Three Review questions

One.

1. $ (\ln 5 \cdot 5^x +e x^{e-1}) dx$


2. $e ^{x+1}$; Consider the following limits
\[
F (x+1) = \lim_{t\to +\infty} \exp \frac{\ln\frac{t+x}{t-2}}{t} = \exp (\lim_{t\to +\infty} t{\ln (1+\frac{x+2}{t-2})}) =
\exp (\lim_{t\to +\infty} {\frac{(x+2) t}{t-2}}) = E^{x+2}.
\]

3. Can go


4. $\frac {5\pi} 2$ (geometric meaning)



5. $\FRAC12 \ln 2$ (integral equals $\FRAC12 \lim\limits_{b\to +\infty} \ln\frac{x^2}{1+x^2}\bigg|_1^b$, notice that the entire definite integral is calculated after the Leibniz formula to limit, Instead of each part of the limit, you get the wrong result without convergence)


Two.

1. D 2. C (note that points where the derivative is zero and that the derivative does not exist are possible extreme points, and 0 points are points where the derivative does not exist)

3. B (Note the derivative $f (x) +\frac{1}{f (x)}\geq 2\sqrt{f (x) \frac{1}{f (x)}}=2$, so there is only a unique root)



4. D



Three.

1.
\[
\frac{d x}{d T} = \frac{1}{1+t^2},
\]
And
\[
\FRAC{DY}{DT} = \frac12 \frac{y^2-e^t}{1-ty},
\]
So
\[
\FRAC{DY}{DX} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac12 \frac{(y^2-e^t) (1+t^2)}{1-ty}.
\]




2.
\[
\mbox{Original}=\lim_{x\to 0}\exp \frac{\ln (1-x+\sin x)}{x^2}
= \exp (\lim_{x\to 0} \frac{\ln (1-x+\sin x)}{x^2}) = \exp (\lim_{x\to 0} \frac{\sin x-x}{x^2}) =e^0 =1
\]




3. Because it is a segmented point, by definition
\[
F ' (0) =\lim_{x\to 0} \frac{\frac{\int_0^x (e^{t^2}-1) DT}{x^2}-f (0)}{x}=\lim_{x\to 0} \frac{{\int_0^x (e^{t^2}-1) DT}}{ X^3}
=\lim_{x\to 0} \frac{e^{x^2}-1}{3x^2} =\frac13.
\]




4. Assuming that the $x >2$, $t =\frac 1x$
\[
\mbox{Original}=-\int \frac{t}{\sqrt{1-(2t) ^2}} DT =\frac14 \sqrt{1-(2t) ^2}+c=\frac{\sqrt{x^2-4}}{4x}+c,
\]
Suppose $x <-2$, $t =-x$.
\[
\mbox{original}=-\int \frac{dt}{t^2 \sqrt{t^2-4}}=-\frac{\sqrt{t^2-4}}{4t}+c=\frac{\sqrt{x^2-4}}{4x}+c.
\]


(triangular transformations can be used, but more examples can be found in the fourth chapter of the book, indefinite integral exercises)




5.
\[
\int_0^1 f ' (x) F ' (x) dx =\frac12 F ' (x) ^2\bigg|_0^1 = \FRAC12 (f ' (1) ^2-f ' (0) ^2),
\]
And because
\[
F ' (x) = E^{-X^2/2},
\]
That
\[
\int_0^1 f ' (x) F ' (x) dx = \frac{1-e}{2e}.
\]





6. Note The $0<t<1$, so that the intersection of the $y =t$ with the $y =x^2$ is $ (\sqrt T, T) $. To $x $ as an integral variable, the area is divided into two sections to calculate
\[
A= \int_0^{\sqrt T} (t-x^2) dx +\int_{\sqrt t}^1 (x^2-t) dx =\frac43 T^{3/2}-t+\frac13.
\]
Of course, you can also use $y $ as an integral variable,
\[
a= \int_0^t \sqrt{y} dy + \int_t^1 (1-\sqrt{y}) dy=\frac43 T^{3/2}-t+\frac13.
\]

(Note that the boundary of the graphics area is $x =0,x=1,y=t$)




Four.

1. Consider the function
\[
\varphi (x) = [F (x)-F (a)][g (b)-G (x)]
\]
As well as the use of the Lowe theorem, because of $\varphi (a) =\varphi (b) =0$, there exists $\xi\in (A, a) $ makes
\[
\varphi (\XI) = f ' (\xi) (g (b)-G (\xi)-G ' (\xi) (f (\xi)-F (a)) = 0,
\]
And because $g ' (x) \neq 0$, so $g (b) \neq g (\XI) $, so the upper formula gets the conclusion.



2. Because
\[
\int_0^\pi f (|\cos x|) dx = \int_0^{\frac \pi 2} f (\cos x) dx + \int_{\frac \pi 2}^{\pi} f (-\cos x) dx,
\]
In the last integral of the formula, the $u =\pi-x$
\[
\int_{\frac \pi 2}^{\pi} f (-\cos x) dx =-\int_{\frac \pi 2}^{0} f (-\cos (\pi-u)) du
=\int^{\frac \pi 2}_{0} f (\cos (U)) du= \int_0^{\frac \pi 2} f (\cos x) dx,
\]
So
\[
\FRAC12 \int_0^\pi F (|\cos x|) dx= \int_0^{\frac \pi 2} f (\cos x) dx = \int_0^{\frac \pi 2} f (|\cos x|) dx.
\]




Five.
The intersection of two equations is $ (0,0), (2a-2p, 2\sqrt{p (A-p)}), (2a-2p,-2\sqrt{p (A-P)}) $, the area is
\[
a=2 \int_0^{2a-2p} \sqrt{2p x} DX
= \frac{16}3 P^{1/2} (a-p) ^{3/2}.
\]
The maximum area of the =\frac is slightly $p a4$

Three Review questions