Find the area in the planar bounded by the curves y = x -1 and (y - 1)² = 4(x + 1).

$\frac{64}{3}$

Eliminating y we get the points of intersection as

$(x-2)^2=4(x+1)$

or x² - 8x = 0

∴ x = 0, 8   ∴ y = -1, 7

Hence the points of intersection are (0, –1) and (8, 7). The vertex of the parabola is at the point (–1, 1).

The line cut axes at (1, 0) and (0, –1)

$A=\int_{-1}^{7}(x_1-x_2)dy=\int_{-1}^{7}[(y+1)-\{\frac{(y-1)^2}{4}-1\}]dy=\{\frac{1}{2}y^2+y-\frac{(y-1)^3}{12}+y\}_{-1}^{7}=\frac{64}{3}$