# Thread: integral calculus

1. ## integral calculus

find the area bounded by the curve y^2=2x+3 and y=x

2. Here's the graph. You should be able to find your limits from it.

3. Hello, alderon!

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

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

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

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

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

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

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

Got it?

4. ## i got it

thanks... now the solution is complete