Prove that the curves y²= 4x and x²= 4y divide the area of the square bonded by x = 0, x = 4, y = 4, and y = 0 into three equal parts.

Solution

The point of intersection of the

Parabolas y² = 4x and x² = 4y are ( 0, 0 ) and ( 4, 4 )

Now the area of the region OAQBO bounded by curves y² = 4x and x² = 4y,

$\int_{0}^{4} (2 \sqrt{x} - \frac{x^{2}}{4}) dx = {[2 \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{3}}{12}]}_{0}^{4} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} sq . units . . . . . . . . . . . . . . (i)$

Again, the area of the region OPQAO bounded by the curve x² = 4y , x = 0,

x = 4 and the x - axis,

$\int_{0}^{4} \frac{x^{2}}{4} dx = {[\frac{x^{3}}{12}]}_{0}^{4} = (\frac{64}{12}) = \frac{16}{3} sq . units . . . . . . . . . . . . . . (ii)$

Similarly, the area of the region OBQRO bounded by the curve y² = 4x, the y-axis, y = 0 and y = 4

$\int_{0}^{4} \frac{y^{2}}{4} dy = {[\frac{y^{3}}{12}]}_{0}^{4} = \frac{16}{3} sq . units . . . . . . . . . . . . (iii)$

From (i), (ii), and (iii) it is concluded that the area of the region OAQBO =

area of the region OPQAO = area of the region OBQRO, i. e., area bounded

by parabolas y² = 4x and x² = 4y divides the area of the square into

three equal parts.

Some More Questions From Application of Integrals Chapter

A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ` 100 and that on a bracelet is ` 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced.

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Application Of Integrals

Prove that the curves y²= 4x and x²= 4y divide the area of the square bonded by x = 0, x = 4, y = 4, and y = 0 into three equal parts.

Some More Questions From Application of Integrals Chapter

Solve the following L.P.P. graphically :
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0
and x, y ≥ 0

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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For Daily Free Study Material Join wiredfaculty

Application Of Integrals

Prove that the curves y²= 4x and x²= 4y divide the area of the square bonded by x = 0, x = 4, y = 4, and y = 0 into three equal parts.

Some More Questions From Application of Integrals Chapter

Solve the following L.P.P. graphically :Minimise Z = 5x + 10ySubject to x + 2y ≤ 120Constraints x + y ≥ 60 x – 2y ≥ 0and x, y ≥ 0

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Solve the following L.P.P. graphically :
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0
and x, y ≥ 0