Can some explain where the first inequality is derived from?

Question

Suppose that f is Riemann integrable on [a,b]. And f is bounded by M. Then, $$\left|\int_a^b f\right| \le \int_a^b|f| \le M(b-a)$$

Can some explain where the first inequality is derived from?

Answer 1

It's the triangle inequality, applied to the Riemann sums. The absolute value is continuous, so limits are no issue.

Answer 2

We have the pointwise inequality $-|f|\leq f\leq |f|$. Since the integral is monotone this implies:

$$-\int_a^b |f|\leq\int_a^b f\leq \int_a^b |f|$$

Which is equivalent to $\left|\int_a^b f\right|\leq \int_a^b |f|$.

Answer 3

Let $P=\{x_0,x_1,\ldots, x_n\}$ be a partition of $[a,b]$. Then by the usual triangle inequality,

$$\left|\sum_{k=1}^n(x_k-x_{k-1})f(x_k)\right|\le \sum_{k=1}^n(x_k-x_{k-1})\left|f(x_k)\right|.$$

This holds for any partition. Since $f$ is integrable, we can take a sequence of partitions $P_n$ with norm converging to $0$, and it follows from the above that

$$\left|\int_a^b f\right|\le \int_a^b|f|$$

Answer 4

If $|f|$ is Riemann-integrable on $[a,b]$, then: $$ -|f| < f < |f|\\ -\int_{[a,b]}|f|<\int_{[a,b]} f <\int_{[a,b]}|f|\\ \left|\int_{[a,b]}f\right|<\int_{[a,b]}|f| $$