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20

Note that $$\int_0^1 f(x)^2 (1 - f(x))^2 \; dx = \int_0^1 \left(f(x)^2 - 2 f(x)^3 + f(x)^4\right) \; dx = 0$$ so $f(x)(1-f(x))$ must always be $0$. Since $f$ is continuous, the only solutions are $f(x)=0$ and $f(x)=1$.


12

Calculate the difference of first and second integral then second and third. $$\int_0^1f^2(x)-f^3(x)dx=0$$ $$\int_0^1f^3(x)-f^4(x)dx=0$$ Subtract both equations: $$\int_0^1f^2(x)-2f^3(x)+f^4(x)dx=0$$ $$\int_0^1f^2(x)\left[1-2f(x)+f^2(x)\right]dx=0$$ $$\int_0^1f^2(x)(1-f(x))^2dx=0$$ Look at the integrand it consists of the product of two squares. It can ...


8

From $$ F(x)= \int_x^1\sqrt{1+t^2}\:dt, \quad x \in \mathbb{R}, $$ one gets $$ F'(x)=\left(\color{red}{-}\int_1^x\sqrt{1+t^2}\:dt\right)'=\color{red}{-}\sqrt{1+x^2} $$ giving $$ F'(1)=-\sqrt{2}. $$


6

In fact, you're not meant to find something easier to integrate. After integrating by parts twice (keep integrating $e^{-2t}$ and deriving the trigonometric part), you'll end up with something like $$\int_0^5 e^{-2t}\sin t\,dt=\text{something }+k\int_0^5e^{-2t}\sin t\,dt$$ with $k\ne 1$. At that point, you just need to do this $$(1-k)\int_0^5 e^{-2t}\sin ...


5

If we make the substitution $t=\frac1x$ and later $x^{10}=u$, then we find that $$\begin{align}\int_0^1\frac{1+x^8}{1+x^{10}}dx&=\int_{\infty}^1\frac{1+t^{-8}}{1+t^{-10}}\frac{(-dt)}{t^2}=\int_1^{\infty}\frac{t^8+1}{t^{10}+1}dt\\ ...


4

This integral is the imaginary part of \begin{align*}\int_0^5\mathrm e^{(-2+\mathrm i)t}\,\mathrm d \mkern1mu t&=\frac1{-2+\mathrm i}\mathrm e^{(-2+\mathrm i)t}\biggr\rvert_0^5=-\frac{2+ \mathrm i}{5}\bigl(\mathrm e^{-10+5\mathrm i}-1)\\ &=-\frac15\bigl(2(\mathrm e^{-10}\cos 5-1)-\mathrm e^{-10}\sin 5+\mathrm i(\mathrm e^{-10}\cos 5-1+2\mathrm ...


4

By indeterminate coefficients: By educated guess, you try a solution of the form $$e^{-2t}(a\cos(t)+b\sin(t)).$$ Deriving, you get $$e^{-2t}(-2a\cos(t)-2b\sin(t)-a\sin(t)+b\cos(t)).$$ The unknown coefficients are obtained by solving $$-2a+b=0,\\-2b-a=1,$$ giving $$a=-\frac15,b=-\frac25.$$ The definite integral easily follows.


4

Hint. A possible route. One may set $$ f(s):=4\int_0^\infty \frac{e^{-sx}+sx-1}{x(e^{2x}-e^{-2x})}dx, \quad s>0. \tag1 $$ In order to get rid of the $x$ in the denominator, we may differentiate under the integral sign getting $$ f'(s)=4\int_0^\infty \frac{1-e^{-sx}}{e^{2x}-e^{-2x}}dx, \quad s>0. \tag2 $$ Then expanding the latter integrand and ...


3

Consider $$\int_0^3 \sqrt{1+x} \, dx = \lim_{n \to \infty} \frac{3}{n}\sum_{k=1}^n\sqrt{1 + \frac{3k}{n}} = \lim_{n \to \infty} \frac{3}{n^{3/2}}\sum_{k=1}^n\sqrt{n + 3k}.$$ Using the binomial expansion, we have $$(n + 3k - 3)^{3/2} = (n + 3k)^{3/2} \left(1 - \frac{3}{n+3k} \right)^{3/2} \\ = (n + 3k)^{3/2}\left(1 - \frac{3}{2}\frac{3}{n+3k} + ...


3

You have to choose the partition points $x_k$ such that $\sqrt{1+x_k}$ becomes manageable. Therefore put $$x_k:=u_k^2-1\quad(0\leq k\leq n)$$ whereby the $$u_k:=1+{k\over n}\quad(0\leq k\leq n)$$ are equally spaced. The $x_k$ are then inequally spaced between $0$ and $3$, but in any case the differences $x_k-x_{k-1}$ tend to $0$ when $n\to\infty$. A typical ...


