How do you find the area bounded by the curves #y = -4sin(x)# and #y = sin(2x)# over the closed interval from 0 to pi?

1 Answer
Mar 15, 2016

Answer:

Evaluate

#int_0^π|-4sin(x)-sin(2x)|dx#

Area is: #8#

Explanation:

The area between two continuous functions #f(x)# and #g(x)# over #x in[a,b]# is:

#int_a^b|f(x)-g(x)|dx#

Therefore, we must find when #f(x)>g(x)#

Let the curves be the functions:

#f(x)=-4sin(x)#

#g(x)=sin(2x)#

#f(x)>g(x)#

#-4sin(x)>sin(2x)#

Knowing that #sin(2x)=2sin(x)cos(x)#

#-4sin(x)>2sin(x)cos(x)#

Divide by #2# which is positive:

#-2sin(x)>sin(x)cos(x)#

Divide by #sinx# without reversing the sign, since #sinx>0# for every #x in(0,π)#

#-2>cos(x)#

Which is impossible, since:

#-1<=cos(x)<=1#

So the initial statement cannot be true. Therefore, #f(x)<=g(x)# for every #x in[0,π]#

The integral is calculated:

#int_a^b|f(x)-g(x)|dx#

#int_0^π(g(x)-f(x))dx#

#int_0^π(sin(2x)-(-4sin(x)))dx#

#int_0^π(sin(2x)+4sin(x))dx#

#int_0^πsin(2x)dx+4int_0^πsin(x)#

#-1/2[cos(2x)]_0^π-4[cos(x)]_0^π#

#-1/2(cos2π-cos0)-4(cosπ-cos0)#

#1/2*(1-1)-4*(-1-1)#

#8#