What is f(x) = int (x^2-2x)(e^x-sinx) dx if f(1 ) = 2 ?

1 Answer
Mar 15, 2016

f(x)=x^2(e^x+cos x)-2x(2e^x+sinx+cosx)+4e^x+2sinx-2cosx+2-e+3cos1

Explanation:

Using the rule
int udv=uv-int vdu

u=x^2-2x => du=(2x-2)dx
dv=(e^x-sinx)dx => v=e^x+cosx

f(x)=(x^2-2x)(e^x+cosx)-int (2x-2)(e^x+cosx)dx
Once again using the rule above mentioned:

u=2x-2 => du=2dx
dv=(e^x+cosx)dx => v=e^x+sinx

->f(x)=(x^2-2x)(e^x+cosx)-[(2x-2)(e^x+sinx)-2int (e^x+sinx)dx
f(x)=(x^2-2x)(e^x+cosx)+(-2x+2)(e^x+sinx)+2(e^x-cosx)+const.
f(x)=x^2(e^x+cosx)-2x(2e^x+sinx+cosx)+4e^x+2sinx-2cosx+const.

Finding const.

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
f(x)=x^2(e^x+cos x)-2x(2e^x+sinx+cosx)+4e^x+2sinx-2cosx+2-e+3cos1