3

$\left\lfloor \sec^{-1}x\right\rfloor=\begin{cases} 0 & \text{ for } x<\sec(1)\approx1.851\\ 1 & \text{ for } x\ge\sec(1) \end{cases}$ $\left\lfloor\cot^{-1}x\right\rfloor=\begin{cases} 1 ...


3

We assume $x>0$ and $s>0$. Then by differentiating the following identity with respect to $s$, $$ f(s)=\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk $$ one may write $$ f'(s)=-\int_0^\infty \exp(−sk)\sin(kx)\,dk=-\frac{x}{x^2+ s^2} $$ giving $$ f(s)=-\arctan \left( \frac{s}x\right)+C. $$ Observing that, as $s \to \infty$, $f(s) \to 0$, we then obtain ...


3

$$h\int_0^T (Q-at)dt=h\left(\int_0^T Qdt-\int_0^T atdt\right)=h\left(Qt]_0^T-\frac{at^2}{2}]_0^T\right)=h\left(QT-\frac{aT^2}{2}\right)=hT\left(Q-\frac{aT}{2}\right)=hT\left(aT-\frac{aT}{2}\right)=\frac{haT^2}{2}=\frac{hQT}{2}$$


3

You rather have $$ h\int_0^T (Q-at)dt=h \left[Qt-\frac{at^2}2\right]_0^T=h\left(QT-\frac{aT^2}2 \right)=h\left(aT^2-\frac{aT^2}2 \right)=h\frac{aT^2}2, $$ then put $aT=Q$ to get the announced result.


3

The answer is $$f(n, a) = \frac{2\pi}{a\sqrt{1-a}}\left(\frac{1-\sqrt{1-a}}{\sqrt{a}}\right)^n \tag{1}$$ for $0 < a < 1$ and $n > 0$ an even integer; we also have $$f(0, a) = \frac{2\pi}{a \sqrt{1-a}} - \frac{2\pi}{a}$$ which (mysteriously) happens to equal what $\sqrt{a}f(1, a)$ would be if we put $n=1$ in formula $(1)$. First, consider the ...


3

You need to be a bit careful in multiplying the numerator and denominator by $\sec x$ or by $\sec x+\tan x$, because these expressions are not defined in the middle of the range of integration, at $\frac12\!\pi$. However, you can write the integral as $$\int_0^\pi\frac{\mathrm d x}{1+\sin ...


3

Handling $z$ term is easy as it is just a polynomial. Handling $\theta$ is easy too. Now to handle the $r$ term. You miss out an $r$. remember when using cylindrical coordinate. We have to use $r\,dr\,d\theta$ rather than just $dr\,d\theta$. This simplifies the problem.


2

By using the formula $(uv)'=u'v+uv'$, you just have $$ \left(e^{x^2}\times\frac{2}{\sqrt{\pi}}\int_0^{{x}}e^{-t^{2}}dt\right)'=2xe^{x^2}\left(\frac{2}{\sqrt{\pi}}\int_0^{{x}}e^{-t^{2}}dt\right)+\left(e^{x^2}\times\frac{2}{\sqrt{\pi}}e^{-x^{2}}dt\right)=2xy+\frac{2}{\sqrt{\pi}} $$ since $$ \left(\int_0^{{x}}e^{-t^{2}}dt\right)'=e^{-x^{2}}. $$


2

Let $x=r\cos\theta$ and $y=r\sin\theta$ so that $$ C_1:\,r=a\cos\theta\qquad C_2:\,r=2a\cos\theta\qquad\theta\in\left[-\frac{\pi}{2},\,\frac{\pi}{2}\right] $$ So the integral becomes \begin{align} \iint_{C_2\setminus C_1} y^2\mathrm d x\,\mathrm d y&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{a\cos\theta}^{2a\cos\theta}r^3\sin^2\theta\,\mathrm d ...


2

By expanding $(1-a\cos^2 x)^{-1}$ as $\sum_{k=0}^{\infty}(a\cos^2 x)^k$, you obtain $$ f(n,a)=\sum_{k=0}^{\infty} a^k \int_{0}^{2\pi} dx \cos(nx) \cos^{2k+2} x. $$ Each cosine can be written as $\cos u = \frac{1}{2}(e^{iu} + e^{-iu})$, and terms of the form $\int_{0}^{2\pi}e^{imu}$ vanish unless $m=0$, in which case they evaluate to $2\pi$. Counting the ...


