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Can anyone prove this? I can understand this intuitively, but can't prove it mathematically. Please help me.

$$ \left|\int_a^b f(x)\,dx\,\right| \le \int_a^b \lvert f(x)\rvert \,dx $$

egreg
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SS_7
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    Hint: $-|f(x)|\le f(x)\le |f(x)|$. But, please, add your thoughts about the problem. This is a main fact about integrals and you should have it in your textbook. – egreg Dec 08 '18 at 12:04
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    Is the claim even true? Consider $f(x)=1$, $a=1$, $b=0$. – Hagen von Eitzen Dec 08 '18 at 12:06
  • @ egreg Thank you. Actually, my textbook put the same hint but I can't understand how to use it... Whenever I solve questions about this(calculating area) I always regard RHS same or bigger than LHS true as RHS means raising the graph to 1 or 2 quadrants and calculating the area. – SS_7 Dec 08 '18 at 12:13
  • Hi Sarah, see the application of Egreg's hint ;) –  Dec 08 '18 at 13:33
  • What is the name of this innequality? – Phosphene May 26 '22 at 19:05

2 Answers2

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Using the hint from the comments sections we can provide a nice proof.

If we assume $f$ is a Riemann-integrable function, this inequality is true, we will provide a proof.

Proof

By properties of absolute value, we get $$-|f(x)|\le f(x)\le |f(x)|.$$ Since $f$ is continuous, we know that $|f|$ is also continuous, hence $|f|$ is Riemann integrable. We can integrate each side of this inequality and we get $$-\int_a^b |f(x)|dx\le \int_a^bf(x)dx\le \int_a^b|f(x)|dx$$ This instantly gives us that: $$\left| \int_a^b f(x)dx\right|\le \int_a^b|f(x)|dx. $$ $\square$

Alternatively, you are considering a continuous version of the triangle inequality. The very definition of an integral is the limit of discrete sums of (Riemann) intervals. To properly prove this from the definition we must go back to the definition of integration: For any Riemann sum we get from the usual triangle inequality for the absolute value:

$$\left|\sum_{k=1}^nf(c_i)(x_i-x_{i-1})\right|\leq\sum_{k=1}^n|f(c_i)|(x_i-x_{i-1})\,\,,\,$$

$$\{a=x_0<x_1<...<x_n=b\}\,\,,\,\,c_i\in[x_{i-1},x_1]$$

Pass now to the limit $\,n\to\infty\,$ while the maximal length of the subintervals goes to zero (this is what is done to get the Riemann integral from Riemann sums). With courtesy of: The triangle inequality for integrals

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    Hello Algebra geek, so is that it? This is the complete proof for method $1$? Where is the proof that $|f|$ is integrable on $[a,b]$? In your proof, you deduced that $|f|$ is integrable because it is continuous and that is sufficient? – CountDOOKU May 08 '21 at 03:45
  • Continuous functions on a closed, bounded interval are riemann integrable. See for instance https://mathcs.org/analysis/reals/integ/proofs/r_cont1.html –  May 08 '21 at 20:48
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Hint: a) If $f$ is a step function, then the inequality follows very easily. b) If $f$ is a regulated function (or a continuous function), use the fact that you can approximate $f$ w.r.t. $\| \cdot \|_\infty$ by a sequence $\{f_n\}$ of step functions. Now use part a).