What is the definite integral of #3 sin2x# from #x = 0# to #x =2#?

Question

2089 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-25T04:44:29

#int_0^2 3sin(2x)dx#

Factor out the constant

#3int_0^2 sin(2x)dx#

Make a substitution

Let #u=2x#

#du=2 dx#

#(du)/2=(2 dx)/2#

#(du)/2=dx#

#3int sin(u)*(du)/2#

#=3*1/2int sin(u)du#

#=3/2int sin(u)du#

Recalculate the boundaries

Upper Bound

#u=2x=2(2)=4#

Lower Bound

#u=2x=2(0)=0#

#3/2int_0^4 sin(u)du#

#=3/2[-cos(u)]_0^4#

#=3/2[-cos(4)-(-cos(0))]#

#=3/2[-cos(4)-(-1)]#

#=3/2[-cos(4)+1]#

#=[(-3cos(4)+3)/2]#

#=2.480465431# sq units