Question

Compute the integral involving absolute value

Answer 1

\[2 \int\limits\limits_{0}^{3}\left| x^2-\left| x \right| -6\right|dx+\int\limits\limits_{3}^{4}\left| x^2-\left| x \right| -6\right|dx\]

Answer 2

because its an even function right?

Answer 3

i dont know how to deal with the absolute signs..

Answer 4

I'll get you started...drop the |x| in the middle (make it just x)...since \(x\ge0\)

Answer 5

how do u know that ?

Answer 6

look at your limits of integration

Answer 7

ah okay.. hm, then what happens when i drop it?

Answer 8

you get the integrand

\[|x^2-x-6|\]

find the roots of \[x^2-x-6\]

Answer 9

(x+2)(x-3), then?

Answer 10

that tells you what to do with the absolute value

Answer 11

the roots are -2 and 3 (ignore -2)

is \[x^2-x-6\] positive or negative on (0,3)?

Answer 12

negative?

Answer 13

yes..so \[|x^2-x-6|=-(x^2-x-6)\] on (0,3)

Answer 14

oh, so thats how u decide the sign for the absolute value?

Answer 15

\[|x|=\left\{\begin{array}{ll}x & x\ge 0\\ -x& x<0 \\ \end{array}\right.\]

Answer 16

by definition

Answer 17

i see, how about this \[\int\limits_{0}^{4 \pi}\left| sinx \right|dx\]

Answer 18

again..find the zeros of \(\sin(x)\) on \((0,4\pi)\)

Answer 19

find out where the sine is positive and negative on the above interval

Answer 20

hm, sinx on 0 is 0, sinx on 4pi is >0

Answer 21

you can cheat a little with that one though

\[\int\limits_{0}^{4 \pi}\left| sinx \right|dx=4\int\limits_{0}^{\pi}|\sin(x)|dx=4\int\limits_{0}^{\pi}\sin(x)dx\]

Answer 22

cheat? how come?

Answer 23

just use the periodic nature of the sine function

Answer 24

i see, okay got it(: thanks! :D

Answer 25

P.S. If in one period it's positive and one period it's negative, break the integral into two :)
\[\int_{-1}^1|x|dx=\int_{-1}^0-xdx+\int_0^1xdx\](This is just an example)