Question

Find the area bounded by the Curve y"2=4x and the line 2x + y = 4​

Answer 1

Step-by-step explanation:

The curves meet where

y² / 4 = x = ( y + 4 ) / 2

=> y² - 2y - 8 = 0

=> ( y - 4 ) ( y + 2 ) = 0

So where y = -2 and where y = 4.

For a "small piece" of area, fix y and then take the horizontal rectangle of height dy between the two curves.  Its width is

( y + 4 ) / 2 - y² / 4  =  ( 8 + 2y - y² ) / 4.

So the area we need is

\begin{gathered}\frac14\int_{-2}^4 (8 + 2y - y^2)\,dy\\= \frac14 \left[ 8y + y^2 - \frac{y^3}{3} \right]_{-2}^4\\= \left(8 + 4 - \frac{16}{3}\right) - \left(-4 + 1 + \frac{2}{3}\right)\\=12 - \frac{16}{3}+3-\frac{2}{3}\\=15 -\frac{18}{3}\\= 15 - 6\\=9\end{gathered}41∫−24(8+2y−y2)dy=41[8y+y2−3y3]−24=(8+4−316)−(−4+1+32)=12−316+3−32=15−318=15−6=9