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

1 Answer
Jan 14, 2017

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.)