The area of the region S = ( x , y ) : 3 x 2 ≤ 4 y ≤ 6 x + 24 is______.

We have, S = ( x , y ) : 3 x 2 ≤ 4 y ≤ 6 x + 24 Then, y = 3 x 2 4 4 y = 6 x + 24 Solving both equations, we get - 2 , 3 and 4 , 12 . Required area A = ∫ − 2 4 6 x + 24 4 − 3 x 2 4 d x ⇒ A = 1 4 ∫ − 2 4 6 x + 24 − 3 x 2 d x ⇒ A = 3 4 x 2 + 6 x − x 3 4 − 2 4 ⇒ A = 12 + 24 − 16 - 3 + 12 - 2 ⇒ A = 27 sq . units

Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesEasy2 minPYQ_2021

MathematicsEasynumerical

The area of the region $S = \{(x, y) : 3 x^{2} \leq 4 y \leq 6 x + 24\}$ is______.

Answer:

27.00

Solution:

We have,

$S = \{(x, y) : 3 x^{2} \leq 4 y \leq 6 x + 24\}$

Then,

$y = \frac{3 x^{2}}{4}$

$4 y = 6 x + 24$

Solving both equations, we get $(- 2, 3)$ and $(4, 12)$ .

Required area

$A = \int_{- 2}^{4} (\frac{6 x + 24}{4} - \frac{3 x^{2}}{4}) d x$

$\Rightarrow A = \frac{1}{4} \int_{- 2}^{4} (6 x + 24 - 3 x^{2}) d x$

$\Rightarrow A = {(\frac{3}{4} x^{2} + 6 x - \frac{x^{3}}{4})}_{- 2}^{4}$

$\Rightarrow A = (12 + 24 - 16 - 3 + 12 - 2)$

$\Rightarrow A = 27 sq . units$

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by two curves

⏱ 2mℹ️ Source: PYQ_2021

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