# Math Help - Integral

1. ## 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

2. 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.

3. 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 $\int \sqrt{x} ( 2-x)~dx$

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

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

we also have

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

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

the integral

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

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

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

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

$= \frac{32\sqrt{2}}{15} + \frac{6}{5}$

4. @ 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?

5. 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 !

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

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

$

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

$

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

6. 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 $x\ge 0$ and |x|= -x if x< 0.
If $x\ge 2$, then $x- 2\ge 0$ and so |x- 2|= x- 2. If x< 2, then x- 2< 0 and so |x- 2|= -(x-2)= 2- x.

7. Originally Posted by 11rdc11
Yes you are correct but I believe this integral diverges.
??? $\sqrt{x}|x-2|$ is continuous and bounded over the finite interval $1\le x\le 4$. It can't possibly diverge!