What is #f(x) = int xe^x-x dx# if #f(-1) = 1 #?

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Answer 1 · 2018-07-30T15:49:50

#f(x) = 2/e+ e^x(x-1)+(3-x^2)/2#

#f(x) = 1+ int_(-1)^x (te^t-t)dt#

as the integral is linear:

#f(x) = 1+ int_(-1)^x te^tdt- int_(-1)^x tdt#

Solve the two integrals separately:

# int_(-1)^x tdt = [t^2/2]_(-1)^x #

# int_(-1)^x tdt = x^2/2-1/2 #

and:

#int_(-1)^x te^tdt = int_(-1)^x t d/dt(e^t) dt#

integrating by parts:

#int_(-1)^x te^tdt = - int_(-1)^x e^t dt + [te^t]_( -1)^x#

#int_(-1)^x te^tdt = - [e^t]_(-1)^x + xe^x+1/e#

#int_(-1)^x te^tdt = - e^x+1/e + xe^x+1/e#

#int_(-1)^x te^tdt = e^x(x-1)+2/e#

Putting together the partial results:

#f(x) = 1+ e^x(x-1)+2/e-x^2/2+1/2#

#f(x) = 2/e+ e^x(x-1)+(3-x^2)/2#