if $f(x)\leq M$ for all $x\in [a,b]$ then $\int_a^b f(x)dx\leq M(b-a)$

Question

I am having trouble with the following problem.

Let $f$ be Riemann integrable functions on $[a,b]$. Show that if $f(x)\leq M$ for all $x\in [a,b]$ then $$\int_a^b f(x)dx\leq M(b-a)$$ Any suggestions or answer would be great help!

Answer 1

Let $P\in\mathcal{P}_{[a,b]}$, i.e., a partition of the interval $[a,b]$. We know that, $f(x)\leq M$.

Take a lower sum. In this way, the infimum of $f$, called $m_{{k}_f}$, (the infimum is in the interval $[x_{k1},x_k]$) meets the next inequality (this inequality holds for every $k$) $$m_{{k}_{f}}\leq M $$Then $$m_{{k}_{f}}(x_k-x_{k-1})\leq (x_k-x_{k-1})M $$ Taking the sum $$\displaystyle\sum_{k=1}^{n}m_{{k}_{f}}(x_k-x_{k_1})\leq\displaystyle\sum_{k=1}^{n}M(x_k-x_{k-1})=M(b-a) \ \text{because $x_0=a$ and $x_n=b$}$$Taking limit when $n\rightarrow \infty$ then, because $f$ is integrable $$\displaystyle\int_{a}^{b}f(x) \, dx=\lim\limits_{n\rightarrow \infty} \displaystyle\sum_{k=1}^{n}m_{{k}_{f}}(x_k-x_{k-1})\leq\lim\limits_{n\rightarrow \infty}M(b-a)=M(b-a)$$Thus, $$\displaystyle\int_{a}^{b}f(x) \, dx\leq M(b-a) $$