CBSE Class 12 · 2005 · Area Under CurvesLong Answer

Find the area bounded by the curve

x^{2} = 4 y

and the straight line

4 y - 2 = x

Answer: $\frac{9}{8}$ square units

Solution

The parabola is

y = \frac{x ^{2}}{4}

and the line is

y = \frac{x + 2}{4}

.
For their intersection,

\frac{x ^{2}}{4} = \frac{x + 2}{4} \Rightarrow x^{2} - x - 2 = 0 \Rightarrow (x - 2) (x + 1) = 0

, so

x = - 1, 2

.
Between these values the line lies above the parabola, so

A = \int_{- 1}^{2} (\frac{x + 2}{4} - \frac{x ^{2}}{4}) d x = \frac{1}{4} [\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3}]_{- 1}^{2} .

= \frac{1}{4} [(2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3})] = \frac{1}{4} [\frac{10}{3} + \frac{7}{6}] = \frac{1}{4} \cdot \frac{27}{6} = \frac{9}{8} .

Hence the required area is

\frac{9}{8}

square units.

CBSE Class 12 Board · 2005 · Subjective