Question

Find the area bounded by the curves y^2=4x and y=2x-4.

Answer 1

Curve $y^2 = 4x$ is parabola and equation $y=2x-4$ represent the straight line.

These two curves will intersect at $(4,4) , (1,-2)$ .

To find the intersection point, simply solve the equation, $(2x-4)^2 = 4x$

$4x^2+16-16x=4x$

$\implies x^2-5x+4=0$

$(x-1)(x-4)=0$

$x = 1, 4$

so $y=-2, 4$

Area under the curve is given by,

$A = \int_{-2}^{4} |(\frac{y+4}{2}) - \frac{y^2}{4} |dy$

$A = \int_{-2}^{0} |(\frac{y+4}{2}) - \frac{y^2}{4} |dy+ \int_{0}^{4} |(\frac{y+4}{2}) - \frac{y^2}{4} |dy$

$A = | \frac{1}{2}(\frac{y^2}{2}+4y)-\frac{1}{4}\frac{y^3}{3} |_{-2}^{0} + | \frac{1}{2}(\frac{y^2}{2}+4y)-\frac{1}{4}\frac{y^3}{3} |_{0}^{4}$

Putting limits,

$A = | \frac{1}{2}(\frac{4}{2}-8)-\frac{1}{4}\frac{(-8)}{3} | + | \frac{1}{2}(\frac{16}{2}+16)-\frac{1}{4}\frac{64}{3} | = |\frac{14}{6}|+|\frac{20}{3}| = \frac{54}{6}$ square units.