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

1 Answer
Rui D.
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#

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