Area Under Curves — CBSE Class 12 · 2009 Board Question with Solution

Prove that the curves

y^{2} = 4 x

and

x^{2} = 4 y

divide the area of the square bounded by

x = 0

x = 4

y = 4

and

y = 0

into three equal parts.

Answer: Each part $= \frac{16}{3}$ square units; hence the three parts are equal.

Solution

The two parabolas

y^{2} = 4 x

(i.e.

y = 2 x

) and

x^{2} = 4 y

(i.e.

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

) cross at

(0, 0)

and

(4, 4)

, dividing the square

0 \leq x \leq 4, 0 \leq y \leq 4

into three regions I (above

y = 2 x

), II (between the curves) and III (below

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

I = \int_{0}^{4} (4 - 2 x) d x = 16 - \frac{4}{3} \cdot 8 = \frac{16}{3} .

II = \int_{0}^{4} (2 x - \frac{x ^{2}}{4}) d x = \frac{4}{3} \cdot 8 - \frac{64}{12} = \frac{128 - 64}{12} = \frac{16}{3} .

III = \int_{0}^{4} \frac{x ^{2}}{4} d x = \frac{64}{12} = \frac{16}{3} .

Since

I = II = III = \frac{16}{3}

square units, the two curves divide the square into three equal parts. Hence proved.

CBSE Class 12 Board · 2009 · Subjective