# Tag Info

0

Some tricks to help you: a. count squares b. calculate the area of known figures, such as triangles and rectangles. For example, the integral in (b) can be calculate as the same of the area of a trapezoid($(1+3)\cdot 2/2$), a rectangle ($1\cdot 3$), and a triangle ($3\cdot 2/2$) to give a grand total of $10$. That should cover it, good luck.

0

see the solution on your graph

2

$$\frac{d f}{d t} = A f^2 +g(t)$$ Let $f(t)=-\frac{1}{A\:h(t)}\frac{d h}{dt}$ which leads to : $$\frac{d^2h}{dt^2}+A\,g(t)h(t)=0$$ It is well-known that this kind of second order linear ODE can be analytically solved for some forms of functions $g(t)$ and cannot for many others. So, it is impossible to answer to the question if the function $g(t)$ is not ...

4

No, this isn't true. As an example: $$f(x)=\begin{cases}1 &\text{if } x\in (0,1/2] \\ 0&\text{else}\end{cases}$$ and $$g(x)=\begin{cases}1 &\text{if } x\in (1/2,1) \\ 0&\text{else}\end{cases}$$

3

Counterexample: Consider $(0,1)$ with Lebesgue measure and the characteristic functions $\chi_{(0,\frac{1}{2})}$ and $\chi_{(\frac{1}{2},1)}$. Then $\chi_{(0,\frac{1}{2})}\chi_{(\frac{1}{2},1)} = 0$ but the product of their integrals is $\frac{1}{4}$.

1

$$\int_a^b{f(x)dx}=\int_a^{b}{f(x)[H(x-a)-H(x-b)]dx}=\int_0^{\infty}{f(x)[H(x-a)-H(x-b)]dx}$$

3

Change coordinates: $u=x+y$, $v=x-y$. Jacobian = $1/2$. Then the integral is $$\int_{-(b-a)}^{b-a} dv \, \int_{2 a}^{2 b} du \, \arctan{u}$$ The integral of arctan is easy using integration by parts: $$\int du \, \arctan{u} = u \arctan{u} - \frac12 \log{(1+u^2)} +C$$ Hopefully you can take it from there.

2

The denominator of your second integral should be $-3+5\cos 2x$, because from the identities $\sin x\cos x=\frac{\sin 2x}{2}$ and $\sin ^{2}x=\frac{ 1-\cos 2x}{2}$ we obtain \begin{equation*} \frac{1+\sin x\cos x}{1-5\sin ^{2}x}=\frac{2+\sin 2x}{-3+5\cos 2x}. \end{equation*} To evaluate \begin{equation*} \int \frac{2+\sin 2x}{-3+5\cos 2x}dx ...

1

one form of the fundamental theorem of calculus is $$d\left(\int_a^b f(t) \, dt \right) = f(b)\, db - f(a) \, da \tag 1$$ if you keep the lower limit fixed at $a$ and the upper limit is a variable $x^2,$ then $(1)$ becomes $$d\left(\int_a^{x^2} \sin^3 t\, dt \right) = \sin^3\left(x^2\right)\, d\left(x^2\right) - \sin^3(a) \times 0 = ... 3 Hint: Use the Chain Rule$$f(g(x))' = f'(g(x)) g'(x)$$where$$f(x) = \int_a^{x} \sin^3 t \,\,dt$$and g(x) = x^2. 1 When you have to solve something of the form:$$\int \dfrac{dt}{(t-a)(t-b)} $$Then you can have$$\dfrac{A}{t-a} + \dfrac{B}{t-b} = \dfrac{1}{(t-a)(t-b)}$$Wich means that$$\left\{\begin{array}{rcl}A\ +B\ & = & 0 \\ Ab + Ba & = & -1 \end{array}\right. $$Once you find A and B then$$\int \dfrac{dt}{(t-a)(t-b)} = ...

