How do you evaluate the definite integral #int (sint+cost)dt# from #[0,pi/4]#?
1 Answer
Nov 6, 2016
Explanation:
We have to know that:
#int(f(x)+g(x))dx=intf(x)dx+intg(x)dx# #intsintdt=-cost+C# #intcostdt=sint+C#
Combining these and keeping the bounds of the integral, we see that:
#int_0^(pi/4)(sint+cost)dt=[-cost+sint]_0^(pi/4)#
Evaluating at
#=[-cos(pi/4)+sin(pi/4)]-[-cos0+sin0]#
#=(-sqrt2/2+sqrt2/2)-(-1+0)#
#=0+1#
#=1#
