# Integral

• Dec 9th 2009, 12:24 AM
math252
Integral
Find the integral of sqrt(x)|2-x| dx from 1 to 4.

What do I do with the absolute value? Is it true that I separate it from integral of sqrt(x)(2-x) from 1 to 2 + integral of sqrt(x)(x-2) from 2 to 4?

If that is true, why does the 2-x change into x-2 for the 2nd part of the integral?

Thanks
• Dec 9th 2009, 12:47 AM
11rdc11
Quote:

Originally Posted by math252
Find the integral of sqrt(x)|2-x| dx from 1 to 4.

What do I do with the absolute value? Is it true that I separate it from integral of sqrt(x)(2-x) from 1 to 2 + integral of sqrt(x)(x-2) from 2 to 4?

If that is true, why does the 2-x change into x-2 for the 2nd part of the integral?

Thanks

Yes you are correct but I believe this integral diverges.
• Dec 9th 2009, 12:49 AM
simplependulum
Quote:

Originally Posted by math252
Find the integral of sqrt(x)|2-x| dx from 1 to 4.

What do I do with the absolute value? Is it true that I separate it from integral of sqrt(x)(2-x) from 1 to 2 + integral of sqrt(x)(x-2) from 2 to 4?

If that is true, why does the 2-x change into x-2 for the 2nd part of the integral?

Thanks

we have $\displaystyle \int \sqrt{x} ( 2-x)~dx$

$\displaystyle = \int 2x^{\frac{1}{2}} - x^{\frac{3}{2}} ~dx$

$\displaystyle = \frac{4}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} + C$

we also have

$\displaystyle (2-x) \geq 0$ for $\displaystyle x \leq 2$

$\displaystyle (2-x) \leq 0$ for $\displaystyle x \geq 2$

the integral

$\displaystyle \int_1^4 \sqrt{x} | 2-x|~dx$

$\displaystyle = (\int_1^2 + \int_2^4 ) \sqrt{x} | 2-x|~dx$

$\displaystyle = \int_1^2 \sqrt{x} ( 2-x) ~dx - \int_2^4 \sqrt{x} ( 2-x)~dx$

$\displaystyle = [\frac{4}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} ]_1^2 - [\frac{4}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} ]_2^4$

$\displaystyle = \frac{32\sqrt{2}}{15} + \frac{6}{5}$
• Dec 9th 2009, 01:48 AM
math252
@ 1st post:
I believe this integral converges...

@ 2nd post:
Aren't you not supposed to ignore the absolute value? the end value I got was, {4/3[x^(3/2)]-2/5[x^(5/2)]} from 1 to 2 PLUS {2/5[x^(5/2)]-4/3[x^(3/2)]} from 2 to 4. I was told it was this way because of the absolute value...can somebody clarify/explain this?
• Dec 9th 2009, 02:45 AM
simplependulum
Quote:

Originally Posted by math252
@ 1st post:
I believe this integral converges...

@ 2nd post:
Aren't you not supposed to ignore the absolute value? the end value I got was, {4/3[x^(3/2)]-2/5[x^(5/2)]} from 1 to 2 PLUS {2/5[x^(5/2)]-4/3[x^(3/2)]} from 2 to 4. I was told it was this way because of the absolute value...can somebody clarify/explain this?

OH , i think there is a step which is very important i have skipped it !

$\displaystyle = (\int_1^2 + \int_2^4 ) \sqrt{x} | 2-x|~dx$
$\displaystyle = \int_1^2 \sqrt{x} | 2-x|~dx + \int_2^4 \sqrt{x} | 2-x|~dx$

$\displaystyle = \int_1^2 \sqrt{x} ( 2-x ) ~dx + \int_2^4 \sqrt{x} [-( 2-x)]~dx$
$\displaystyle = \int_1^2 \sqrt{x} ( 2-x) ~dx - \int_2^4 \sqrt{x} ( 2-x)~dx$
• Dec 9th 2009, 05:33 AM
HallsofIvy
Quote:

Originally Posted by math252
Find the integral of sqrt(x)|2-x| dx from 1 to 4.

What do I do with the absolute value? Is it true that I separate it from integral of sqrt(x)(2-x) from 1 to 2 + integral of sqrt(x)(x-2) from 2 to 4?

If that is true, why does the 2-x change into x-2 for the 2nd part of the integral?

Thanks

Because |x|= x if $\displaystyle x\ge 0$ and |x|= -x if x< 0.
If $\displaystyle x\ge 2$, then $\displaystyle x- 2\ge 0$ and so |x- 2|= x- 2. If x< 2, then x- 2< 0 and so |x- 2|= -(x-2)= 2- x.
• Dec 9th 2009, 05:35 AM
HallsofIvy
Quote:

Originally Posted by 11rdc11
Yes you are correct but I believe this integral diverges.

??? $\displaystyle \sqrt{x}|x-2|$ is continuous and bounded over the finite interval $\displaystyle 1\le x\le 4$. It can't possibly diverge!