# Tag Info

1

Say that $|f(x)| \le M$ is not satisfied a.e. and pick $\epsilon > 0$ and $\delta > 0$ such that if $G = \{|f(x)| \ge M + \epsilon\}$ then $\mu(G) \ge \delta$. Now notice that $$0 = \lim_{n \to \infty}\mu(\{|f(x) - f_n(x)| \ge \epsilon\}) \ge \lim_{n \to \infty}\mu(\{|f(x)| - |f_n(x)| \ge \epsilon\}) \ge \mu(G) \ge \delta.$$ You can apply the ...

0

The answer will be zero. We can do this problem to ways: First by the Second Fundamental theorem of calculus which states that $\frac{d}{dx} \int_b^xf(t)dt=f(x)$ for your exercise we $$\int_0^ax\sqrt{a^2-x^2}dx =\sqrt{a^2-a^2} =0$$ Secondly we can use $u$- substitution and set up the integral as $$\frac{2}{2}\int_0^a \sqrt{a^2-x^2}dx$$ here you can ...

5

The way the limits in red were changed is simply that the piecewise definition of $H$, stated earlier in the question, says that $H(x)=0$ when $x<0$. Thus $$\int_{-\infty}^0 (\text{anything}\times H(x))\, dx = 0.$$ The idea that $\delta(0)=\infty$ should not be taken too literally. Notice that $$\int_{-\infty}^\infty 3.4\delta(x) f(x)\,dx = 3.4f(0), ... 0 This a simple sum of a geometric series - and using the technique to calculate such sum is well described here: https://www.google.com/?gws_rd=ssl#q=calculate+geometric+series where: a_1 = 1, r=1/2, and n goes to infinity. 1 I only got asymptotic inequality. Numerical exploration tells us that the quantity in question - properly rescaled monotonically approaches the left limit in the inequality. Asymptotics. After expanding e^n=\sum_{k=0}^{n}\frac{n^k}{k!}+\frac1{n!}\int_0^ne^t(n-t)^ndt, making a substitution t=nx in the integral and rearranging the terms, the original ... 1 The proof is quite similar to the proof of part a). We consider a sequence x_n \to x_0 such that x_n \neq x_0 for all n > 0, and the difference quotient$$\frac{F(x_n) - F(x_0)}{x_n - x_0} = \int_Y \underbrace{\frac{f(x_n,y) - f(x_0,y)}{x_n - x_0}}_{g_n(y)}\,d\mu(y).By the mean value theorem, for every n, and every y, there is a \xi_n(y) ... 0 Hint: Your error is that you have used the ''primitive'' of \delta(t-\tau) with respect to t, that is u(t-\tau) but the integral is respect to \tau and the ''primitive'' is -u(t-\tau). 0 In This Answer, I provided a primer on The Dirac Delta. The Dirac Delta is not a function. It is a Generalized Function or Distribution. The "symbol," \int_{-\infty}^{\infty} f(\tau)\,\delta(t-\tau)\,d\tau, is not an integral, although it does share certain properties with the integral. It is, rather, a linear functional that maps a test function f ... 1 Use the chain rule: \begin{align} y & = \int_6^{x^3} \sin^3(5t)\,dt \\[12pt] y & = \int_6^u \sin^3(5t)\,dt \\[6pt] u & = x^3 \\[10pt] \text{Therefore } \frac{dy}{du} & = \sin^3(5u) \\[6pt] \text{and } \frac{du}{dx} & = 3x^2 \\[10pt] \text{so } & \underbrace{\frac{dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx}}_\text{chain rule} = ... 2 The fundamental theorem of calculus: if F(x)=\int_{a}^{x}f(t)dt then F'(x_0)=f(x_0). We have F(x)=\int_{6}^{x^3}\sin ^3 (5t)dt. Let \alpha = x^3. then F(\alpha ^ {\frac{1}{3}})=\int_{6}^{\alpha}\sin ^3 (5t)dt Now derive: (F(\alpha ^ {\frac{1}{3}}) )'=\frac{1}{3}\alpha ^ {\frac{-2}{3}}F'(\alpha ^{\frac{1}{3}})  That was from chain rule. But ... 1 Let us define the function f(\tau)=u(\tau)-u(\tau-4) = \begin{cases} 1 & 0<\tau<4\\ 0 & \tau<0\text{ or }\tau>4. \end{cases} $$(One way of seeing the second equality is that u(\tau) is zero for \tau<0 and one for t>0, and u(\tau-4) is zero for \tau<4 and one for \tau>4. Thus subtracting them, ... 1 Suppose we seek to evaluate$$\int_0^{\pi/2} \frac{1}{4\sin^2(x)+5\cos^2(x)} dx = \frac{1}{4} \int_0^{2\pi} \frac{1}{4\sin^2(x)+5\cos^2(x)} dx.$$Introduce z=\exp(ix) so that dz=iz \; dx to get$$\frac{1}{4}\int_{|z|=1} \frac{1}{4(z-1/z)^2/(-4)+5(z+1/z)^2/4} \frac{dz}{iz} \\ = \int_{|z|=1} \frac{1}{-4(z-1/z)^2+5(z+1/z)^2} \frac{dz}{iz} \\ = \int_{|z|=1} ...

