Let $f:[-3,1] \rightarrow R$ be given as $f(x)= \begin{cases}\min \{(x+6), & \left.x^{2}\right\}, & -3 \leq x < 0\\ \max \{\sqrt{x}, & \left.x^{2}\right\}, & 0 \leq x \leq 1\end{cases}$ If the area bounded by $y=f(x)$ and $x$-axis is $A$, then the value of $6 A$ is equal to ____.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Area Under Curves

Let

f : [- 3, 1] \to R

be given as

f (x) = {min {(x + 6), max {x, x^{2}}, x^{2}}, - 3 \leq x < 0 0 \leq x \leq 1

If the area bounded by

y = f (x)

and

x

-axis is

A

, then the value of

6 A

is equal to ____.

Answer: 41

[- 3, 0)

the graphs of

x + 6

and

x^{2}

meet where

x^{2} - x - 6 = 0

, that is at

x = - 2

within this range. At

x = - 3

the values are

3

and

9

, so the smaller is

x + 6

[- 3, - 2]

, and on

(- 2, 0)

the smaller is

x^{2}

.
On

[0, 1]

the inequality

x \geq x^{2}

holds, so the larger is

x

.
The function is non-negative throughout, so the area is

A = \int_{- 3}^{- 2} (x + 6) d x + \int_{- 2}^{0} x^{2} d x + \int_{0}^{1} x d x = \frac{7}{2} + \frac{8}{3} + \frac{2}{3} = \frac{41}{6}

and

6 A = 41

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer