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- #1

- Jan 17, 2013

- 1,667

\(\displaystyle \big | \int^{b}_{a} f(t) \, dt \big | \leq \int^{b}_{a} |f(t)|\, dt\)

1- How to prove the inequality ,what are the conditions ?

2- Does it work for improper integrals ?

- Thread starter ZaidAlyafey
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

\(\displaystyle \big | \int^{b}_{a} f(t) \, dt \big | \leq \int^{b}_{a} |f(t)|\, dt\)

1- How to prove the inequality ,what are the conditions ?

2- Does it work for improper integrals ?

- Feb 13, 2012

- 1,704

If f(*) is Riemann integrable on (a,b) then the integral is...I know we have the following

\(\displaystyle \big | \int^{b}_{a} f(t) \, dt \big | \leq \int^{b}_{a} |f(t)|\, dt\)

1- How to prove the inequality ,what are the conditions ?

$$\int_{a}^{b} f(t)\ dt = \lim_{n \rightarrow \infty\ , \text{max} \Delta t_{i} \rightarrow 0} \sum_{i=0}^{n-1} f(t_{i})\ \Delta t_{i}\ (1)$$

For any finite sum is...

$$ |\sum_{i=0}^{n-1} a_{i}| \le \sum_{i=0}^{n-1} |a_{i}|\ (2)$$

... and that proves the item 1...

Kind regards

$\chi$ $\sigma$

- Thread starter
- #3

- Jan 17, 2013

- 1,667

That is the triangle inequality on the elements of the sequence , right ?$$ |\sum_{i=0}^{n-1} a_{i}| \le \sum_{i=0}^{n-1} |a_{i}|\ (2)$$

- Feb 13, 2012

- 1,704

The so called 'triangle inequality' holds in general for vectors or complex numbers and extablishes that...That is the triangle inequality on the elements of the sequence , right ?

$$ |\sum_{i=1}^{n} a_{i}| \le \sum_{i=1}^{n} |a_{i}|\ (1) $$

See here...

Triangle Inequality -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$

- Admin
- #5

- Jan 26, 2012

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- Jan 31, 2012

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If $f(x)$ and $g(x)$ are integrable on $[a,b]$ (by which I mean Riemann integrable) and $f(x) \le g(x)$ for all $x \in [a,b]$, then $\int_{a}^{b} f(x) \ dx \le \int_{a}^{b} g(x) \ dx$.

Now if $f(x)$ is integrable on $[a,b]$, so is $|f(x)|$.

And $-|f(x)| \le f(x) \le |f(x)|$ for all $ x \in [a,b]$.

So $- \int_{a}^{b} |f(x)| \ dx \le \int_{a}^{b} f(x) \ dx \le \int_{a}^{b} |f(x)| \ dx $.

- Feb 13, 2012

- 1,704

The answer to point 2 is slighly more complex. If we consider an improper integral in (a,b) where a is a singularity of f(*), then we intend...

\(\displaystyle \big | \int^{b}_{a} f(t) \, dt \big | \leq \int^{b}_{a} |f(t)|\, dt\)

1- How to prove the inequality ,what are the conditions ?

2- Does it work for improper integrals ?

$$\int_{a}^{b} f(t)\ dt = \lim_{x \rightarrow a+} \int_{x}^{b} f(t)\ dt\ (1)$$

The problem in such a case is that in can be that $\int_{a}^{b} f(t)\ dt$ converges and $\int_{a}^{b} |f(t)|\ dt$ diverges. An interesting example of such a case is...

$$\int_{0}^{1} \frac{\sin \frac {1}{t}}{t}\ dt\ (2)$$

Kind regards

$\chi$ $\sigma$

- Thread starter
- #8

- Jan 17, 2013

- 1,667

This is just like absolute convergence in series . If the integral absolutely convergent then it is convergent . If it is ''absolutely divergent" then the integral may or may not converge.The answer to point 2 is slighly more complex. If we consider an improper integral in (a,b) where a is a singularity of f(*), then we intend...

$$\int_{a}^{b} f(t)\ dt = \lim_{x \rightarrow a+} \int_{x}^{b} f(t)\ dt\ (1)$$

The problem in such a case is that in can be that $\int_{a}^{b} f(t)\ dt$ converges and $\int_{a}^{b} |f(t)|\ dt$ diverges. An interesting example of such a case is...

$$\int_{0}^{1} \frac{\sin \frac {1}{t}}{t}\ dt\ (2)$$

Kind regards

$\chi$ $\sigma$