# integral calculus

• Jul 31st 2007, 05:54 PM
alderon
integral calculus
find the area bounded by the curve y^2=2x+3 and y=x
• Jul 31st 2007, 06:12 PM
galactus
Here's the graph. You should be able to find your limits from it.
• Jul 31st 2007, 06:34 PM
Soroban
Hello, alderon!

Quote:

Find the area bounded by the curves: .$\displaystyle \begin{array}{cccc}y^2 & = & 2x+3 & {\color{blue}[1]} \\ y & = & x & {\color{blue}[2]}\end{array}$
We'll do this one "sideways".

Equation $\displaystyle {\color{blue}[1]}$ is: .$\displaystyle x \;=\;\frac{1}{2}y^2 - \frac{3}{2}$

Code:

                |    /                 |    *                 |*::/               *::|::/           *:::::|:/         *:::::::|/     - - *:-:-:-:-*- - - -         *::::::/|           *:::/ |               *  |             /  |*                 |    *

Substitute $\displaystyle {\color{blue}[2]}$ into $\displaystyle {\color{blue}[1]}$: .$\displaystyle y^2 \;=\;2y + 3\quad\Rightarrow\quad y^2 - 2y - 3 \;=\;0$

. . which factors: .$\displaystyle (y + 1)(y - 3) \;=\;0$

. . and has roots: .$\displaystyle y \:=\:\text{-}1,\,3$

The area is: .$\displaystyle A \;=\;\int_{-1}^3\left[y - \left(\frac{1}{2}y^2 - \frac{3}{2}\right)\right]\,dy$

Got it?

• Aug 1st 2007, 08:13 PM
alderon
i got it
thanks... now the solution is complete