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$$|\int^{b}_{a}f(x)dx| \leq \int^{b}_{a} |f(x)|dx$$

We know for a fact that $a \leq |a|$.

Also, prove using riemann sum.

$$\lim_{n \to \infty} \sum^{n}_{i=1}f(x^*)\Delta x = \int^{b}_{a}f(x)dx$$ Thus, $$|\lim_{n \to \infty} \sum^{n}_{i=1}f(x^*)\Delta x| = |\int^{b}_{a}f(x)dx|$$

By limit laws, we can pull the limit outside.

$$ \lim_{n \to \infty} |\sum^{n}_{i=1}f(x^*)\Delta x| = |\int^{b}_{a}f(x)dx|$$

Thus,

$$ \lim_{n \to \infty} |\sum^{n}_{i=1}f(x^*)\Delta x| = \int^{b}_{a}|f(x)dx|$$

Have I done all the steps right? I want to show that its $\leq$ but I cannot think of a way how?

  • 0
    Triangle inequality – 2017-02-16
  • 0
    Your last step is wrong. Rest are fine. – 2017-02-16
  • 0
    Can you be more specific as to why its wrong? – 2017-02-16

1 Answers 1

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By the triangle inequality we have,

$$|b_1+b_2| \leq |b_1|+|b_2|$$

This implies

$$|b_1+b_2+b_3| \leq |b_1+b_2|+|b_3|$$

$$\leq |b_1|+|b_2|+|b_3|$$

Etc.

It follows that,

$$\sum_{i=1}^{n} |b_i| \geq |\sum_{i=1}^{n} b_i|$$

If you want to more rigorously show this use induction.

Now choose $b_i=f(x^{*}) \Delta x$. Assuming $b>a$, then $\Delta x>0$. Then we have,

$$\sum_{i=1}^{n} |f(x^{*})| \Delta x \geq |\sum_{i=1}^{n} f(x^{*}) \Delta x|$$

In the limit, as $n \to \infty$ then,

$$\int_{a}^{b} |f(x)| dx \geq |\int_{a}^{b} f(x) dx |$$