1

Assuming that you integrate over the straight line between $A$ and $B$: First, find a parametrization of the curve. Here, $(x,y)=(t,t)$, $1\leq t\leq 2$ will do. Then $dx=dt$ and $dy=dt$, so your integral, according to the definition of curve integral, transforms into $$\int_1^2 12t^2t+\frac{t}{t}+4t+2t+4t^3+\log_{e} t\,dt.$$ Can you take it from here? ...

1

Another simplest way : From your given equation we have , $$2(y\,dy+x\,dx)=2\sqrt{x^2+y^2}\,dx$$ $$\implies \frac{d(x^2+y^2)}{\sqrt{x^2+y^2}}=2\,dx$$ $$\implies 2\sqrt{x^2+y^2}=2x+2C \implies \sqrt{x^2+y^2}=x+C$$

1

I write this answer just to confirm that your way of working this problem is correct. Edit I just noted that the question asks for $a\in\mathbf R$ and not $a>0$ (as I first read). Note, however, that the integral is even, so you actually only have to consider $a>0$. The final result is that the integral is convergent if and only if $|a|>1$. This ...

1

Note that $f(x) = g'(x)$, and $g(x)>0$ for $x>1$ since $f>0$, so $g(x)\le[f(x)]^2 \implies \frac{g'(x)}{\sqrt{g(x)}}\ge 1$ for $x>1$. Integrating from $1$ to $x$ gives $$2\left(\sqrt{g(x)} - \sqrt{g(1)}\right)\ge x-1 \implies \frac{1}{2}(x-1)\le\sqrt{g(x)}\le f(x).$$

