Definite Integrals Part 2

For |x+1| critical point, x+1=0⇒x=−1
For |x+2| critical point, x+2=0⇒x=−2
For |x+3| critical point, x+3=0⇒x=−3
For |x+4| critical point, x+4=0⇒x=−4
For |x+5| critical point, x+5=0⇒x=−5 Here maximum function is represent by the dotted line.
∴ Point of intersection A of line y=−x−1 and y=x+5 :
−x−1=x+5
⇒2x=−6
⇒x=−3
∴y=−3+5=2
∴‌ Point ‌A=(−3,2)
∴

∫

−6

0f(x)dx
=

−3

∫

−6

(−x−1)dx+

∫

−3

0(x+5)dx
=

−3

(−‌ x2 2 −x)−6

+[‌

x2

2

+5x]−30
=[(−‌

9

2

+3)−(−‌

36

2

+6)]+[0−(‌

9

2

−15)]
=(−‌

3

2

+12)+‌

21

2

=‌

21

2

=21