Thread: Area between 2 curves (well...3)

1. Area between 2 curves (well...3)

$y = 4x^2$
$y = 8x^2$
$2x+y=6$

x > or equal to 0

I have found the intersection points of 4x^2 and 2x+y=6 to be at (-1.5,9) and (1,4) and the intersection points of 8x^2 and 2x+y=6 to be at (-1,4) and (.75,4.5)

I set up my integral to be from -1.5 to 1 $8x^2-4x^2$ = 35/6

I also tried setting up my integral from 0 to 1 $8x^2-4x^2$ = 4/3

But...this is wrong...can anyone help plz? thx

2. Hello, silencecloak!

I think you need a better sketch . . .

Find the area of the region enclosed by: . $\begin{array}{c}y \:= \:4x^2 \\y \:= \:8x^2 \\y\:=\:\text{-}2x+6 \\ x \:\geq\:0 \end{array}$

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: $x = \tfrac{3}{4}\,\text{ and }\,x = 1$

The first area is between the two parabolas:

. . $A_1 \;=\;\int^{\frac{3}{4}}_0\bigg[8x^2-4x^2\bigg]\,dx$

The second is between the line and the lower parabola:

. . $A_2 \;=\;\int^1_{\frac{3}{4}}\bigg[(-2x + 6) - 4x^2\bigg]\,dx$

Got it?

3. Originally Posted by silencecloak
$y = 4x^2$
$y = 8x^2$
$2x+y=6$

x > or equal to 0

I have found the intersection points of 4x^2 and 2x+y=6 to be at (-1.5,9) and (1,4) and the intersection points of 8x^2 and 2x+y=6 to be at (-1,4) and (.75,4.5)

I set up my integral to be from -1.5 to 1 $8x^2-4x^2$ = 35/6

I also tried setting up my integral from 0 to 1 $8x^2-4x^2$ = 4/3

But...this is wrong...can anyone help plz? thx
did you note that the problem states that $x \geq 0$ ?

$
A = \int_0^{\frac{3}{4}} 8x^2 - 4x^2 \, dx + \int_{\frac{3}{4}}^1 (6-2x) - 4x^2 \, dx
$