The area (in sq. units of the region <math> <mfenced open="{" close="}" separators="|"> <mrow> <mrow> <mo>(</mo> <mi>x</mi> <mo>,</mo> <mi>y</mi> <mo>)</mo> <mo>∈</mo> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>∣</mo> <mn>4</mn> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>≤</mo> <mi>y</mi> <mo>≤</mo> <mn>8</mn> <mi>x</mi> <mo>+</mo> <mn>12</mn> </mrow> </mrow> </mfenced> </math> is:

x = 1 <!- [endif]-><!- [endif]-> 4 x 2 = 8 x + 12 <!- [endif]-><!- [endif]->Area x = - 1,3 <!- [endif]-><!- [endif]-> = ∫ - 1 3 ? ( 8 x + 12 ) - 4 x 2 d x <!- [endif]-><!- [endif]-> = 8 x 2 2 + 12 x - 4 x 3 3 - 1 3 <!- [endif]-><!- [endif]->

The area (in sq. units of the region <math> <mfenced open="{" close="}" separators="|"> <mrow> <mrow> <mo>(</mo> <mi>x</mi> <mo>,</mo> <mi>y</mi> <mo>)</mo> <mo>∈</mo> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>∣</mo> <mn>4</mn> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>≤</mo> <mi>y</mi> <mo>≤</mo> <mn>8</mn> <mi>x</mi> <mo>+</mo> <mn>12</mn> </mrow> </mrow> </mfenced> </math> &nbsp;is:

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Maths Class 12th Maths Application of Integrals

The area (in sq. units of the region $\{(x, y) \in R^{2} ∣ 4 x^{2} \leq y \leq 8 x + 12\}$ is:

Option 1 -

$\frac{124}{3}$

Option 2 -

$\frac{125}{3}$

Option 3 -

$\frac{128}{3}$

Option 4 -

$\frac{127}{3}$

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Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

$x = 1$

$4 x^{2} = 8 x + 12$

Area $x = - 1,3$

$= \int_{- 1}^{3} ? [(8 x + 12) - 4 x^{2}] d x$

$= {[\frac{8 x^{2}}{2} + 12 x - \frac{4 x^{3}}{3}]}_{- 1}^{3}$

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The area (in sq. units of the region $\{(x, y) \in R^{2} ∣ 4 x^{2} \leq y \leq 8 x + 12\}$ is: