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Firstly, we know $f$ is continuous, since $\int_0^x f$ is a continuous function when $f$ is Riemann-integrable, or even Lebesgue-integrable (which I'm assuming is the case since otherwise this doesn't make sense). Now, we then have by the FTC that $f$ is differentiable, and it follows that $f'(x)=f(x)$ for all $x$ also due to the FTC. Since $f(0)=0$, we are ...

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$$y=\frac{x^3}{x^2+1}\iff x^3-yx^2-y=0$$ You finish because the equation being of the third degree, has necessarily a real root.

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The function $\;\tan\frac\pi{n+2}\;$ is monotone descending and positive, and also $\;\lim\limits_{n\to\infty}\tan\frac\pi{n+2}=0\;$ , and since $$\sin\frac{n\pi}3=\begin{cases}\pm\frac{\sqrt3}2\;,&n=1,2\pmod 3\\{}\\0\,,&n=0\pmod 3\end{cases}$$ and its series is bounded: $$\left|\sum_{n=1}^N\sin\frac{n\pi}3\right|=\left|\frac{\sqrt3}2+\frac{\... 6 Suppose f is continuous. The fundamental theorem of calculus tells us that it is actually differentiable, and moreover by differentiating both sides we get$$f'(x) = f(x)$$So f(x) is a function which is its own derivative, and hence is of the form f(x) = ce^{x}. What is the constant c? Simply compute$$f(0) = \int_{0}^{0} f(x) = 0$$to see that ... 6 Show that$$\lim_{x\to\infty}f(x)=\infty\quad\text{and}\quad \lim_{x\to-\infty}f(x)=-\infty.$$Then use the intermediate value theorem. 5 It is known as the Shafer-Fink inequality. Here you may find different proofs and some improvements, too. 5 Definitely by parts, as substitution of x won't get you anywhere. Let u=e^x and dv=e^x in \int udv=uv-\int vdu and we have$$ \int xe^xdx=xe^x-\int e^x dx=xe^x-e^x+c $$As a general tip, you will usually want to use parts if you have an exponential, which doesn't get any nastier as you anti-differentiate, and a polynomial which will disappear ... 5 You can do it by parts$$\int { x{ e }^{ x }dx } =\int { xd{ e }^{ x } } =x{ e }^{ x }-\int { { e }^{ x }dx } ={ e }^{ x }\left( x-1 \right) +C$$4 Yes you can solve it by substitution (which is not trivial in this case) but you can choose:$$u = e^x(x-1) \rightarrow du=xe^xdx\rightarrow dx=\dfrac{du}{xe^x}$$Replacing in the integral, you get:$$\int{xe^x \mathrm dx}= \int{du=u+C=e^x(x-1)}+C$$NOTE the easiest way to solve the integrals in this form (P_1(x)e^x) is by using integration by parts 4 If \lim na_n=0 then clearly a_n\sim \dfrac{a_n}{1+na_n} and the two series \sum a_n and \sum \dfrac{a_n}{1+na_n} have the same nature. Thus \sum \dfrac{a_n}{1+na_n} is divergent. If \lim na_n=\ell with \ell>0 or \ell=+\infty then \frac{a_n}{1+na_n}\sim\frac{k}{n} where k=1 if \ell=+\infty and k=\ell/(1+\ell) otherwise. But \sum\... 4 Since x\left(1-x\right)<1  if x\in\left(0,1\right)  we have$$\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(-x^{2}+x+1\right)}{x}dx=\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\log\left(x\right)x^{k-1}\left(1-x\right)^{k}dx -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\log\left(1-x\right)x^{k-1}\left(1-x\...

4

You may write $(-1)^n$ as $\cos(\pi n)$, so the sequence given by: $$a_n = (-1)^n\sin\left(\frac{\pi}{3}n\right) = \frac{1}{2}\left[\sin\left(\frac{4\pi}{3}n\right)-\sin\left(\frac{2\pi}{3}n\right)\right]$$ has partial sums bounded by $\frac{\sqrt{3}}{2}$ in absolute value and $$\sum_{n\geq 1} a_n \tan\left(\frac{\pi}{n+2}\right)$$ is conditionally ...

3

Directly ratio test: $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{|\alpha-1|(n+1)!}{(n+2)!-(n+1)!+1}\frac{(n+1)!-n!+1}{n!}=$$ $$=|\alpha-1|\frac{\left(n\cdot n!+1\right)(n+1)}{(n+1)(n+1)!+1}=|\alpha-1|\frac{n+\frac1{n!}}{(n+1)+\frac1{n!}}\xrightarrow[n\to\infty]{}|\alpha-1|$$ and the series converges exactly when $\;|\alpha-1|<1\;$

3

Hint: $$\int \frac{x^3t}{(x^2+t^2)^2}dt$$ $$=x^3\int \frac{t}{(x^2+t^2)^2}dt$$ $$u=(x^2+t^2)\Rightarrow du=2tdt$$ $$=x^3\frac{1}{2}\int \frac{1}{u^2}du$$ $$=x^3\frac{1}{2}\int u^{-2}du$$ It's pretty simple from here, you can simply use the power rule, substitute $u$ back in and simplify to have the indefinite integral. From there just compute the bounds.