-1

hint: $\dfrac{1}{t^2-t-20} = -\dfrac{1}{9}\cdot \dfrac{1}{t+4} + \dfrac{1}{9}\cdot \dfrac{1}{t-5}$

4

Note that the signs you wrote are wrong. $$\frac{1}{t^2-t-20}=\frac{1}{(t\color{red}{-}5)(t\color{red}{+}4)}=\frac 19\left(\frac{1}{t-5}-\frac{1}{t+4}\right)$$

4

We have $$\int\frac{1+\sin x\cos x}{1-5\sin^2x}\mathrm{d}x =\int\frac{\mathrm{d}x}{1-5\sin^2x}+\int\frac{\sin x\cos x}{1-5\sin^2x}\mathrm{d}x$$ and $$\int\frac{\sin x\cos x}{1-5\sin^2x}\mathrm{d}x =-\frac{1}{10}\int\frac{\mathrm{d}\left(1-5\sin^2x\right)}{1-5\sin^2x} =-\frac{1}{10}\ln\left(1-5\sin^2x\right).$$ Then, $$\int\frac{\mathrm{d}x}{1-5\sin^2x} ... 0 The t in these substitutions is t=\tan(x/2), not t=\tan x so that$$\sec^2(x/2)= \tan^2 (x/2)+1=t^2+1$$1 According to your definition this certainly can not be true. If f_{max}\in C then so is a(f_{max}) for any a\in \mathbb{R}. But even if you fix that problem I'm pretty sure such a function can't exist. 3 No, for the same reason that the result doesn't work for functions (just define an antiderivative, and so on). The result for functions is discussed in this question. Similarly for series and sequences, basically by using the integral test. 1 We know that the volume of the solid generated by rotating the curve y=f(x)\space  about the x-axis is given as follows$$V=\int_{x_1}^{x_2}\pi y^2dx$$Hence, the volume of the solid generated by rotating the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \space about the x-axis is obtained by setting y^2=\frac{b^2}{a^2} (a^2-x^2) \space in the above ... 0 For complete safety, you can write$$\int x\sqrt{1-x^2}\,dx=\int x(t)\sqrt{t}\,\frac{-dt}{2\,x(t)}=-\frac12\int \sqrt{t}\,dt.$$Had the x's not canceled each other, you should have had to substitute x by \sqrt{1-t} in the end. 0 An idea: We have$$\int\sqrt x\;dx=\frac23 x^{3/2}+C\implies \int f'(x)\sqrt{f(x)}dx=\frac23 (f(x))^{3/2}+C$$Now, just observe that$$\;x=-\frac12\left(1-x^2\right)'\;$$1 You can do as the following : Since xdx=-\frac{1}{2}dt,$$\int x\sqrt{1-x^2}dx=\int\left(\sqrt{1-x^2}\right)\cdot \color{red}{xdx}=\int \sqrt t\cdot\color{red}{\frac{-1}{2}dt}$$0 A piecewise constant function lies below f if and only if it is nonpositive everywhere. So the supremum of the lower sums of f is zero. Among piecewise constant functions lying above f, for any \delta>0 we have a function which is 1 on [-\delta,\delta] and zero elsewhere. The integral of this function is 2 \delta, which can be made ... 0 Fix 1> \epsilon > 0, and consider the step function$$ \varphi(x) = \begin{cases} 1 &: -\epsilon \leq x \leq \epsilon \\ 0 &: \text{ otherwise} \end{cases} $$Then$$ \int_{-1}^1 \varphi(x)dx = 2\epsilon $$Hence$$ \int_{-1}^{1^{\ast}} f(x)dx = 0 $$Now for any step function \psi \leq f, then choose a partition -1 < x_1 < x_2 < ... 6 Let u=\cos x. Then \mathrm du=-\sin x\mathrm dx. Therefore$$\begin{split}\int\frac{\sin^3x}{(\cos x)^{4/3}}\mathrm dx&=\int\frac{u^2-1}{u^{4/3}}\mathrm du\\&=\int \left(u^{2/3}-u^{-4/3}\right)\mathrm du\\&=\frac35u^{5/3}+3u^{-1/3}+c\\&=\frac35(\cos x)^{5/3}+3(\cos x)^{-1/3} +c.\end{split}$$0 Hint: prove that, if g is continuous on [0,1], hence continuous at x_0, there exists \delta>0 such that, for every x\in[x_0-\delta,x_0+\delta], g(x)>g(x_0)/2 (a variation of the permanence of sign theorem). Then… 0 Observe that the quadratic  x^2 -3x +3  has discriminant$$b^2-4ac=(-3)^2-4(1)(3)=9-12=-3$$Hence no real solutions, which means it never crosses the x-axis, and since it opens up, this means it is always positive. Using this, and the following formula for \arctan \frac{1}{u}$$\arctan \frac{1}{u}=\frac{1}{2}\pi - \arctan u, u>0 $$We can rewrite ... 2 The crux to completing your work is to prove the following$$\int_0^1 \frac{\log^2(1-x)}{x^2} \,\mathrm{d}x = -2 \int_0^1 \frac{\log(1-x)}{x}\,\mathrm{d}x$$You are close though, let me show you how to finish your work. I really liked this problem. Summarizing your work, you have proven that$$ \int_0^\infty \left(\frac{\log x}{1-x}\right)^2 ...

