Hello, r_maths!
Did you find the limits of integration?
I sketched the curves.
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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
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**
From the graph, we know they intersect at x = -½
Then assuming (2x + 1) ≠ 0, we can divide by it.