Review question two

A short answer, some process omitted, just give ideas and results

One

1.5 (the derivative of the odd function is even function, the previous conclusion of the workbook)


2.1


3. $ (f ' (x) \arctan f (x) +\frac{f (x) F ' (x)}{1+f^2 (x)}) dx$



4. $\frac{\pi}{16}$ (first using parity, get the original $=\frac14 \INT_0^{\PI/2} (\sin 2x) ^2 D (2x) $,

Note that $ \INT_0^{\PI/2} (\sin 2x) ^2 d (2x) = \INT_0^{\PI/2} (\cos 2x) ^2 D (2x) =\frac{\pi}{4}$, the book concludes with an example)



5. $x ^2 +x^2 \ln x+c$




Two

1. D 2. A ($f (X_0) $ is a constant that is differential to a constant of 0) 3. D 4. A




Three.


1.
\[
Y ' = \frac{x}{|x|\sqrt{1-x^2}}.
\]



2.
Take the logarithm on both sides
\[
Y\ln x =x \ln y
\]
Take differential on both sides
\[
Dy \ln x +\frac yx dx =\ln y dx +\frac xy dy
\]
So
\[
dy= \frac{\ln y \frac yx}{\ln x-\frac x y}dx.
\]



3.
\[
\mbox{original}= \lim_{x\to 0} \frac{1-\frac{\sin x}{x}}{1-\cos x} =\lim_{x\to 0}
\frac{x-\sin x}{x (1-\cos x)}=2\lim_{x\to 0}\frac{x-\sin x}{x^3}=2\lim_{x\to 0} \frac{1-\cos x}{3x^2}=\frac13.
\]




4.
\[
\begin{aligned}
\mbox{original}= \int \frac{\sin x^2 \cos x^2}{1+\sin x^2} d (x^2) = \int \frac{\sin x^2}{1+\sin x^2}d (\sin x^2)
\\= \int (1-\frac 1{1+\sin x^2}) d (\sin x^2) =\sin x^2-\ln|\sin x^2|+c.
\end{aligned}
\]




5. According to the problem set, the curve equation on both sides of the $x $ derivative
\[
2 e^x + 2\sin y y ' = 0,
\]
The tangent slope of the curve over $ (0,\frac \PI 3) $ is
\[
Y ' \bigg|_{x=0,y=\frac \pi 3}=-\frac{2\sqrt 3}{3},
\]
The slope of the line perpendicular to it is $\frac{\sqrt 3}{2}$, that is, the line is
\[
y= \frac{\sqrt 3}{2} (x+1) +1.
\]



6. =\sqrt{x+1}$ the $t
\[
\mbox{Original}= 2\int_1^2 (t^2-1) ^2 DT =\frac{76}{15}.
\]




7. Note $x that y$ is a parametric equation about $t $, based on the calculation formula of arc length
\[
s= \int_{1}^{\frac \pi 2} \sqrt{x ' (t) ^2 +y ' (t) ^2} dt = \int_1^{\frac \pi 2} \frac{1}{t} dt =\ln \frac \pi 2.
\]



Four.

1. When the $x <1$, $1-x>0$, consider
\[
\varphi (x) =e^x (1-x).
\]
Since $\varphi ' (x) =0$ introduced $x =0$, it is easy to verify $\varphi ' (0) <0$, so $x =0$ is the maximum point. Also noted that
\[
\lim_{x\to-\infty} \varphi (x) =0.
\]
Therefore, according to the nature of the limit, there is $M >0$, so that when $x <-m$ $|\varphi (x) |<\frac12$. Also notice that $\varphi (1) =0$, and $\varphi (x) $ is $ (-\infty, 1]$ 's continuous function.
Therefore, depending on the nature of the continuous function on the closed interval, there is a maximum value on the $[m,1]$, at which point the maximum value is $\varphi (0) =1$, and when the $x <-m$, $|\varphi (x) |<\frac12$. Sum up
\[
\varphi (x) =e^x (1-x) \leq 1, \qquad x<1,
\]
Also because $1-x>0$, that is
\[
E^x \leq \frac{1}{1-x}.
\]



2. Consider the function
\[
\varphi (x) =\int_1^x F (t) DT,
\]
At the $x =1$, Taylor expands to get
\[
\varphi (x) = f (1) (x-1) + \frac{1}{2} f ' (1) (x-1) ^2 + \frac{1}{6} f ' (\xi) (x-1) ^3.
\]
So that $x =0$ and $x =2$ get
\[
\varphi (0) =-F (1) + \frac{1}{2} f ' (1)-\frac{1}{6} f ' (\xi_1)
\]
And
\[
\varphi (2) =f (1) +\frac F ' (1) +\frac16 F ' (\xi_2).
\]
The above two-type subtraction
\[
\FRAC16 (f "(\xi_2) +f" (\xi_1)) = \varphi (2)-\phi (0)-2 f (1) = \int_0^2 f (t) dt-2f (1).
\]
Suppose $f ' (\xi_2) >f ' (\xi_1) $, then
\[
\frac13 F ' (\xi_1) < \FRAC16 (f ' (\xi_2) +f "(\xi_1)) =\int_0^2 F (t) dt-2f (1) < \frac13 F ' (\xi_2),
\]
Since the $f ' (x) $ is continuous according to the question, the existence of $\xi\in (\xi_1,\xi_2) $ makes it possible to obtain the value of a continuous function
\[
\frac13 F ' (\xi) =\int_0^2 F (t) dt-2f (1).
\]
If the $f "(\xi_2) =f" (\xi_1) $, then $\xi_1$ or $\xi_2$ is the request.



Five. According to test instructions
\[
V (\xi) = \pi \int_0^\xi \frac{x}{(1+x^2) ^2} dx=-\frac12 \frac{\pi}{1+x^2} \bigg|_{0}^\xi =\frac \pi 2-\frac{\pi}{2 (1+\ xi^2)}.
\]
and
\[
\FRAC12 \lim_{\xi \to +\infty} =\frac \pi 4.
\]
Solution $V (a) =\frac \pi 4$ $a =1$.

Review question two