What is #f(x) = int -e^(2x)-2e^x-x dx# if #f(-3 ) = 1 #?

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Answer 1 · 2017-07-04T22:21:18

#f(x) = (11e^6+4e^3+1)/(2e^6) -(e^(2x)+4e^x+x^2)/2 #

Choose #x= -3# as lower limit of integration and pose:

#f(x) = 1+ int_(-3)^x (-e^(2t)-2e^t-t)dt#

Clearly:

#f(-3) = 1+int_(-3)^(-3) (-e^(2t)-2e^t-t)dt = 1#

Using now the linearity of integrals:

#f(x) = 1 -int_(-3)^x e^(2t)dt -2int_(-3)^xe^tdt - int_(-3)^xtdt#

Now:

#int_(-3)^x e^(2t)dt = [e^(2t)/2]_(-3)^x = e^(2x)/2-e^-6/2 = e^(2x)-1/(2e^6)#

#int_(-3)^xe^tdt = [e^t]_(-3)^x = e^x-e^(-3) = e^x-1/e^3#

#int_(-3)^xtdt = [t^2/2]_(-3)^x = x^2/2 -9/2#

Then:

#f(x) = 1-e^(2x)/2-2e^x-x^2/2 +1/(2e^6)+2/e^3+9/2#

#f(x) = (11e^6+4e^3+1)/(2e^6) -(e^(2x)+4e^x+x^2)/2 #

graph{ (11e^6+4e^3+1)/(2e^6) -(e^(2x)+4e^x+x^2)/2 [-10, 10, -5, 5]}