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10

Hint: $$\int r^3\sqrt{8-r^2}dr= \frac12\int r^2\sqrt{8-r^2}d(r^2)=\frac12\int t\sqrt{8-t}\,dt.$$ You can integrate by parts on the second factor. This substitution will work for all integrals of the form $$\int P_o(r)(a^2-r^2)^\alpha dr$$ where $P_o$ is a polynomial with odd powers, and successive integrations by part will progressively lower the ...


9

Substitute $r=2\sqrt{2} \sin\theta$; $dr=2\sqrt{2}cos\theta\ d\theta$ $$(2\sqrt{2})^5\int\sin^3\theta\ \cos^2 \theta \ d\theta=(2\sqrt{2})^5\int\sin^3\theta\ (1-\sin^2 \theta )\ d\theta$$$$(2\sqrt{2})^5\int(\sin^5 \theta- \sin^3 \theta )\ d\theta=(2\sqrt{2})^5\left(\int((1-\cos^2\theta)d(\cos\theta)-\int(1-\cos^2\theta)^2 d(\cos\theta)\right)$$


5

Take $\log\left(x\right)=v$. We have \begin{align} I&=\int_{0}^{\infty}\frac{\exp\left(-x^{n}\right)-\exp\left(-x^{m}\right)}{x\log\left(x\right)}\ dx\\[10pt] &=\int_{-\infty}^{\infty}\frac{\exp\left(-e^{vn}\right)-\exp\left(-e^{vm}\right)}{v}\ dv\\[10pt] &=\int_{0}^{\infty}\frac{\exp\left(-e^{vn}\right)-\exp\left(-e^{vm}\right)}{v}\ dv-\int_{0}^{...


4

You cannot, because your inequality does not hold: just take $a=4$ and $b=3.5$. If we take the function $$ \psi: \xi \mapsto \int_{0}^{+\infty}\cos(\xi x)e^{-x^4}\,dx $$ this is its graph over $[0,10]$: $\hspace{2cm}$ and it is not monotonic in absolute value, even if fast-decaying.


3

Let $u=8-r^2, du=-2rdr$ to get $\displaystyle\int r^3\sqrt{8-r^2}\,dr=-\frac{1}{2}\int r^2\sqrt{8-r^2}(-2r)dr=-\frac{1}{2}\int(8-u)\sqrt{u}du=-\frac{1}{2}\int(8u^{1/2}-u^{3/2})du$ $\displaystyle=-\frac{1}{2}\left[\frac{16}{3}u^{3/2}-\frac{2}{5}u^{5/2}\right]+C=-\frac{8}{3}(8-r^2)^{3/2}+\frac{1}{5}(8-r^2)^{5/2}+C$


3

Since the number $\sqrt{2}-1$ is invariant under the transformation $x \mapsto \frac{1-x}{1+x}$, it is natural to make that substitution. Doing so, we find: $$\begin{align} I &= \int_0^{\sqrt{2}-1} \frac{\ln(1+x^2)}{1+x} dx \\&= \int_{\sqrt{2}-1}^1 \cfrac{\ln(1+x^2)+\ln\left(\frac2{(1+x)^2}\right)}{1+x}dx \\&=\frac12 I + \frac12 \int_{\sqrt{2}-1}...


2

You can use $\displaystyle\int_0^2\int_0^{\sqrt{4-2y}}\frac{1}{\sqrt{2y-y^2}}dx\,dy=\int_0^2\frac{\sqrt{4-2y}}{\sqrt{2y-y^2}}dy=\int_0^2\frac{\sqrt{2}}{\sqrt{y}}dy=\lim_{t\to0^+}\int_t^2\frac{\sqrt{2}}{\sqrt{y}}dy$ $\hspace{.6 in}\displaystyle=\lim_{t\to0^+}\sqrt{2}\left[2\sqrt{y}\right]_t^2=\lim_{t\to0+}2\sqrt{2}(\sqrt{2}-\sqrt{t})=4$ Alternate method: ...


2

The reason that the dog bone contour is inapplicable here is that branch cuts from, say, $0$ to $\infty$ and $1$ to $\infty$ do not "collapse" into the "slit" from $0$ to $1$. To see this, we note that for $z=x+iy$, $x>1$ and $y\to 0+$ we have $\arg(z)=0$ and $\arg(1-z)=-\pi$, while for $x>1$ and $y\to 0^-$, we have $\arg(z)=2\pi$ and $\arg(z)=\pi$. ...


2

First Integral $$\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{xy}x\,dz\,dy\,dz=\int_{0}^{1}\int_{0}^{1-x}x^2y\,dy\,dx=\frac{1}{2}\int_{0}^{1}x^2 (1-x)^2 dx=\frac12\beta(3,3)=\frac1 {60}$$ Second Integral $$x=r\cos\theta\quad,\quad y=r\sin\theta$$ $$\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|=r$$ $$\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{3}r^2\,z\,dz\,dr\,d\...


1

We have to prove that the Fourier cosine transform of $e^{-x^k}$ is non-negative and decreasing on $\mathbb{R}^+$. We have: $$ \int_{0}^{+\infty}\cos(\xi x)e^{-x}\,dx = \frac{1}{1+\xi^2},\qquad \int_{0}^{+\infty}\cos(\xi x)e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\,e^{-\xi^2/4} \tag{1} $$ hence the claim is expected to hold by some kind of interpolation argument. ...


1

the answer of Behrouz Maleki is correct (+1). I add a figure that illustrate the solution. The region of integration is a triangle in the plane $(u,v)$ delimited by the lines $$ v=s \qquad u=t \qquad v=u $$


1

Hint $$I=\int\limits_s^t\int\limits_s^u e^{-\lambda(t-v)}(u-v)^{-\beta-1}dvdu=\int_{s}^{t} \int_{v}^{t}e^{-\lambda(t-v)}(u-v)^{-\beta-1}dudv$$ $$I=-\frac{1}{\beta}\int_{s}^{t}(t-v)^{-\beta}e^{-\lambda(t-v)}dv=-\frac{1}{\beta}\int_{s-t}^{0}(-v)^{-\beta}e^{\lambda v}dv=-\frac{1}{\beta}\int_{0}^{t-s}v^{-\beta}e^{-\lambda v }dv$$


1

Since you want the sine series (if I understand you correctly), you first extend your function to an odd function: $$ \tilde f(x)= \begin{cases} -x & -2\pi<x<-\pi\\ x & -\pi\leq x\leq \pi\\ -x & \pi<x\leq 2\pi \end{cases} $$ Then calculate the Fourier coefficients of this $4\pi$-periodic function as usual: $$ a_n=\frac{2}{4\pi}\int_{-2\...


1

First, as your friend likely noticed, we can rewrite the main curve as $y=2\cosh 10x$. Also, we are going from $x=0$ to $x=6$, so we integrate from $0$ to $6$. This solid of revolution can be solved using method of rings, where our radius is simply $y=2\cosh 10x$, so we get: $$\int_0^6 \pi(2\cosh 10x)^2dx$$ Now, if we calculate this out, we get the same ...


1

Since this is between $y=\frac 5 x$ and $y=0$ and we're rotating it around $y=6$, we need to use method of cylinders. Therefore, we're going to take the inverse of this function and say $x=\frac 5 y$. Now, at $x=1$, we have $y=5$ and at $x=5$, we have $y=1$. Therefore, from $y=1$ to $y=5$, the height of the cylinder is $\frac 5 y-1$ (remember to subtract $1$...


1

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1

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