# Integrate the following equations using U-Substitution

## 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)
= u3/2-3u1/2

Thus we have:
3
∫(u3/2-3u1/2) du
0

Finally:

(2u5/2)/5-(2u3/2) from 0 to 3.
Thus I get: -4.15.

However the correct answer is positive 4.15.

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SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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)
= u3/2-3u1/2

Thus we have:
3
∫(u3/2-3u1/2) du
0

Finally:

(2u5/2)/5-(2u3/2) from 0 to 3.
Thus I get: -4.15.

However the correct answer is positive 4.15.
Be careful.

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

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

Dick
Homework Helper

## 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)
= u3/2-3u1/2

Thus we have:
3
∫(u3/2-3u1/2) du
0

Finally:

(2u5/2)/5-(2u3/2) from 0 to 3.
Thus I get: -4.15.

However the correct answer is positive 4.15.