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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Area Under Curves

Let the area of the bounded region \ (x, y): 0 ≤ 9 x ≤ y^2, y ≥ 3...

Let the area of the bounded region {(x,y):0≤9x≤y2,y≥3x−6}\left\{(x, y): 0 \leq 9 x \leq y^2, y \geq 3 x-6\right\} be A. Then 6 A is equal to ________

Correct answer

0

Step-by-step solution

0≤9x≤y2&y≥3x−60 \leq 9 x \leq y^2 \& y \geq 3 x-6A= Required Area =[∫01(−3x)dx−∫01(3x−6)dx]A=−3(x3232)∣01−(3x22−6x)∣01A=−2[1−0][32−6]A=−2−32+6=52 Sq. unit ∴6A=6×52=15\begin{aligned} & A=\text{ Required Area }=\left[\int_0^1(-3 \sqrt{x}) d x-\int_0^1(3 x-6) d x\right] \\ & A=-\left.3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)\right|_0 ^1-\left.\left(\frac{3 x^2}{2}-6 x\right)\right|_0 ^1 \\ & A=-2[1-0]\left[\frac{3}{2}-6\right] \\ & A=-2-\frac{3}{2}+6=\frac{5}{2} \text{ Sq. unit } \\ & \therefore 6 A=6 \times \frac{5}{2}=15\end{aligned}$$
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