Choose x= -3x=−3 as lower limit of integration and pose:
f(x) = 1+ int_(-3)^x (-e^(2t)-2e^t-t)dtf(x)=1+∫x−3(−e2t−2et−t)dt
Clearly:
f(-3) = 1+int_(-3)^(-3) (-e^(2t)-2e^t-t)dt = 1f(−3)=1+∫−3−3(−e2t−2et−t)dt=1
Using now the linearity of integrals:
f(x) = 1 -int_(-3)^x e^(2t)dt -2int_(-3)^xe^tdt - int_(-3)^xtdtf(x)=1−∫x−3e2tdt−2∫x−3etdt−∫x−3tdt
Now:
int_(-3)^x e^(2t)dt = [e^(2t)/2]_(-3)^x = e^(2x)/2-e^-6/2 = e^(2x)-1/(2e^6)∫x−3e2tdt=[e2t2]x−3=e2x2−e−62=e2x−12e6
int_(-3)^xe^tdt = [e^t]_(-3)^x = e^x-e^(-3) = e^x-1/e^3∫x−3etdt=[et]x−3=ex−e−3=ex−1e3
int_(-3)^xtdt = [t^2/2]_(-3)^x = x^2/2 -9/2∫x−3tdt=[t22]x−3=x22−92
Then:
f(x) = 1-e^(2x)/2-2e^x-x^2/2 +1/(2e^6)+2/e^3+9/2f(x)=1−e2x2−2ex−x22+12e6+2e3+92
f(x) = (11e^6+4e^3+1)/(2e^6) -(e^(2x)+4e^x+x^2)/2 f(x)=11e6+4e3+12e6−e2x+4ex+x22
graph{ (11e^6+4e^3+1)/(2e^6) -(e^(2x)+4e^x+x^2)/2 [-10, 10, -5, 5]}
