Question

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

Answer 1

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

Explanation:

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]}