0

This is a homogeneous equation so your substitution will turn into a separable equation which you can solve. It's probably easiest to proceed by rewriting the equation as $${ dy \over dx} = {{\sqrt{x^2 + y^2} - x} \over y}$$ Both the numerator and the denominator of the right-hand side are homogeneous in $(x,y)$ of the same degree, so your substitution $y = ... 0 Better than either way would be to estimate the integral from$0$to$\frac12, using two equal partitions as before. The reason is that the errors arise from omitting the segments of the circle cut off by the slanting sides at the top of the trapezia. The greater the slant (as with the sides on the right), the longer the side, and so the greater the omitted ... 3 You have a mistake. The composition is $$\psi (x) = \left( {F \circ g} \right)(x)$$ and hence you have \eqalign{ & \psi '(x) = F'(g(x))g'(x) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = g'(x)f(g(x)) \cr} 2 Your function is\psi(x)=F(g(x))-F(a)$Where$F(x)$is a primitive of$f(x)$so the derivative is: $$\psi'(x)=F'(g(x))g'(x)=f(g(x))g'(x)$$ -1 You did the composition wrong! The integral is$F\circ g(x)$and not$g\circ F(x)$. 0 I suggest that either they use some other sense of integration (see edit below), or there is an error. As far as I can see, the integral (forgetting different scaling constants) $$\int_0^{+\infty}w\cos(w)\coth(w)\,dw$$ is divergent. One argument, that could be done more rigorous, is that $$\coth(w)\approx 1$$ where the$\approx$is really exponentially ... -1 i have got this here $$x- \left( 1/2\, \left( y \left( x \right) \right) ^{2}+{\it \_C1} \right) {{\rm e}^{y \left( x \right) }}=0$$ 1 Your rearrangement is incorrect. You have $$\frac{dx}{dy} = \frac{1}{dy/dx} = x+ye^y,$$ and then you can now integrate in the usual way: $$x'-x=ye^y \\ (e^{-y}x)'=y \\ e^{-y}x = \frac{1}{2}y^2+A,$$ and so $$x = \frac{1}{2}y^2e^{y}+Ae^{y},$$ which cannot be solved for$y$in terms of elementary functions. 0 This is how it works: $$\left\{ \matrix{ x = \sin (u) \hfill \cr x = 1\, \to \,\,\,\,\,\,\,u = {\pi \over 2} \hfill \cr x = - 1 \to u = - {\pi \over 2} \hfill \cr} \right.$$ and also $$\sqrt {1 - {x^2}} = \sqrt {1 - {{\sin }^2}(u)} = \sqrt {{{\cos }^2}(u)} = \left| {\cos (u)} \right| = \cos (u)$$ The last equality holds since$ - {\pi ...

2

\begin{align*} \int_{-1}^{1}\sqrt{1-x^2}\,dx &= 2\int_{0}^{1}\sqrt{1-x^2}\,dx \\&= 2\int_{0}^{\pi/2}\cos^2\theta\,d\theta\\&=\int_{0}^{\pi/2}(\cos(2\theta)+1)\,d\theta\\&=\int_{0}^{\pi/2}1\,d\theta=\color{red}{\frac{\pi}{2}}\end{align*} as expected. We exploited the parity of the function $\sqrt{1-x^2}$, the substitution $x=\sin\theta$, ...

2

Hint $$\int_{-1}^1\sqrt{1-x^2}dx=\int_{\arcsin(-1)}^{\arcsin(1)}\sqrt{1-\sin^2(x)}\cos(x)dx=...$$

2

The equation $$y'=1+x+y+xy$$ can be put in the "standard" form: $$\frac{dy}{dx}-(x+1)y=1+x$$ One integrant factor for the ODE is $$\mu (x)=\exp\left[-\int (1+x)dx\right]$$

1

HINT: we have $$y'=1+x+y+xy=1+x+y(1+x)=(1+x)(1+y)$$ from here we get $$\frac{dy}{dx}=(1+x)(1+y)$$ and $$\frac{dy}{1+y}=(1+x)dx$$ if $$y\ne -1$$

1

$$y' = 1 + x + y + xy = 1 + x + (1+x) y\Rightarrow y' - (1+x) y = 1+x.$$ Now you can use integration factor with $P(x) = -(1+x)$ and $Q(x) = 1+x$.

0

The Green-Gauss theorem states $$\int\!\!\!\int\limits_A {\left( {{{\partial Q} \over {\partial x}} - {{\partial P} \over {\partial y}}} \right)da} = \int\limits_{\partial A} {Pdx + Qdy}$$ Choose $Q=0$. Then you have $$\int\!\!\!\int\limits_A { - {{\partial P} \over {\partial y}}da} = \int\limits_{\partial A} {Pdx}$$ Now in order to relate this to ...

