1. ## Integration

2. Originally Posted by r_maths
Hello,

all your considerations are OK.

To integrate the square root you have to use integration by substitution.

I've attached an image with my calculations:

3. Hello, r_maths!

Did you find the limits of integration?

I sketched the curves.
. . . . . . . . . . . . . . . . . . . ______
The upper curve is: .y .= .√2x + 1
. . This is the upper half of a parabola with vertex (-½,0), opening to the right.

The lower curve is: .y .= .(2x + 1)(x - 2)/6
. . This is an up-opening parabola with x-intercepts: -½ and 2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _____
To find their intersections: .(2x + 1)(x - 2)/6 .= .√2x + 1

Square both sides: .(2x + 1)²(x - 2)²/36 .= .2x + 1

Divide by (2x + 1): . (2x + 1)(x - 2)²/36 .= .1 .**

. . which simplifies to the cubic: .2x³ - 7x² + 4x - 32 .= .0

. . which factors: .(x - 4)(2x² + x + 8) .= .0

. . and has one real root: .x .= .4

Therefore, the limits are: .x = -½ and x = 4

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
From the graph, we know they intersect at x = -½
Then assuming (2x + 1) ≠ 0, we can divide by it.

4. The limits were kindly given from the question.