How do you find the area of the region bounded by the curves #y=sin(x)#, #y=e^x#, #x=0#, and #x=pi/2# ?

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Answer 1 · 2014-09-22T03:58:42

Over the interval #[0,pi/2]# the function #y=e^x# is greater that the function #y=sin(x)#. This is why #y=sin(x)# is subtracted from #y=e^x#.

#int_0^(pi/2)e^x-sin(x)dx#

#=[e^x-(-cos(x))]_0^(pi/2)#

#=[e^x+cos(x)]_0^(pi/2)#

#=[e^(pi/2)+cos(pi/2)-(e^0+cos(0))]#

#=[e^(pi/2)+0-(1+1)]#

#=[e^(pi/2)-(2)]#

#=[e^(pi/2)-2]#

#=2.81048-># Solution