3

By replacing $x$ with $\arctan t$, $$I = \int_{0}^{\pi/2}\frac{dx}{4\sin^2 x+5\cos^2 x} = \int_{0}^{+\infty}\frac{dt}{4t^2+5}=\color{red}{\frac{\pi}{4\sqrt{5}}}.$$

1

Notice, $$\int_{0}^{\pi/2}\frac{dx}{4\sin^2x+5\cos^2 x}$$ $$=\int_{0}^{\pi/2}\frac{dx}{\cos^2 x\left(4\frac{\sin^2x}{\cos^2 x}+5\right)}$$ $$=\int_{0}^{\pi/2}\frac{\sec^2 x\ dx}{5+4\tan^2 x}$$ $$=\frac{1}{4}\int_{0}^{\pi/2}\frac{d(\tan x)}{\left(\frac{\sqrt 5}{2}\right)^2+(\tan x)^2}$$ $$=\frac{1}{4}\frac{2}{\sqrt 5}\left[\tan^{-1}\left(\frac{2\tan x}{\sqrt ... 2 Let$$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{1}{4\sin^2 x+5\cos^2 x}dx$$Now Divide both \bf{N_{r}} and \bf{D_{r}} by \cos^2 x\;, We get$$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{\sec^2 x}{4\tan^2 x+5}dx\;,$$Now Put \displaystyle 2\tan x=t\;, Then \displaystyle 2\sec^2 xdx =dt\Rightarrow \sec^2 xdx = \frac{1}{2}dt and Changing ... 1 Hint: 4\sin^2x+5\cos^2x=4+\cos^2x 1$$\int_\alpha^{T+\alpha-\beta}\cos\left(\frac{2\pi k}{T}x\right)dx=\frac{1}{\frac{2\pi k}{T}}\sin\left(\frac{2\pi k}{T}x\right)\vert_0^{T-\beta}, \ \ \ \alpha\in\mathbb R$$3 I guess you need to assume \alpha \ne \beta. Then if one defines$$g(x)=\alpha \int_{a}^{x}f(x)dx+\beta \int_{x}^{b}f(x) dx$$then g(c)=0 for all c \in [a,b]. Thus g is a constant and by fundamental theorem of calculus g is differentiable. So g'(x)=0 for all x. Thus (\alpha-\beta)(f(x))=0 and we are done 2 Hint: Differentiate your equation with respect to c and use the fundamental theorem of calculus. For \alpha -\beta \neq 0, you will then see the claim to be true. 4 If \alpha \neq \beta, and we differentiate with respect to c, we get (\alpha-\beta)f(c) = 0, from which we get f(c) = 0, hence f = 0 on [a,b]. If \alpha = \beta, all we can conclude is that \alpha \int_a^b f(x) dx = 0, and so either \alpha = 0, or f has zero average over [a,b]. 0 Let y = x-n \Rightarrow dx = dy \Rightarrow \displaystyle \int_{n}^{n+1} \dfrac{\sin x}{x} dx = \displaystyle \int_{0}^1 \dfrac{\sin(y+n)}{y+n}dy. We treat this as Lebesgue integral, and the function \dfrac{\sin(y+n)}{y+n} is dominated by 1 which is integrable over [0,1] because \sin(y+n) \leq 1 \leq y+n, y \geq 0 \Rightarrow \displaystyle ... 4 There are a couple of answers already but I'll provide the simple one that was alluded to in comments. For all x\in[n,n+1], |\frac{\sin x}x|\le\frac1n and so |\int_n^{n+1}\frac{\sin x}x\,\mathrm dx|\le\frac1n\to0. 0 If you permit special functions, \int \frac{\sin x}{x}dx = \text{Si}(x). Further, there is the well known identity \displaystyle\lim_{x \to \infty} \text{Si}(x) = \frac{\pi}{2}. From this, we get$$\lim_{n\rightarrow\infty} \int_n^{n+1}\frac{\sin x} x \, dx = \lim_{n \to \infty} [\text{Si}(n+1)-\text{Si}(n)]$$We can see that the growth of n in the ... 0 \int e^{-3 x}\cos 6 xdx = e^{-3x} \cdot \frac{\sin6x}{6} - (-e^{-3x} \cdot - \frac{\cos6x}{6} - \int -3e^{-3x} \cdot - \frac{\cos6x}{6}dx) = e^{-3x} \frac{\sin6x}{6} - e^{-3x} \frac{\cos6x}{6} + \int e^{-3x} \frac{\cos6x}{2}dx then subtract \int e^{-3x} \frac{cos6x}{2}dx from both sides to get: \frac{1}{2}\int e^{-3x}cos6xdx= e^{-3x} \frac{sin6x}{6} ... 0 HINT: Let$$I = \int e^{-3x}\cos6x\ \mathrm{d}x$$Notice that integral is in your last step. Replace it with an I and then solve for I. 0 Clever complex analysis solves this easily, see if you follow: \cos x=\Re(e^{ix}) We'll use this fact and change the integrand to e^{3x}e^{ix}. We'll get,$$\int e^{(3+i)x}dx=\frac{e^{(3+i)x}}{3+i}=(\frac{3}{10}-\frac{i}{10})e^{3x}e^{ix}=(\frac{3}{10}-\frac{i}{10})e^{3x}(\cos x+ i \sin x)=\frac{3}{10}e^{3x}(\cos x +i \sin x)-\frac{i}{10}e^{3x}(\cos x ...

