Integrate the following equations using U-Substitution

[Xetman]

Homework Statement

Integrate √(x²(x+3)) from -3 to 0.

0
∫√(x²(x+3)) dx
-32. The attempt at a solution
Here is what I did:
√(x²(x+3))
= x√(x+3)

Let u=x+3, du=dx, x=u-3
Insert the bounds and change the bounds to: 0 to 3.

x√(x+3)=
(u-3)√(u)
= u^3/2-3u^1/2

Thus we have:
3
∫(u^3/2-3u^1/2) du
0

Finally:

(2u^5/2)/5-(2u^3/2) from 0 to 3.
Thus I get: -4.15.

However the correct answer is positive 4.15.
I know the answer should be positive but what I don't understand is why I'm getting a negative answer. Please help.
Thanks in advance.

 

[SammyS]

Homework Statement

Integrate √(x²(x+3)) from -3 to 0.

0
∫√(x²(x+3)) dx
-3

2. The attempt at a solution
Here is what I did:
√(x²(x+3))
= x√(x+3)

Let u=x+3, du=dx, x=u-3
Insert the bounds and change the bounds to: 0 to 3.

x√(x+3)=
(u-3)√(u)
= u^3/2-3u^1/2

Thus we have:
3
∫(u^3/2-3u^1/2) du
0

Finally:

(2u^5/2)/5-(2u^3/2) from 0 to 3.
Thus I get: -4.15.

However the correct answer is positive 4.15.
I know the answer should be positive but what I don't understand is why I'm getting a negative answer. Please help.
Thanks in advance.

Be careful.

\displaystyle \sqrt{x^2} = |x|

In fact, for x<0, \displaystyle\ \sqrt{x^2} = -x\ .

 

[Dick]

Homework Statement

Integrate √(x²(x+3)) from -3 to 0.

0
∫√(x²(x+3)) dx
-3


2. The attempt at a solution
Here is what I did:
√(x²(x+3))
= x√(x+3)

Let u=x+3, du=dx, x=u-3
Insert the bounds and change the bounds to: 0 to 3.

x√(x+3)=
(u-3)√(u)
= u^3/2-3u^1/2

Thus we have:
3
∫(u^3/2-3u^1/2) du
0

Finally:

(2u^5/2)/5-(2u^3/2) from 0 to 3.
Thus I get: -4.15.

However the correct answer is positive 4.15.
I know the answer should be positive but what I don't understand is why I'm getting a negative answer. Please help.
Thanks in advance.

It's positive because if x is negative then sqrt(x^2)=(-x). Right?

 

[Xetman]

Oh yeah. Silly mistake XD thanks.