How would you integrate the absolute value of a function?

Answer 1

By finding out where the function is positive and where it is negative,
A useful trick here: Example: Int | (sin x - cos x) |

(sin x - cos x) = sqrt(2) * (sin x cos pi/4 - cos x sin pi/4)

= sqrt(2) * sin (x - pi/4)

Answer 2

For example x^3
lets assume you need area for this equation from -1 to 1
If you look at in drawn, the area above and below the line are the same so if you did this function as an integral from -1 to 1 you'd get 0 but if you take the absolute integral of this function it would work like this
y = |x^3|
y = |x^4/4| [-1,1
1/4 - [1/4] but look at this change all negatives from set turn into positives
[1^4]/4 "+" [(-1)^4]/4
1/4 "+" 1/4 = 1/2 not 1/4 - 1/4 = 0

Answer 3

well split the function depending upon the limits.....

Integral of |sinx| form -pi/4 to +pi/4 will be

intg(Sinx)from(0 to pi/4) -intg(Sinx)from(-pi/4 to 0)...

as sine is +ve in 1st quad ie from 0 to pi/4
and -ve in 4th quad ie from -pi/4 to 0

Remove the modulus before u integarate...

Answer 4

integ abs(3x) dx

By definition abs(x) = x if x >= 0
= -x if x < 0

1) x >= 0 => abs(3x) = 3x

So, integ 3x dx = (3x^2)/2 + C

2) The close argument will be use to solve this case
x < 0 => abs(-3x) = -3x

So, integ abs(-3x) dx = (-3X^2)/2 + C