Hello, silencecloak!
I think you need a better sketch . . .
Find the area of the region enclosed by: . $\displaystyle \begin{array}{c}y \:= \:4x^2 \\y \:= \:8x^2 \\y\:=\:\text{}2x+6 \\ x \:\geq\:0 \end{array}$
The area is in quadrant 1, between two parabolas and a slanted line. Code:

 \ * y = 8x²
 \
 \ * y = 4x²
 *
 \ *
 *:\ *
 *:::\ *
 *::::::*
 *:::::* \
*++\
 ¾ 1 \ y = 2x+6
We need two integrals . . .
The line intersects the parabolas at: $\displaystyle x = \tfrac{3}{4}\,\text{ and }\,x = 1$
The first area is between the two parabolas:
. . $\displaystyle A_1 \;=\;\int^{\frac{3}{4}}_0\bigg[8x^24x^2\bigg]\,dx$
The second is between the line and the lower parabola:
. . $\displaystyle A_2 \;=\;\int^1_{\frac{3}{4}}\bigg[(2x + 6)  4x^2\bigg]\,dx$
Got it?