Textbook Fifth Chapter

Section I.


1-4: Slightly

5. Note: It is generally proposed that a $1/n$, and then observe
\[
\lim_{n\to \infty} (\frac{n}{n^2+1} +\frac{n}{n^2+2^2}+\cdots + \frac{n}{n^2+n^2})
=\lim_{n\to \infty} (\frac{1}{1+1/n^2} +\frac{1}{1+ (2/n) ^2} +\cdots +\frac{1}{1+ (n/n) ^2}) \frac{1}{n}
\]
By definition of definite integral, the $\delta x_i =\frac1n$, $\xi_i=i/n$, and $f (\XI) =\frac{1}{1+ (\xi_i) ^2}$, i.e.
\[
\mbox{native}= \int_0^1 \frac{1}{1+x^2} dx.
\]



6. The integral mean value theorem is exactly like the exercise book. Slightly




7. Contradiction: Suppose $f (x) \not \equiv 0$, that is, the presence $x _0\in (A, A, b) makes $f (X_0) >0$. By the guaranteed number of continuous functions can be obtained, the existence of $\delta>0$ makes when $x \in[x_0-\delta, x_0+\delta]$, $f (x) >0$, so according to the continuous function closed interval must be obtained minimum value is introduced, when $x \in [x_0-\ delta,x_0+\delta]$ $f (x) >m>0$, i.e.
$\int_{x_0-\delta}^{x_0 +\delta} f (x) dx> 2m\delta >0$, and because $f (x) \geq 0$, so $\int_a^{x_0-\delta} f (x) Dx\geq 0$ and $\ int_{x_0+\delta}^ b F (x) Dx\geq 0$, so
\[
\int_a^b f (x) dx = \int_a^{x_0-\delta} f (x) dx + \int_{x_0-\delta}^{x_0+\delta} f (x) dx +\int_{x_0+\delta}^ b F (x) dx>0 ,
\]
With the problem set contradiction.


section II

1. Note that the derivation of the integral upper limit function and the principle of derivative of complex function


2.

(1) Basic integration table, the original function is $\arcsin (\frac x2) $


(2) slightly


(3) Direct application of triangular formula, $\cot^2 x = \csc^2 x-1$. If you forget, then the unfamiliar structure is written sine cosine, it is easy to see.
\[
\cot^2 x =\frac{\cos^2 x}{\sin^2 x} = \frac{1-\sin^2 x}{\sin^2 x} =\csc^2 x-1.
\]


(4) Notice the absolute value and the symmetry of the $\sin$ function on the geometric $\pi$
\[
\INT_0^{2\PI} \sin x dx =2\int_0^\pi \sin x dx =-2\cos x\bigg|_0^\pi =4.
\]


(5) Piecewise function to definite integral, type of courseware example



3. It is noted that the integral value is zero when the upper and lower bounds of the definite integral are equal, so it is the limit of the $0/0$ type, the direct use of the law of Lofa, paying special attention to the problem of complex function in the derivation of integral upper limit function


4. When $0\leq X\leq 1$,
\[
\phi (x) =\int_0^x t^2 dt = \frac13 x^3.
\]
And when $1< x \leq 2$,
\[
\phi (x) = \int_0^1 t^2 dt +\int_1^x t dt= \frac12 x^2-\FRAC16.
\]
So
\[
\phi (x) = \begin{cases}
\frac13 x^3, & 0 \leq x \leq 1,
\\
\FRAC12 x^2-\FRAC16, & 1<x\leq 2.
\end{cases}
\]
When $ < x <1$ and $1<x<2$, it is easy to see $\phi (x) $ continuous. The continuity of the $x =1$ is discussed below,
\[
\lim_{x\to 1^-} \phi (x) = \lim_{x\to 1^-} \frac13 x^3 =\frac13,
\]
and
\[
\lim_{x\to 1^+} \phi (x) = \lim_{x\to 1^+} (\frac12 x^2-\frac16) =\frac13,
\]
So $\phi (x) $ is continuous at the $x =1$.
(Note that the segment point must be defined)


5. Proof: As for any $x _0\in (A, b) $
\[
F ' (x_0) = \frac{f (x_0) (x_0-a)-\int_0^{x_0} f (t) dt} {(x_0-a) ^2},
\]
According to the mean value theorem of integrals there are
\[
\INT_A^{X_0} f (t) dt = f (\xi) (x_0-a), \qquad \xi\in (A,X_0).
\]
And because $f ' (x) \leq 0$, so $f (X_0) \leq F (\xi) $, so
\[
F (x_0) (x_0-a)-F (\xi) (x_0-a) \leq 0
\]
Certificate.



6. (1)
\[
F ' (x) = f (x) +\frac{1}{f (x)} \geq 2 \sqrt{f (x)} \frac{1}{\sqrt{f (x)}}=2
\]
(Description: $a ^2+b^2 \geq 2ab$)



(2) as $f (x) >0$
\[
F (a) = \int_b^a \frac{1}{f (t)}dt <0
\]
And
\[
F (b) = \int_a^b f (t) DT >0,
\]
According to the 0-point theorem, there is a $\xi \in (A, b) $ to make the $F (\XI) =0$. There is only one, assuming there are two roots, namely $F (\xi_1) =f (\xi_2) =0$, according to the Lowe theorem exists $\xi_3\in (\xi_1,\xi_2) $ to make $F ' (\xi_3) =0$, which with the conclusion of (1) $F ' (x) \geq 2$ contradiction.



7. Make
\[
\varphi (x) = (a+x) \int_a^x F (t) Dt-2 \int_a^x t F (t) DT,
\]
The
\[
\varphi ' (x) = \int_a^x f (t) dt-f (x) (X-A),
\]
According to the mean value theorem of integrals, existence of $\xi \in (a,x) $ makes
\[
\varphi ' (x) = f (\xi) (x-a)-F (x) (X-A) = (f (\xi)-f (x)) (x-a) <0
\]
So $\varphi (x) $ is monotonically incremented on $ (a, a), so $\varphi (b) >\varphi (a) =0$, which is the certificate.

Textbook Fifth Chapter