# Tag Info

## Hot answers tagged calculus

7

Rewrite as $$\int(1-\cos^2 x) \cos^2 x \sin x \ dx$$ and let $t=\cos x\Rightarrow dt = -\sin x \ dx$.

7

Recall that $\sin(a-b)=\sin a\cos b -\cos a\sin b$. Let $a=x+h$ and $b=x$. Remark: You used precisely the same approach, except that instead of using $\sin(x+h-x)$, as above, you used $\sin(x+h-h)$.

5

$$x^5-x^4+4x-4=x^4(x-1)+4(x-1)=(x-1)(x^4+4)$$ Again, $$x^4+4=(x^2)^2+2^2=(x^2+2)^2-2\cdot x^2\cdot 2=(x^2+2)^2-(2x)^2$$

5

Hint: Due to parity, $\displaystyle\int_{-\infty}^\infty f(x)~dx=2\cdot\int_0^\infty f(x)~dx.~$ Then, let $t=\dfrac1{1+x^2}$ , and recognize the expression of the beta function in the new integral.

4

Your method is plenty rigorous. If $f(x)\to 1$ as $x\to\infty$, then of course the sequence $f(n)$ also tends to $1$ as $n\to\infty$. Indeed, for any $\epsilon$ there exists an $M$ such that if $x>M$ then $f(x)$ is within $\epsilon$ of $1$. Thus for any $\epsilon$, pick an integer $N>M$, and we will have $f(n)$ within $\epsilon$ of $1$ whenever ...

4

$$\lim_{x\to 0}\frac{\sqrt{36+x}-6}{x}\frac{\sqrt{36+x}+6}{\sqrt{36+x}+6}=$$ $$=\lim_{x\to 0}\frac{36+x-36}{x(\sqrt{36+x}+6)}=\lim_{x\to 0}\frac{1}{\sqrt{36+x}+6}=1/12$$

4

$$\lim_{y\to32}\frac{\sqrt[5]{y^2} - 3\sqrt[5]{y} + 2}{y - 4\sqrt[5]{y^3}}$$ We set $y^{\frac{1}{5}}=x$ When $y \to 32, x \to 2$ So,we have: $$\lim_{x \to 2} \frac{x^2-3x+2}{x^5-4x^3}=\lim_{x \to 2} \frac{(x-1) (x-2)}{x^3(x^2-4)}=\lim_{x \to 2} \frac{(x-1)(x-2)}{x^3(x-2)(x+2)}=\lim_{x \to 2} \frac{x-1}{x^3(x+2)}=\frac{1}{8 \cdot 4}=\frac{1}{32}$$

3

Oh, I think I am late, but decided not to erase it. Let $m = 2l$. We claim that the quantity $$\lfloor mk / n \rfloor \equiv \lfloor 2lk / n \rfloor \pmod 2$$ depends only on $r_{k} = lk \text{ mod } n$. Indeed, write $lk = nq + r_{k}$. Then $$\lfloor 2lk / n \rfloor = \lfloor 2(nq + r) / n \rfloor = 2q + \lfloor 2r_{k} / n \rfloor.$$ So it follows ...

3

There is another way to do this, one that I find highly interesting. Credit must go to @juantheron, whose post in another question first mentioned this. Start by defining $\displaystyle x = \sqrt{y^2+a^2} \implies x^2 = y^2 + a^2 \implies xdx = ydy \implies \frac{dx}{y} = \frac{dy}{x}$ Using the simple algebra of proportions, we can see that ...

3

It seems that you are asking about the interval of convergence of the series $$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\tag{1}$$ The ratio test says that the series $$\sum_{k=0}^\infty a_k\tag{2}$$ converges if there exists an $r\lt1$ and a $k_0$ so that for all $k\ge k_0$, we have $$\left|\frac{a_{k+1}}{a_k}\right|\le r\tag{3}$$ Hint: In $(1)$, we have $$... 3 Hint: Substitite$$\sec (x)= \cosh (u) \tan(x) = \sinh(u)\sec^2 (x) \, \mathrm dx = \cosh (u) \,\mathrm du\mathrm dx = \frac1{\cosh(u)} \, \mathrm du$$Edit: Why was this downvoted? Here is the entire process$$\begin{align} \int \sec^3 x \, \mathrm dx &{}= \int \cosh^2 u\,\mathrm du \\ &= \frac{1}{2}\int ( \cosh 2u +1) ...

