Question

Evaluate the following term `int_0^(3pi/2) 5|sinx|dx` .How would i do this using FTC2(F(b)-F(a))?

Answer 1

We need to split up the integral.

Explanation:

Recall that `absu = {(u,"if",u >= 0),(-u,"if",u<0):}`

so `abs(sinx) = {(sinx,"if",sinx >= 0),(-sinx,"if",sinx < 0):}`.

We are integrating on `[0,(3pi)/2]` and we know that

`{(sinx >=0,"if" ,0 <=x <= pi),(sinx < 0,"if",pi < x <= (3pi)/2):}`

Therefore,

`abs(sinx) = {(sinx,"if",0 <= x <= pi),(-sinx,"if",pi < x < (3pi)/2):}`.

`int_0^((3pi)/2) 5abs(sinx) dx = 5int_0^((3pi)/2) abs(sinx) dx`

`= 5[int_0^pi sinx dx + int_pi^((3pi)/2) -sinx dx]`

`= 5[int_0^pi sinx dx - int_pi^((3pi)/2) sinx dx]`

Now use the fact that `int sinx dx = -cosx +C` to find each of the integrals.

`= 5[{:-cosx]_0^pi +{:cosx]_pi^((3pi)/2)]`

`= 5[(-cospi+cos0)+(cos((3pi)/2)-cospi)]`

`= 5[(-(-1)+1+0-(-1)]`

`= 5[3] = 15`

Bonus method

Some people prefer to integrate `abs(f(x))` by simply integrating from one zero to the next without first adjusting the sign. Any integral that comes out negative, we make positive.

The notation for this technique is

`int_0^((3pi)/2) 5abs(sinx) dx = abs(int_0^pi 5sinx dx)+abs(int_pi^((3pi)/2) 5sinx dx)`

The first of these two integrals will be positive and the second will be negative. (That's why the first method changed the sign for the second integral before integrating.)