What is #f(x) = int (x^2-2x)(e^x-sinx) dx# if #f(1 ) = 2 #?
1 Answer
Mar 15, 2016
Explanation:
Using the rule
#u=x^2-2x# =>#du=(2x-2)dx#
#dv=(e^x-sinx)dx# =>#v=e^x+cosx#
Once again using the rule above mentioned:
#u=2x-2# =>#du=2dx#
#dv=(e^x+cosx)dx# =>#v=e^x+sinx#
Finding
#f(1)=2#
#1(e+cos1)-2(cancel(2e)+cancel(sin1)+cos1)+cancel(4e)+cancel(2sin1)-2cos1+const.=2#
#e-3cos1+const.=2# =>#const.=2-e+3cos1~=0.903#
So