2

Split the integral, do a substitution on the second one, use $f(x)f(-x)=1$, and recombine to get $$\int_{-1}^{1} f(x) g(x)dx = \int_0^1 \left( f(x) + \frac{1}{f(x)} \right) g(x) dx.$$ Next show that $y+1/y \geq 2$ for $y > 0$. Can you finish?


2

Break the integral in $[-1,0]$ and $[0,1]$. Since $g$ is even, we have $\int_0^1 g(t)dt=1/2$. Now, $$\int_{-1}^1f(t)g(t)dt=\int_{-1}^0f(t)g(t)dt+\int_0^1f(t)g(t)dt=\int_0^1f(-t)g(-t)dt+\int_0^1f(t)g(t)dt=\int_0^1(f(t)+f(-t))g(t)dt=\int_0^1(f(t)+\frac{1}{f(t)})g(t)dt,$$ where the next-to-last equality uses that $g$ is even and the last equality uses ...


2

That is the Integral Chebyshev inequality. The following proof is from http://imar.ro/journals/Mathematical_Reports/Pdfs/2010/2/Niculescu.pdf (Theorem 3): If $f$ and $g$ are both increasing (or both decreasing) then $$ \tag{*} 0 \le \bigl(f(x) - f(y) \bigr) \cdot \bigl(g(x) - g(y) \bigr) $$ for all $x, y \in [a, b]$. It follows that $$ 0 \le \int_a^b ...


2

Hint. Integrating by parts twice gives $$ \begin{align} \int_0^5e^{-2t}\sin t\:dt&=\left. -\frac12e^{-2t}\sin t\right|_0^5+\frac{1}{2}\int_0^5e^{-2t}\cos t\:dt\ \end{align} $$ and $$ \begin{align} \int_0^5e^{-2t}\cos t\:dt&=\left. -\frac12e^{-2t}\cos t\right|_0^5-\frac{1}{2}\int_0^5e^{-2t}\sin t\:dt\ \end{align} $$ Can you take it from here?


2

As property of definite integrals it is known that for any function integrand $f(x)$ $$\int_0^a{f(x)}\text{d}x = \int_0^a{f(a-x)}\text{d}x$$ Geometrically this means the area under the curve when flipped about the line $ x= \dfrac{a}{2}$ as a rigid figure cannot change. The property can be stated symbolically as in one particular generalisation where ...


2

Another approach. We have $$\sum_{k\geq0}\frac{\left(-1\right)^{k}x^{k}}{k!}=e^{-x} $$ hence $$e^{-x}-1+x=\sum_{k\geq2}\frac{\left(-1\right)^{k}x^{k}}{k!} $$ and so $$I=4\int_{0}^{\infty}\frac{e^{-x}-1+x}{x\left(e^{2x}-e^{-2x}\right)}dx=4\sum_{k\geq2}\frac{\left(-1\right)^{k}}{k!}\int_{0}^{\infty}\frac{x^{k-1}}{e^{2x}-e^{-2x}}dx $$ now note that $$ ...


2

I don't know if there exists a closed form in terms of some special known functions. One may recall that the following standard power series $$ \sin u = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} u^{2n+1}, \quad u \in \mathbb{R},\tag1 $$ has an infinite radius of convergence allowing us to obtain $$ \int_0^\infty \sin(xe^{-x})\:dx=\sum^{\infty}_{n=0} ...


2

$$F(x)=\int_x^1\sqrt{1+t^2}dt$$ or $$F(x)=\int_x^1f(t)dt$$ Note that if, $$F(x)=\int_{g(x)}^{h(x)}f(t)dt$$ then $$F'(x)=f(h(x))h'(x)-f(g(x))g'(x)$$ So, here, $$F'(x)=f(1)\frac{d1}{dx}-f(x)\frac{dx}{dx}=-f(x)$$ Thus, $$F'(1)=-f(1)=-\sqrt2$$


2

\begin{align*} C(x+n,y+n) &= \int_{-\infty}^\infty \frac{dt}{(1+t^2)^n(1+it)^x(1-it)^y} \\ &= \int_{-\infty}^\infty \frac{du/\sqrt{n}}{(1+u^2/n)^n(1+iu/\sqrt{n})^x(1-iu/\sqrt{n})^y} \qquad (\textrm{let }u=t/\sqrt{n}) \\ &= \int_{-\infty}^\infty \frac{1}{\sqrt{n}} e^{-u^2} + O(1/n) \qquad (\textrm{standard limit for }e) \\ &= ...



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