Intergration problem, absolute value

**Pettzon** · September 7th 2011, 08:32 AM

This is the problem:

$\int_{0}^{2}|x^{2}-x|$

Solution:

$|x^2-x| = x|x-1|$

$\int_{0}^{2}|x^{2}-x| = \int_{0}^{1} x(x-1) + \int_{1}^{2} x(x-1)$

$=/ \\ \int x(x-1) = x^2/2(x-1) - x^3/6 = x^3/2 - x^2/2 - x^3/6 \\ /\\ = [x^3/2 - x^2/2 - x^3/6]_{0}^{1} + [x^3/2 - x^2/2 - x^3/6]^{2}_{1} = 1$

This is the correct answer, however I was wondering why:

$\int_{0}^{1} x(x-1) + \int_{1}^{2} x(x-1) \neq \int_{0}^{1} x^2-x + \int_{1}^{2} x^2-x$

I got $\int_{0}^{1} x^2-x + \int_{1}^{2} x^2-x = 2/3$ , am I right?

**Prove It** · September 7th 2011, 08:34 AM

Originally Posted by Pettzon

This is the problem:

$\int_{0}^{2}|x^{2}-x|$

Solution:

$|x^2-x| = x|x-1|$

$\int_{0}^{2}|x^{2}-x| = \int_{0}^{1} x(x-1) + \int_{1}^{2} x(x-1)$

$=/ \\ \int x(x-1) = x^2/2(x-1) - x^3/6 = x^3/2 - x^2/2 - x^3/6 \\ /\\ = [x^3/2 - x^2/2 - x^3/6]_{0}^{1} + [x^3/2 - x^2/2 - x^3/6]^{2}_{1} = 1$

This is the correct answer, however I was wondering why:

$\int_{0}^{1} x(x-1) + \int_{1}^{2} x(x-1) \neq \int_{0}^{1} x^2-x + \int_{1}^{2} x^2-x$

I got $\int_{0}^{1} x^2-x + \int_{1}^{2} x^2-x = 2/3$ , am I right?

For starters, I think you'll find that $\displaystyle |x^2 - x| = |x||x - 1|$ , not $\displaystyle x|x-1|$ ...

**Plato** · September 7th 2011, 08:37 AM

Originally Posted by Pettzon

This is the problem:

$\int_{0}^{2}|x^{2}-x|$

$\int_{0}^{2}|x^{2}-x|=\int_{0}^{1}(x-x^2)+\int_{1}^{2}(x^2-x)$

**Pettzon** · September 7th 2011, 08:56 AM

Originally Posted by Prove It

For starters, I think you'll find that $\displaystyle |x^2 - x| = |x||x - 1|$ , not $\displaystyle x|x-1|$ ...

$|x^2 - x| = |x(x-1)| = x|x-1|$ ?

Originally Posted by Plato

$\int_{0}^{2}|x^{2}-x|=\int_{0}^{1}(x-x^2)+\int_{1}^{2}(x^2-x)$

Of course, thank you!

**Prove It** · September 7th 2011, 09:32 AM

Originally Posted by Pettzon

$|x^2 - x| = |x(x-1)| = x|x-1|$ ?

Of course, thank you!

Wrong, what if $\displaystyle x < 0$ ? It's impossible for $\displaystyle |x(x - 1)|$ to be negative, but $\displaystyle x|x - 1|$ would be.

The general rule is $\displaystyle |AB| = |A||B|$ , so $\displaystyle |x(x-1)| = |x||x-1|$ .

Anyway, for this integral, you should know that

$\displaystyle \begin{align*}|x^2 - x| &= \begin{cases}x^2 - x\textrm{ if }x^2 - x \geq 0\\x - x^2\textrm{ if }x^2-x<0 \end{cases} \\ &= \begin{cases}x^2 - x\textrm{ if } x \leq 0 \textrm{ or }x \geq 1 \\ x - x^2 \textrm{ if }0 < x < 1\end{cases} \end{align*}$

so that means

$\displaystyle \begin{align*} \int_0^2{|x^2-x|\,dx} &= \int_0^1{|x^2-x|\,dx} + \int_1^2{|x^2-x|\,dx} \\ &= \int_0^1{x - x^2\,dx} + \int_1^2{x^2 - x\,dx} \end{align*}$

**Pettzon** · September 7th 2011, 10:52 AM

Originally Posted by Prove It

For starters, I think you'll find that $\displaystyle |x^2 - x| = |x||x - 1|$ , not $\displaystyle x|x-1|$ ...

Originally Posted by Prove It

Wrong, what if $\displaystyle x < 0$ ? It's impossible for $\displaystyle |x(x - 1)|$ to be negative, but $\displaystyle x|x - 1|$ would be.

The general rule is $\displaystyle |AB| = |A||B|$ , so $\displaystyle |x(x-1)| = |x||x-1|$ .

Anyway, for this integral, you should know that

$\displaystyle \begin{align*}|x^2 - x| &= \begin{cases}x^2 - x\textrm{ if }x^2 - x \geq 0\\x - x^2\textrm{ if }x^2-x<0 \end{cases} \\ &= \begin{cases}x^2 - x\textrm{ if } x \leq 0 \textrm{ or }x \geq 1 \\ x - x^2 \textrm{ if }0 < x < 1\end{cases} \end{align*}$

so that means

$\displaystyle \begin{align*} \int_0^2{|x^2-x|\,dx} &= \int_0^1{|x^2-x|\,dx} + \int_1^2{|x^2-x|\,dx} \\ &= \int_0^1{x - x^2\,dx} + \int_1^2{x^2 - x\,dx} \end{align*}$

Ah, I see. Thanks a bunch!

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