The area of the region S- x-y-3x2- 4y- 6x-24 is

3x^24≤ y≤3x2+6 Point of intersection for two curves y=3x^24 and y=3x2+6 are ( 2, 3) and (4, 12) Area of shaded region ∫ _ 2^4 ( 3x2+6 3x^24 )dx. =3x^24+6x x^34 ...

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Standard XII

Mathematics

Greatest Integer Function

The area of t...

Question

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

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Solution

$\frac{3 x^{2}}{4} \leq y \leq \frac{3 x}{2} + 6$
Point of intersection for two curves
$y = \frac{3 x^{2}}{4}$ and $y = \frac{3 x}{2} + 6$ are $(- 2, 3)$ and $(4, 12)$
Area of shaded region
$4 \int - 2 (\frac{3 x}{2} + 6 - \frac{3 x^{2}}{4}) d x {\begin{matrix} = \frac{3 x^{2}}{4} + 6 x - \frac{x^{3}}{4} \end{matrix} ∣ ∣ ∣}_{- 2}^{4}$
$= 20 - (- 7) = 27$

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The area of the region S={(x,y):3x2≤4y≤6x+24} is

3x24≤y≤3x2+6 Point of intersection for two curves y=3x24 and y=3x2+6 are (−2,3) and (4,12) Area of shaded region 4∫−2(3x2+6−3x24)dx=3x24+6x−x34∣∣∣4−2 =20−(−7)=27

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