Find the area enclosed by the curves 3x² + 5y = 32 and y = |x - 2| is:

$\frac{33}{2}$

The given curves are parabola $x^2=-\frac{5}{3}(y-\frac{32}{5})$ ... (i)

y = x - 2 for x > 2  ... (ii)

y = - (x - 2) for x < 2 ... (iii)

(i) and (ii) meet at P(3, 1) whereas (i) and (iii) meet at Q (-2, 4) and (ii), (iii) meet at R at (2, 0) as shown in the figure.

The vertex of the parabola is at $(0,\frac{32}{5})$ and is downwards.

Required area = shaded area in the figure

$=\int\limits_{-2}^3y\,dx-ΔQNR-ΔPMR=\int\limits_{-2}^3\frac{1}{2}(32-3x^2)dx-\frac{1}{2}(4.4)-\frac{1}{2}(1.1)=\frac{1}{5}[32x-x^3]_{-2}^3-8-\frac{1}{2}$

$=\frac{1}{5}(32×2-35)-\frac{17}{2}=25-\frac{17}{2}=\frac{33}{2}$ sq. units