Area of region under 2 curves. Find the area of the region R enclosed by the line $y=2x−1$ and the parabola $y^2=4x+141$
Here is what I have done:


*

*Write each equation in terms of x. ($x=\frac{y+1}{2}$ and $x=\frac{y^2-141}{4}$)

*Find intersections: y= -11 and 13

*Write the integral (I don't quite know).

*Calculate answer.


If someone could check over my step 1 and 2 that would be awesome. Also If someone could do step 3 and 4 (would be awesome).
 A: I checked $1$ and $2$ and it is correct.
Step 3:
If $f(y)>g(y)$ for $x$-values between $[-11,13]$ then the integral is
$\int_{-11}^{13}{\left(f(y)-g(y)\right)}{dy}$
Since $\frac{y+1}{2}>\frac{y^2-141}{4}$ from $[-11,13]$ we have 
$\int_{-11}^{13}\left(\frac{y+1}{2}-\frac{y^2-141}{4}\right)dy$
$\frac{1}{4}\int_{-11}^{13}\left({2y+2}-{y^2+141}\right)dy$
Now use the following rule that $\int{jy^k}dy=\frac{j}{k+1}y^{k+1}$
$\frac{1}{4}\int_{-11}^{13}\left(-y^2+2y+143\right)dy$
$\frac{1}{4}\left(\int_{-11}^{13}\left(-y^2\right)dy+\int_{-11}^{13}2ydy+\int_{-11}^{13}143dy\right)$
$\frac{1}{4}\left(\int_{-11}^{13}\left(-y^2\right)+\int_{-11}^{13}2y+\int_{-11}^{13}143\right)$
$\frac{1}{4}\left(|_{-11}^{13}\left(-\frac{1}{3}y^3\right)dy+|_{-11}^{13}{y^2}+|_{-11}^{13}143y\right)$
Now use $\int_{a}^{b}f(x)=F(b)-F(a)$ where $F(x)$ is the indefinite integral.  
$\frac{1}{4}\left(\left(-\frac{1}{3}{(13)}^3+\frac{1}{3}{(-11)}^3\right)+\left({13}^{2}-(-11)^{2}\right)+\left(143(13)-143(-11)\right)\right)$
Now do the calculations on your calculator this ensure you did not make any mistakes.
$\frac{1}{4}(\frac{-2197-1331}{3}+48+3432)$
$\frac{1}{4}(\frac{-2197-1331+10440}{3})$
$\frac{6912}{12}=576$
You are correct.