4

You can integrate by parts. A primitive to $1/(1-x)^2$ is $1/(1-x)-1=x/(1-x)$. The derivative of $(\ln x)^2$ is $2(\ln x)/x$. Thus \begin{aligned} \int_0^1 \frac{(\ln x)^2}{(1-x)^2}\,dx &= \Bigl[\frac{x}{1-x}(\ln x)^2\Bigr]_0^1-\int_0^1 \frac{x}{1-x}\frac{2\ln x}{x}\,dx\\ &=-2\int_0^1\frac{\ln x}{1-x}\,dx. \end{aligned}

1

Use the recurrence formula $$\int_{0}^{\pi/2}\sin^n(x)dx = \frac{n-1}n\int_{0}^{\pi/2}\sin^{n-2}(x)dx.$$

2

Comments: Your method seems to be a bit too complicated. It's possible that part of the reason you seem to be confused is that you're writing $D^2$ for two different things: One, the disc of radius $1$ inside $\mathbb{R}^2$; this is the $D^2$ that appears in your statement of the problem and defintions etc. However, in the attempt you write up, you also ...

0

Differentiation with respect to $\theta$ will give $$\psi'(\theta)=\frac{\int_{-\infty}^{\infty}\Delta\exp{\{\Delta\theta-f(\nu)\Delta^2\}}h(\Delta)\ \mathrm{d} \Delta}{\int_{-\infty}^{\infty}\exp{\{\Delta\theta-f(\nu)\Delta^2\}}h(\Delta) \ \mathrm{d}\Delta}.$$

1

$$Q(Q^{-1} (x)) = x \implies 1 = Q'(Q^{-1}(x)) \cdot (Q^{-1}(x))'$$ So, as $Q'(x) = - \frac{1}{2\pi} e^{- \frac{x^2}{2} }$, if we denote $Q^{-1}(x)$ by $\xi(x)$ (for simplicity), we have $$- 2 \pi e^{\frac{ \xi^2 (x) }{2}} = \xi ' (x) .$$ This looks like a rather challenging problem, as it involves solving a nonlinear ordinary differential equation of ...