3

If you don't mind partial fractions, you can simply rewrite your integral as $$\int\frac{\cos x}{\cos^4x}dx=\int\frac{\cos x}{(1-\sin^2x)^2}dx$$ then substitute $u=\sin x$.

3

A man walks into a restaurant, orders food, eats, then leaves. Did the man have to cook his food? Did he pay? Did he sit down? Did he eat the food he ordered or just eat something else entirely. Your problem is known in Psychology as the interpretation and application of one's schema. http://en.wikipedia.org/wiki/Schema_(psychology) Because people are ...

2

Let me start by giving a unified viewpoint, and then I'll reconcile the other things you heard with it. A vector is an element of a vector space*. Can we consider functions as vectors using this? It's a minor matter to show that for a nonempty set $X$ the set of functions $\{f\mid f:X\to \Bbb R \}$ is a vector space under the operations ...

2

Hint : It will be easy if you didn't write $\cos^2x$ as $1-\sin^2x$. Write the integrand as $\sin x(1-\cos^2x)\cos^2x$ then set $t=\cos x$.

2

We integrate $y''$ to get $y'$. So, $y'=x^3+\frac{1}{2}x^2+c_{1}$, where $c_{1}$ is the constant of integration (to be found). Recall that $y'$ is the "slope function", $m$, of the tangent line. In this case, $m=-4$. That is, $y'(x=1)=-4$. We use this condition to find the value of $c_{1}$: $$y'(1)=1+\frac{1}{2}+c_{1}=-4\quad\Rightarrow\quad ... 2 f is continuous and$$\bullet\lim_{x\to-\infty }f(x)=-\infty\bullet \lim_{x\to\infty }f(x)=+\infty.$$By the intermediate value theorem, f(\mathbb R)=\mathbb R, then f is surjective. An other way to show the injectivity, is to remark that$$f'(x)=5x^4>0$$for all x\in\mathbb R\backslash \{0\}, then the function is strickly increasing on ... 2 Square both equations ans use the Pythagorean identity to eliminate the trig terms. You will get the Cartesian equation of (part of) the unit circle 2 We can use the fact that if a_n>0 for all n\ge1 and the sequence \frac{a_{n+1}}{a_n} converges in [0,\infty], then$$ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} $$(see this question). We have that$$ \lim_{n\to\infty}\frac{n+1}n=1 $$and$$ \lim_{n\to\infty}n^{1/n}=1. $$2 Check out these answers on this site: Find \lim_{n\to\infty} \sqrt[n]{n} \lim_{n\to\infty} \sqrt[n]n = 1 There was another beautiful solution on this very site that used the AM-GM inequality but I can's seem to find it now. The gist of it was to write,$$ (n)^{\frac 1 n} = (\sqrt n \cdot \sqrt n \cdot 1 \cdot 1 ... \cdot 1)^{\frac 1 n } \le ...

2

\begin{align} |\sqrt{x} - \sqrt{y}| = \frac{|x - y|}{|\sqrt{x} + \sqrt{y}|} \le \frac{|x - y|}{|\sqrt{x} - \sqrt{y}|} < \frac{\epsilon^2}{|\sqrt{x} - \sqrt{y}|} \end{align}

2

Let $u = 1/t$, then $du = -\dfrac{dt}{t^2}$, then $I = -\displaystyle \int_{1}^{1/x} \dfrac{e^u}{u^2}du = \displaystyle \int_{1}^{1/x} e^ud\left(1/u\right) = x\cdot e^{1/x} - e - F\left(1/x\right)$ by integration by parts.

