Consider the function f(x)=|sin x|. -pi<=x<=pi

Question

a.Rewrite the function without using absolute values

b.Find the anti derivative F for f satisfying F(0)=1

Answer 1

Hello Robert: let's do this one!

We have: f(x) = | sin x | for -Π <= x <= Π

(a) The absolute value means f(x) = sin x when sin x is >= 0

and = - sin x when sin x < 0

In the given range, sin x is >= 0 for 0 <= x <= Π namely, in quadrants I and II

and < 0 for -Π <= x < 0 namely, in quadrants III and IV

Therefore, our definition of the function without using absolute values is as follows:

f(x) = sin x for 0 <= x <= Π

= - sin x for -Π < x < 0

Note that the graph of sin x goes below the x-axis for -Π <= x <= 0. So. the abs value makes it positive and the same as the graph of sin x from o to the x-axis.

You can see a nice diagram here: https://www.geogebra.org/m/WBkz4hfC

Therefore, the area under the curve is just twice the area under the curve from 0 to the x-axis.

(b) F the anti-derivative of f is just the (indefinite) integral wrt x.

Therefore, F(x) = ∫ | sin x | dx

= 2 ∫ sin x dx

= 2 (-cos x) + C where C is arbitrary constant

= -2cos x +C . . . . . . . . . . . . . . . . . . . (1)

To determine the value of C, we use the fact that F(0) = 1 => substitute 0 for x in (1)

F(0) 1 = - 2 cos 0 + C

1 = -2 (1) + C

=> C = 3

Hence, F(x) = -2 cos x + 3