JEE Main 2022 — Definite Integration Question with Solution

For $$\left|x+1\right|$$ critical point, x + 1 = 0 $$\Rightarrow$$ x = $$-$$1

For $$\left|x+2\right|$$ critical point, x + 2 = 0 $$\Rightarrow$$ x = $$-$$2

For $$\left|x+3\right|$$ critical point, x + 3 = 0 $$\Rightarrow$$ x = $$-$$3

For $$\left|x+4\right|$$ critical point, x + 4 = 0 $$\Rightarrow$$ x = $$-$$4

For $$\left|x+5\right|$$ critical point, x + 5 = 0 $$\Rightarrow$$ x = $$-$$5

Here maximum function is represent by the dotted line.

$$\therefore$$ Point of intersection A of line y = $$-$$x $$-$$1 and y = x + 5 :

$$-x-1=x+5$$

$$\Rightarrow 2x=-6$$

$$\Rightarrow x=-3$$

$$\therefore$$ $$y=-3+5=2$$

$$\therefore$$ Point $$A=(-3,2)$$

$$\therefore$$ $${\int }_{-6}^{0}f(x)dx$$

$$={\int }_{-6}^{-3}(-x-1)dx+{\int }_{-3}^0(x+5)dx$$

$$={\left(-\frac{x^2}{2}-x\right)}_{-6}^{-3}+{\left[\frac{x^2}{2}+5x\right]}_{-3}^0$$

$$=\left[\left(-\frac{9}{2}+3\right)-\left(-\frac{36}{2}+6\right)\right]+\left[0-\left(\frac{9}{2}-15\right)\right]$$

$$=\left(-\frac{3}{2}+12\right)+\frac{21}{2}$$

$$=+\frac{21}{2}+\frac{21}{2}$$

$$=21$$