Let the area of the region { ( x , y ) : 0 ? x ? 3 , 0 ? y ? min { x 2 + 2 , 2 x + 2 } } be A . Then 12A is equal to [2024]

Question

Let the area of the region ${(x, y) : 0 \leq x \leq 3, 0 \leq y \leq \min {x^{2} + 2, 2 x + 2}}$ be $A .$ Then 12A is equal to ___________ . [2024]

Smart Achievers · Accepted Answer

(164)

Required area, $A = \int_{0}^{2} (x^{2} + 2) d x + \int_{2}^{3} (2 x + 2) d x$

$= {[\frac{x^{3}}{3} + 2 x]}_{0}^{2} + {[2 \frac{x^{2}}{2} + 2 x]}_{2}^{3}$

$\Rightarrow A = \frac{8}{3} + 4 + (9 + 6 - 4 - 4) = \frac{41}{3}$

$∴$ $12 A = 164 sq. units$

Solution

Q.
Let the area of the region ${(x, y) : 0 \leq x \leq 3, 0 \leq y \leq \min {x^{2} + 2, 2 x + 2}}$ be $A .$ Then 12A is equal to ___________ . [2024]

Ans.
(164)

Required area, $A = \int_{0}^{2} (x^{2} + 2) d x + \int_{2}^{3} (2 x + 2) d x$

$= {[\frac{x^{3}}{3} + 2 x]}_{0}^{2} + {[2 \frac{x^{2}}{2} + 2 x]}_{2}^{3}$

$\Rightarrow A = \frac{8}{3} + 4 + (9 + 6 - 4 - 4) = \frac{41}{3}$

$∴$ $12 A = 164 sq. units$

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Solution

Q. Let the area of the region {(x,y):0≤x≤3,0≤y≤min{x2+2,2x+2}} be A. Then 12A is equal to ___________ . [2024]

Ans. (164) Required area, A=∫02(x2+2) dx+∫23(2x+2) dx =[x33+2x]02+[2x22+2x]23 ⇒A=83+4+(9+6-4-4)=413 ∴ 12A=164 sq. units

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Q.
Let the area of the region ${(x, y) : 0 \leq x \leq 3, 0 \leq y \leq \min {x^{2} + 2, 2 x + 2}}$ be $A .$ Then 12A is equal to ___________ . [2024]

Ans.
(164)

Required area, $A = \int_{0}^{2} (x^{2} + 2) d x + \int_{2}^{3} (2 x + 2) d x$

$= {[\frac{x^{3}}{3} + 2 x]}_{0}^{2} + {[2 \frac{x^{2}}{2} + 2 x]}_{2}^{3}$

$\Rightarrow A = \frac{8}{3} + 4 + (9 + 6 - 4 - 4) = \frac{41}{3}$

$∴$ $12 A = 164 sq. units$