1

the characteristics are given by $$\frac{dx}{dt} = x, \frac{dy}{dt} = y, x(0) = 1, y(0) = b$$ which gives $$x = e^t, y = be^t, b = \frac yx, t=\ln x.$$ along the characteristics we have $$\frac{du}{dt} = u(1+b)e^t, u = 1, t=0 \to \int_1^u\frac{du}{u} = (1+b)\int_0^t e^t\, dt$$ integrating we get $$\ln u=(1+b)(e^t-1). u(x, y) = ... 0 we will use the facts$$\cosh x + \sinh x = e^x, \cosh x - \sinh x = e^{-x} $$repeatedly.$$\begin{align}2\cosh(x+y)&=e^{x+y} + e^{-x -y}\\ &= e^xe^y + e^{-x}e^{-y}\\ &=(\cosh x + \sinh x)(\cosh y + \sinh y) +(\cosh x - \sinh x)(\cosh y - \sinh y)\\ &= 2\left(\cosh x\cosh y+\sinh x \sinh y\right)\end{align}$$dividing out by 2 gives ... 1 Since$$ {\rm erfi}(x)=\frac{2}{\sqrt{\pi}} e^{x^2} D(x), \qquad D(x)= e^{-x^2} \int_0^{x} e^{t^2} \, dt, $${\rm erfi}(x) will be very much larger in magnitude than D(x) for x of large magnitude (in fact, D(x)\sim 1/(2x) as |x|\to\infty, {\rm erfi}(x)\sim e^{x^2}/(\sqrt{\pi} x) as |x|\to\infty.) So, for many large x, the value of D(x) ... 6 You can also do this:$$\begin{align*}\Gamma(t) &= \int_{x=0}^\infty x^{t-1} e^{-x} \, dx \\ &= \int_{x=0}^1 x^{t-1} e^{-x} \, dx + \int_{x=1}^\infty x^{t-1} e^{-x} \, dx \\ &= \int_{x=0}^1 x^{t-1} e^{-x} \, dx + \int_{u=1}^0 (u^{-1})^{t-1} e^{-1/u} (-u^{-2}) \, du \\ &= \int_{x=0}^1 x^{t-1} e^{-x} + x^{-t-1} e^{-1/x} \, dx, \end{align*}$$... 10 Yes, although you probably won't like this much: consider$$ \int_0^1 2x \left(-2\log{x}\right)^{n-1} \, dx. $$Because x (\log{(1/x)})^n \to 0 as x \to 0, the integrand is continuous and bounded on (0,1]. Hence this is a proper integral. Now, set x^2 = e^{-t}, so$$ 2x \, dx = - e^{-t} \, dt, and  -2\log{x} = -2\log{(e^{-t/2})} = t . Then the ... 3 We have \begin{align} I & =\int_0^{2\pi}e^{\cos(x)}\cos(nx-\sin(x))dx = \int_0^{2\pi} e^{\cos(x)}e^{inx-i\sin(x)}dx = \int_0^{2\pi} e^{inx}e^{\cos(x)-i\sin(x)}dx\\ & = \int_0^{2\pi}e^{inx}e^{e^{-ix}}dx \end{align} Setting z=e^{ix}, we have dz = ie^{ix}dx = izdx. Hence, \begin{align} I & = \oint_{\vert z \vert=1} z^n e^{1/z} = \oint_{\vert z ... 0 From2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) = \dot{r}^2-v_0^2,$$set r = 1/u, (this is the most important thing to remember for problems with 1/r^2 forces) then$$ \dot{r} = -\frac{\dot{u}}{u^2},$$so$$2GM\left(u_0-u\right) = \frac{\dot{u}^2}{u^2} + v_0^2. $$Then, rearranging,$$ 1 = \frac{\dot{u}^2}{u^2(v_0^2+2GM(u_0-u))} $$Now you just have to ... 2$$\ \int_{r_0}^{r}\sqrt{\frac{r_0r}{2GMr_0-2GMr+v_0^2r_0r}}dr=\ =\sqrt{\frac{r_0}{v_0^2r_0-2GM}}\int_{r_0}^{r}\sqrt{\frac{r}{\frac{2GMr_0}{v_0^2r_0-2GM}+r}}dr$$Where I assumed that \ v_0^2r_0>2GM. Now set$$\ \beta=\sqrt{\frac{r_0}{v_0^2r_0-2GM}}\ \alpha=\frac{2GMr_0}{v_0^2r_0-2GM}$$So you need to deal with this kind of integral:$$\ ...

1

Whenever I see $\displaystyle\frac{g'(x)}{g(x)}$, my immediate instinct is to write that as $[\ln g(x)]'$. Here it helps, as follows: $f'(x) = [\,\displaystyle\frac{1}{x} + (\ln g)'\,]\,f(x) - c\,(\ln g)'$ $f' = \displaystyle\frac{f}{x} + (f-c)\,(\ln g)'$ Now define a new function $u(x) = f(x) - c$. Then, $u' = \displaystyle\frac{u + c}{x} + (\ln g)'\,u$ ...

2

$$\dfrac1{x^2-x-1} = \dfrac1{(x-1/2)^2-(\sqrt5/2)^2} = \dfrac1{(x-a)(x-b)} = \dfrac1{a-b}\left(\dfrac1{x-a} - \dfrac1{x-b} \right)$$ where $a=\dfrac{1+\sqrt5}2$ and $b = \dfrac{1-\sqrt5}2$.

0

Suppose we're on $[-1,1].$ Then $$\frac{1}{n}\sum_{k=1}^{n}[f(\frac{k}{n})+ f(-\frac{k}{n})] = 2\cdot\frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n}).$$ On the left side we have Riemann sums converging to $\int_{-1}^1 f,$ on the right we have two times Riemann sums converging to $\int_0^1f.$ Thus $\int_{-1}^1 f= 2\int_0^1f.$

1

Hint: From what you got on that calculator, use the following identities to simplify it: $\sin(x)=2\sin(x/2)\cos(x/2)$ $1+\cos(x)=2\cos^2(x/2)$ You'll see that the results are same.

0

Its very likely that the expression was simplified further from $\tan(x/2)$ as that expression has many identities, hence why its revered for its integral substitution properties. Also, your answer is correct. Your method is very brute force, but if you know what your doing, your not overkilling the problem that much.

0

In your first expression, $n$ depends on $x$. So for more clarity, I'd rewrite it as $$dn(x)=-n(x) N \pi \left(r_1+r_2\right)^2\ dx$$ Also note that $N, \pi, r_1$ and $r_2$ are constants. So now we have $$\frac{1}{n(x)}\ dn(x)=-N \pi \left(r_1+r_2\right)^2\ dx$$ $$\int\frac{1}{n(x)}\ dn(x)=-N \pi \left(r_1+r_2\right)^2\int dx$$ $$... 2 A Finding the area of an area divided by two functions has three steps. Finding the right and left edges of the shapes, find the area between the upper line of the shape and the x-axis, and finding the area too much in the previous problem. For the left and right edges of the shape, what you want is an intersection of the functions, so simply put them ... 1 It is a Fresnel integral. The limit exists since$$ I(b)=\int_{0}^{b}\sin t^2\,dt = \frac{1}{2}\int_{0}^{b^2}\frac{\sin x}{\sqrt{x}}\,dx$$converges by Dirichlet's test (integral version), because \sin x is a function with a bounded primitive and \frac{1}{\sqrt{x}} is a monotonic function converging to zero as x\to +\infty. To compute it, we may use ... 1 If you divide the square along its diagonal and use x=1\implies r=\sec\theta and y=1\implies r=\csc\theta, you get \displaystyle\int_0^{\frac{\pi}{4}}\int_0^{\sec\theta}r\cos\theta dr d\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_0^{\csc\theta}r\cos\theta dr d\theta \hspace{.3 in}=\displaystyle\int_0^{\frac{\pi}{4}}\frac{1}{2}\sec\theta ... 0 You want to show that f=0 a.e. on E right? Since you already have f \geq 0, just showing that \mu(\{x\in E: f(x) >0 \}) = 0 would suffice. Now$$\{x\in E: f(x) >0 \} = \bigcup_{n\in\mathbb{N}} \{x\in E: f(x) \geq \frac{1}{n} \}. What is $\mu ( \{x\in E: f(x) \geq \frac{1}{n} \})$ for all $n$? Use Markov's inequality. So what does that ...

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