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.)