3

Hint. As suggested by @Claude Leibovici, one may write, as $k \to \infty$, $$\left|\frac{\binom{2(k+1)}{k+1}x^{k+1}}{\binom{2k}{k}x^{k}}\right|=\frac{2(k+1)(2k+1)}{(k+1)^2}|x| \to 4|x|.$$ Then the radius of convergence is $R=\dfrac14$.

3

Let's say $x=\tan \theta$, so that: $$\theta \geq \frac{3\tan \theta}{1+2\sqrt{1+\tan^2 \theta}}$$ $\sqrt{1+\tan^2\theta}=\sec\theta$ because the range of $\arctan x$ always has $\cos \theta > 0$: $$\theta \geq \frac{3\tan \theta}{1+2\sec\theta}$$ Multiply both the numerator and denominator by $\cos \theta$: $$\theta \geq \frac{3\sin\theta}{\cos\theta+2}$$...

3

$f(x)=sin(x)$ in the interval (-10,10). In fact any open interval of length more than $2\pi$ will work.

3

As @Dr.MV said, we can use the comparison test to solve this problem. $\lvert n \rvert < 1 \rightarrow$ In this case, we have $\lvert n^{k^2} \rvert < \lvert n^k \rvert$, so since $\sum_{k=0}^\infty n^k$ converges, so does $\sum_{k=0}^\infty n^{k^2}$. $\lvert n \rvert \geq 1 \rightarrow$ In this case, we have $\lvert n^{k^2} \rvert \geq \lvert n^k \... 2 Pick any point$x_0$and define$ f(x) = x_0 $for all$x$. 2 We know that$\sin t\approx t$for$t\approx 0$. But instead of a hand-wavy$\approx$we need an explicit bound fro$\frac{\sin t}t$here. Since the sine is concave on$[0,\frac \pi6]$(or even on$[0,\pi]$), we see that the straight line segment from$(0,0)$to a point$(a,\sin a)$with$0<a\le \pi$(and in fact for some larger$a$as well) is always ... 2 You should know that$\dfrac d{dx}\ln|x|= \dfrac 1 x$. To see that that is true, first do it piecewise: for$x>0$, and then for$x<0$. If$x<0$then you have$\ln(-x)$, and$-x$is positive, and you use the chain rule. Once you have done this just once, then remember it for use on subsequent occasions such as the exercise that you quote in your ... 2 Directly using Cauchy-Hadamard formula with the$\;n\,$th root: $$\sqrt[k]{\frac{k^2}{3^k}}=\frac{\sqrt[k]{k^2}}3\xrightarrow[k\to\infty]{}\frac13$$ and thus the radius of convergence is$\;R=3\;$...so you were right! (Though your use of equality signs isn't very careful...). But there is only one for each power series, not several. 2 Since the tangent line is parallel to$y=2x+10$then it's of the form$y=2x + d$for some real number$d$. Now plug this into the equation and see when it obtains a double/triple real root. $$2x + d = -3x^3 + 2x - 1 \implies -3x^3 = d+1 \implies x^3=\frac{d+1}{-3}$$ This equation has a real triple root only when RHS is$0$, as otherwise we get two complex ... 2 The general theorem of means (perhaps more popularly known as Young's inequality as the comment to this answer suggests) implies that if$x, y, r, s$are positive with$r + s = 1$then $$x^{r}y^{s} \leq rx + sy\tag{1}$$ Putting$x = a^{p}, y = b^{q}, r = 1/p, s = 1/q$we get $$ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}$$ 2 Hint. One may evaluate$(1)$with the following steps. From the geometric series evaluation $$\sum_{n=2}^{\infty}\sqrt{\frac{(1-\sin x)^{n-2}}{(1+\sin x)^{n+2}}}=\frac{1}{(1+\sin x)^2}\frac1{1-\sqrt{\frac{1-\sin x}{1+\sin x}}},\quad 0<x<\frac{\pi}2,$$ one may write $$I=\int_0^{\pi/2}\frac{1}{(1+\sin x)^2}\frac1{1-\sqrt{\frac{1-\sin x}{1+\sin x}}}... 1 1) Differentiation is definitely the way to go 2) Recall that \int \frac{1}{x+1} dx = \ln|x+1|... Therefore, the absolute value goes away when you differentiation, implying that \frac{d}{dx} \ln|x+1| = \frac{1}{x+1} 1 you can have$$g(x,y)= \begin{pmatrix} 1 & 1 \\ 1 & a\end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}=\begin{pmatrix} x+y \\ x+ay\end{pmatrix}$$The function F can be written,$$F(x,y)=f(x+y, x+ay)$$I guess you can complete the rest by using chain rule. 1 If f is any function that satisfies |f(x)| \le C x^2 for x close to 0, then f'(0)=0. 1 I will update this answer with more details when I have time, but here's something to think about: Plug in \sin |x^2+x| into the definition of the derivative, \lim_{h \to 0} \dfrac {f(x+h)-f(x)}{h}. Look at the limits as h \to 0^{+}, and as h \to 0^{-}. You are interested in it only when x=0, so you can plug that in for x. Now do the same with ... 1 Assuming that differentiating f is legal, and using the Fundamental Theorem of Calculus (FTC),$$f' (x) = f(x)$$Integrating,$$f (x) = f_0 \exp (x)$$We note that$f (0) = 0$. Hence,$f_0 = 0$and, thus,$f (x) = 0$for all$x \in \mathbb R\$.

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