CBSE Class 12 · 2013 · Area Under CurvesLong Answer

Using integration, find the area bounded by the curve

x^{2} = 4 y

and the line

x = 4 y - 2

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

Solution

The parabola is

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

and the line is

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

.
They meet where

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

, i.e. at

(- 1, \frac{1}{4})

and

(2, 1)

.
Between these points the line lies above the parabola:

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} \cdot \frac{27}{6} = \frac{9}{8} .

Hence the required area is

\frac{9}{8}

square units.

CBSE Class 12 Board · 2013 · Subjective