let $L_1$ be the limit at $-\infty$ and $L_2$ the limit at $+\infty$. By definition, we have some $M_1,M_2$ such that $$x<M_1\implies \left|\frac{f(x)}{x^2}-L_1\right|<1\quad \text{and}\quad x>M_2 \implies \left|\frac{f(x)}{x^2}-L_2\right|<1$$ and thus $$x<M_1\implies |f(x)|<(|L_1|+1)x^2\quad ...


1

Hint: $$\dfrac{\mathrm d}{\mathrm dx}\Big[\cos^2x+\sin^2x\Big]=\dfrac{\mathrm d}{\mathrm dx}\Big[\cos x\cos x\Big]+\dfrac{\mathrm d}{\mathrm dx}\Big[\sin x\sin x\Big].$$ Now use the product rule knowing that: $$\dfrac{\mathrm d}{\mathrm dx}\sin x=\cos x\quad\color{grey}{\text{and}}\quad\dfrac{\mathrm d}{\mathrm dx}\cos x=-\sin x.$$


1

A different manner to answer to the question is to show a graphical representation to the function $f(x)$ for various values of $n$ and a graphical representation of the respective integrals $F(X)$. ( The old saw that “One little picture says more than a long speech”) This makes more understandable the behavior of the function and integral for $n$ tending ...


1

HINT: For even function $f(x)$, $$\int_{-a}^af(x)dx=2\int_0^af(x)dx$$ For odd $$\int_{-a}^af(x)dx=0$$ $$(-x)=x^4,\cos(-x)=\cos x,\arctan\left(-\dfrac x2\right)=-\arctan\dfrac x2$$ $\implies$ if $\displaystyle g(x)=\left(x^4+4\cos x\right)\arctan\frac x2,g(-x)=-g(x)$ If $\displaystyle h(x)=x\cdot\arctan\dfrac x2,h(-x)=h(x)$ So, the given integral ...


3

In cylindrical co-ordinates (height along central axis $x$, distance from central axis $r$ and angle $\theta$), let the shape be bounded by $$0\le x \le x_0$$ $$0\le r \le r_0(x)$$ $$0\le\theta<2\pi$$ So the solid of revolution is given by $$V=\int_0^{x_0}\int_{0}^{r_0(x)}\int_0^{2\pi}dV=\int_0^{x_0}\int_{0}^{r_0(x)}\int_0^{2\pi}r d\theta dr dx$$ ...


-1

\begin{equation} \begin{split} f(x) &= \int^{\ln2}_0 (3e^u - e^{2u} - 2)\sin nu \,du\\ &=3\int^{\ln2}_0 e^u\sin nu \,du-\int^{\ln2}_0e^{2u}\sin nu \,du-2\int^{\ln2}_0\sin nu \,du\\ &=3\int^{\ln2}_0 e^u\sin nu \,du-\int^{\ln2}_0e^{2u}\sin nu \,du+\frac{2}{n}(\cos(n\ln2)-1) \end{split} \end{equation} The first two integrals can be solven by using ...


1

$$ \frac{x^2+ax+b}{x+1}=x+a-1+\frac{b-a+1}{x+1}. $$ Hence $a-1=1$, unless I made some mistake...


0

Let $P(x)$ be the profit function, then $$ \begin{align} P(x)&=x\cdot p(x)-C(x)\\ &=x(4100 − 9x)-(14000 + 500x − 4.8x^2 + 0.004x^3)\\ &=4100x-9x^2-14000 - 500x + 4.8x^2 - 0.004x^3\\ &=-14000+3600x-4.2x^2- 0.004x^3.\\ \end{align} $$ To maximize the profit function, take its first derivative and set equal to zero. $$ \begin{align} ...


1

Rewrite the equation of the ellipse $x^2+4y^2=8$ as $$ \begin{align} x^2+4y^2&=8\\ 4y^2&=8-x^2\\ y^2&=\frac{8-x^2}{4}\\ y&=\left(\frac{8-x^2}{4}\right)^{\frac12}.\tag1 \end{align} $$ The slope of the lines that are tangent to the ellipse is $$ \begin{align} \frac{dy}{dx}&=\frac{d}{dx}\left(\frac{8-x^2}{4}\right)^{\frac12}\\ ...


3

We have the ODE: $$t^{5}\frac{\mathrm{d}y}{\mathrm{d}t}+y^{5}=0$$ We can therefore rearrange this to give: $$t^{5}\frac{\mathrm{d}y}{\mathrm{d}t}=-y^{5}$$ Multiplying both sides by $\frac{1}{t^{5}y^{5}}$, we get: $$\frac{1}{y^{5}}\frac{\mathrm{d}y}{\mathrm{d}t}=-\frac{1}{t^{5}}$$ Multiplying both sides by $\mathrm{d}t$, we get: ...


1

To find the points of intersection, note that $$-\frac 12 = -\frac{x}{2y} \iff x = y$$ This means that the two lines parallel to the given line are tangent to the eclipse when $x = y$. We can find the two points of tangency to the ellipse, $\;(x_1, y_1), (x_2, y_2)\;$ where $\;x_1 = y_1,\;$ $x_2 = y_2\;$ by using the fact that $x = y$, together with the ...


0

So if $(a,b)$ be the point of contact , you have $$-\frac12=-\frac a{2b}\iff a=b\text{ and }a^2+4b^2=8$$ Solve for $a,b$ So, we have $$\frac{y-b}{x-a}=-\frac12$$


0

I now recognize it is quite a silly question: you just write taylor expansion to the first order with integral remainder, change variables in the integral and then add and subtract $\frac{1}{2}f^{\prime\prime}\big(X(s)\big)\big(X(t)-X(s)\big)^2$ writing the subtracted part as $f^{\prime\prime}\big(X(s)\big)\big(X(t)-X(s)\big)^2\int_0^1(1-\theta) d\theta$.


0

Assume for example that $y_1(0) = 1, y_1'(0) = 0, y_2'(0) = 1$. Then $W(0) = 1$ and the solution of the problem $$ W' + pW = 0 \\ W(0) = 1 $$ is unique.


2

The derivation as it stands is not valid. Even though Bilou06 was probably speaking about convergent series, the relation $\sum_n\sum_m a_{n,m}=\sum_m\sum_n a_{n,m}$ can also fail for regularized divergent series. In this case the conclusion is false - an error term is needed. We have two formulas at hand: $$\zeta(-n)=-\frac{B_{n+1}}{n+1}$$ ...


1

If $$ a_n=\frac{(-1)^n}{(-1)^n+\sqrt{n}}, $$ then $$ a_{2n}+a_{2n+1}=\frac{1}{\sqrt{2n}+1}-\frac{1}{\sqrt{2n+1}-1}= \frac{\sqrt{2n+1}-\sqrt{2n}-2}{(\sqrt{2n+1}-1)(\sqrt{2n}+1)}=-\frac{1}{n}+{\mathcal O}(n^{-3/2}). $$ Hence, the series diverges to $-\infty$.



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