2

Yes, it is a mistake, but only the second limit is wrong (it should be $-1$, as you have calculated). Recall that $x \to a^+$ means $x \to a$ from the right. So $x \to \dfrac{1}{\pi}^+$ is the case where $x > \dfrac{1}{\pi}$, so the limit is $0$ (as the author has correctly written) and not $-1$ (as you wrote). The other limit where $x \to ... 2 If there's a root of that equation of the form$1+ib$then: $$(1+ib)^4 - 2(1+ib)^3 -12(1+ib)^2-14(1+ib) + 35 = 0\Longrightarrow \\(1 + 4ib -6b^2 -4ib^3 + b^4) - 2(1 + 3ib - 3b^2 -ib^3) + 12(1 + 2ib - b^2) - 14(1+ib) + 35 =0 \Longrightarrow b^4-12b^2+32 + i(-2b^3+8b) = 0$$ Therefore: $$\begin{cases}b^4 - 12b^2+32 = 0\\-2b^3+8b = 0 \end{cases}$$ From the ... 2 You already know that $$\log(1-x)=-\sum_{k=1}^{\infty} \frac{x^k}{k}=\sum_{k=1}^{\infty} a_kx^k$$ Then, $$a_k=-\frac{1}{k}$$ The ratio test, then, is: $$\biggl|{\frac{a_{k+1}}{a_k}}\biggr|=\frac{\frac{1}{k+1}}{\frac{1}{k}}=\frac{k}{k+1}$$ The convergence radius$R$is given by: $$\lim_{k\rightarrow \infty}\biggl|\frac{a_{k+1}}{a_k}\biggr|=\frac{1}{R}$$ ... 2 Don't let the$(-1)^k$or$(-x)^k = (-1)^kx^k$trouble you. They have the effect of canceling each other out for odd$k$, and besides, for the ratio test, we apply it taking the absolute value of the general term$|a_k|$. $$|a_k| = \frac{(x)^k }{k}$$ $$\frac{a_{k+1}}{a_k} = \frac{\frac{(x)^{k+1}}{k+1}}{\frac{(x)^k }{k}} = \frac{xk}{k+1}$$ 2 First, your claim that$c<b$is simply because$c \in A$, and not because$f(b)>0$. (There may be points$x$such that$f(x) > 0$and$x < c$.) Answering your question: The reason why$b>c+\delta/2$is because otherwise, if$c<b \le c+\delta/2$, then$|b-c|<\delta$, which implies$|f(b)-f(c)|<\varepsilon=|f(c)|/2$. This implies$f(b) ...

2

$$\lim_{n\to\infty }\left(1+\frac{1}{n}\right)^n=\lim_{n\to\infty }e^{n\ln\left(1+\frac{1}{n}\right)}=\lim_{n\to\infty }e^\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}=\lim_{x\to 0}e^{\frac{\ln(1+x)}{x}}.$$ Moreover $$\lim_{x\to 0}\frac{\ln(1+x)}{x}\underset{(*)}{=}\lim_{x\to 0}\frac{1}{1+x}=1$$ Then $$\lim_{x\to 0}e^{\frac{\ln(1+x)}{x}}=e^1=e$$ ...

1

Using Stirling's Asymptotic Approximation, we have \begin{align} \frac1n\log\left(\frac{n!}{n^n}\right) &=\frac1n\left(n\log(n)-n+\frac12\log(2\pi n)-n\log(n)+O\left(\frac1n\right)\right)\\ &=-1+\frac1{2n}\log(2\pi n)+O\left(\frac1{n^2}\right) \end{align} Therefore,  \left(\frac{n!}{n^n}\right)^{1/n}=\frac1e\left(1+\frac{\log(2\pi ...

1

$\bf{My\; Solution::}$ Let $\displaystyle I = \int\frac{1}{\sqrt{x^2-a^2}}dx$ Let $\left(x^2-a^2\right)=y^2\;,$ Then $\displaystyle 2xdx = 2ydy\Rightarrow \frac{dx}{y}=\frac{dy}{x}$ Now Using Ration and Proportion, We Get $\displaystyle \frac{dx}{y}=\frac{dy}{x}=\frac{d(x+y)}{(x+y)}$ So Integral \$\displaystyle I = \int\frac{dx}{y} = ...

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