# Area between a curve, x axis and a line

I need to calculate the area of the plane between the curve

$y=4\sqrt{x}$, the $x$ axis and the line $y=2x + 2$.

Also, I need to calculate the area using vertical rectangle and another time using horizontal rectangle.

I would like to know how I can start this exercice as I am not sure what is that "rectangle method".

The vertical rectangle means integration with respect to $\mathrm{d}x$, and the horizontal rectangle means integration with respect to $\mathrm{d}y$.

$$A_{\text{vertical rectangle}}=\int^1_0\left(2x+2-4\sqrt{x}\right)\mathrm{d}x+\int^0_{-1}\left(2x+2\right)\mathrm{d}x\\ A_{\text{horizontal rectangle}}=\int^4_0\left(\dfrac{y^2}{16}-\left(\dfrac{y}{2}-1\right)\right)\mathrm{d}y$$

• Thanks for figuring out what integral I need to do, but I'm not sure to understand how you got there... I understand the vertical rectangle, but how you got the formula of the horizontal one ? and also how do you get [a, b] ? – student Apr 27 '14 at 17:20
• @Jean-FrançoisSavard The area between $f(x),g(x)$ on the interval $(a,b)$ with $f(x)>g(x)$ is given by $\int^b_a(f(x)-g(x))\mathrm{d}x$. For the vertical rectangles, see that the area is piecewise defined. – user122283 Apr 27 '14 at 17:21
• I'm still not sure how you get to the horizontal one.. And also why is it from 0 to 4 – student Apr 27 '14 at 17:29
• Nevermind, just figured out wasn't sure what "picewise" meant as english is not my main language but now I understand and your explanation was actually great, thanks. – student Apr 27 '14 at 17:43