JEE PYQ: Area Under Curves Question 3
Question 3 - 2021 (17 Mar Shift 2)
Let $f: [-3, 1] \to \mathbb{R}$ be given as $f(x) = \begin{cases} \min{(x+6), x^2}, & -3 \leq x \leq 0 \ \max{\sqrt{x}, x^2}, & 0 \leq x \leq 1 \end{cases}$. If the area bounded by $y = f(x)$ and $x$-axis is $A$, then the value of $6A$ is equal to ____.
Type: Numerical
Show Answer
Answer: 41
Solution
Area $= \int_{-3}^{-2}(x+6),dx + \int_{-2}^{0}x^2,dx + \int_0^1 \sqrt{x},dx = \frac{41}{6}$. So $6A = 41$.
BETA