0

I know that the Riemann integral can't handle integrals like $\int_{\mathbb{Q}}1dx$, because you can always choose a mesh which results in the points selected in each interval of the partition either being 1 or 0. On the other hand, since the rationals are countable, they are a null set w.r.t. Lebesgue measure, so the integral is 0. In other words, the ...

1

Your application of Green’s Theorem is justified. You can think of $r$ and $\theta$ as the labels of axes in a different Cartesian plane. You have to be a little careful about $\mathcal C$ and $\mathcal R$ with this point of view, though—they need to be replaced by their preimages under the polar-to-Cartesian map.

6

This is based on the following formula for the harmonic numbers: $$H_n = \int_0^1 \frac{1-t^{n}}{1-t} \, dt$$ Why is this true? In the integer case, we can just remember the identity $$1-t^{n} = (1-t)(1+t+t^2+\dotsb+t^{n-1});$$ then the integral is $$\int_0^1 \sum_{k=1}^{n} t^{k-1} \, dt = \sum_{k=1}^n \int_0^1 t^{k-1} \, dt = \sum_{k=1}^n \frac{1}{k} = ... 1 To answer your second question: Start from \frac{f(x_0)-f(x_0t)}{1-t} and replace f with its series representation. A little algebra yields$$\frac{\sum a_n x_0^n-\sum a_n x_0^nt^n}{1-t} =\frac{\sum a_n x_0^n(1-t^n)}{1-t} =\sum a_n x_0^n(1+t+\dots+t^{n-1}).$$You can integrate the last expression term-by-term in [0,1] to get$$\sum a_n ...

3

My attempt at (2): \int_0^1 \frac{f(x_0)-f(t)}{1-t}dt = \int_0^1 \frac{\sum_{n=0}^\infty a_n x_0^n - \sum_{n=0}^\infty a_n (x_0t)^n}{1-t}dt = \int_0^1 \frac{\sum_{n=0}^\infty a_n x_0^n( 1 - t^n) }{1-t}dt = \int_0^1 \sum_{n=0}^\infty a_n x_0^n\sum_{j=0}^{n-1}t^j dt = \sum_{n=0}^\infty a_n x_0^n\sum_{j=0}^{n-1} \int_0^1t^j dt = \sum_{n=0}^\infty a_n ...

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