Question

The area enclosed by the curves 3x² + 5y = 32 and y = | x – 2 | is –

• A. $\frac{17}{2}$
• B. $\frac{33}{2}$
• C. $\frac{23}{2}$
• D. None of these

Answer 1

Answer: (B)

$\frac{33}{2}$

Answer 2

3x² + 5y = 32

̃ x² = – $\frac{- 5}{3}$ $\left( y - \frac{32}{5} \right)$ y = |x – 2| = $\left\{ \begin{matrix} x - 2, & if\mspace{6mu} x > 2 \\ 2 - x, & if\mspace{6mu} x < 2 \end{matrix} \right.\$

̃ $\left\{ \begin{matrix} \frac{x}{2} + \frac{y}{- 2} = 1 \\ \frac{x}{2} + \frac{y}{2} = 1 \end{matrix} \right.\$

Here C (–2, 4), D(3, 1), A (2, 0)

E(3, 0), B (–2, 0)

Required area = $\int_{- 2}^{3}{ydx}$ – DABC – DADE

=$\int_{- 2}^{3}\frac{1}{5}$ (32 – 3x²) dx –$\frac{1}{2}$ (BC × AB) –$\frac{1}{2}$ (DE × AE)

= $\frac{1}{5}$ $(32x - x^{3})_{x = - 2}^{3}$– $\frac{1}{2}$ (4 × 4) – $\frac{1}{2}$ (1 × 1)

= $\frac{1}{5}$ (96 – 27 + 64 – 8) – $\frac{17}{2}$ = $\frac{33}{2}$