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Find the area enclosed by the parabola 4y = 3x^2 and the line

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Find the area enclosed by the parabola $4 y = 3 x^{2}$ and the line $2 y = 3 x + 12$

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Benjamin Discussed

Find the area enclosed by the parabola

4 y = 3 x^{2}

and the line

2 y = 3 x + 12

14 mins ago

Discuss this question LIVE

14 mins ago

Text solutionVerified

The area enclosed between the parabola,

4 y = 3 x^{2}

and the line,

2 y = 3 x + 12

is represented by the shaded are

OB A O

The points of intersection of the given curves are

A (- 2, 3)

and

(4, 12)

.
We draw

A C

and

B D

perpendicular to

x

-axis.

∴ A re a OB A O = A re a C D B A - (A re a O D BO + A re a O A CO)

= \int_{- 2}^{4} \frac{1}{2} (3 x + 12) d x - \int_{- 2}^{4} \frac{3 x ^{2}}{4} d x

= \frac{1}{2} [\frac{3 x ^{2}}{2} + 12 x]_{- 2}^{4} - \frac{3}{4} [\frac{x ^{3}}{3}]_{- 2}^{4}

= \frac{1}{2} [24 + 48 - 6 + 24] - \frac{1}{4} [64 + 8]

= \frac{1}{2} [90] - \frac{1}{4} [72]

= 45 - 18

= 27

sq. units

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Question Text	Find the area enclosed by the parabola $4 y = 3 x^{2}$ and the line $2 y = 3 x + 12$
Answer Type	Text solution:1
Upvotes	150