# 英国代考|MATH0004 Analysis 2数学分析2

1. If $f$ is Riemann-integrable and $c \in \mathbb{R}$ then $c f$ is also Riemannintegrable and
$$\int_{a}^{b} c f=c \int_{a}^{b} f$$
{Hint: $\S$ 9.2, Exercise 1.}
2. Prove that if $f$ and $g$ are continuous on $[a, b]$, then
$$\int_{a}^{b}(f+g)=\int_{a}^{b} f+\int_{a}^{b} g$$
{Hint: FTC. For the case of arbitrary Riemann-integrable functions, see 9.8.3.}
3. Corollary $9.4 .8$ has a much easier proof: assuming to the contrary that $f(c)>0$ at some point $\mathrm{c}$, argue that $f(x) \geq \frac{1}{2} f(\mathrm{c})$ on a subinterval of $[a, b]$ and construct a lower sum for $f$ that is $>0$.
4. True or false (explain): If $f \geq 0$ and
$$\int_{a}^{b} f=0$$
then $f$ is Riemann-integrable.

therefore
$$\frac{1}{x}(x-1) \leq \int_{1}^{x} \frac{1}{t} d t \leq 1 \cdot(x-1)$$
by 9.1.8; thus (iv) holds for $x>1$.
Suppose $0<x<1$. For $x \leq t \leq 1$ we have
$$1 \leq \frac{1}{t} \leq \frac{1}{x}$$
therefore
$$1 \cdot(1-x) \leq \int_{x}^{1} \frac{1}{t} d t \leq \frac{1}{x}(1-x)$$

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