Find the area of the region between the curves x=4-y² and x=-3y
I keep getting 56/3 but that doesnt match any other answer. can someone help me out?
First, I think it'll be much easier to represent both as functions of x: $\displaystyle y=4-x^2\,,\,y=-3x$
These two funtions intersect at $\displaystyle x=-1,\,4\Longrightarrow$ the wanted area is the integral:
$\displaystyle \int\limits^4_{-1}(4-x^2+3x)dx$ (why? Chech that the parabola lies above the line in the integration interval...)
I got as answer $\displaystyle \frac{125}{6}=20.833333$
Tonio