Definite Integrals Part 4

Concept:First, solve the integro-differential equation to find $f(x)$. Then use the derivative of $g(x)$ to locate its critical points. Analyse sign changes of $g'(x)$ to find points of local minima and maxima.Explanation:Given: $f(x)=1-2x+\int_{0}^{x} e^{(x-t)} f(t) dt$.Rewrite as $f(x)=1-2x+e^{x}\int_{0}^{x} e^{-t} f(t) dt$.Differentiate both sides with respect to $x$ using Leibniz rule:$f'(x)=-2+[e^{x}\int_{0}^{x} e^{-t}f(t)dt + e^{x}(e^{-x}f(x))]$.But $e^{x}\int_{0}^{x} e^{-t}f(t)dt = f(x)-1+2x$. Substitute:$f'(x)=-2+[f(x)-1+2x]+f(x) = 2f(x)+2x-3$.Thus $\frac{df}{dx} - 2f = 2x-3$.Integrating factor: $e^{\int -2 dx}=e^{-2x}$.Solution: $f(x)e^{-2x}=\int (2x-3)e^{-2x} dx$.Using integration by parts: $f(x)=1-x+Ce^{2x}$.From $f(0)=1$ (original equation at $x=0$), we get $C=0$. So $f(x)=1-x$.Now $g(x)=\int_{0}^{x} (f(t)+2)^{15}(t-4)^{6}(t+12)^{17} dt$.Differentiate: $g'(x)=(f(x)+2)^{15}(x-4)^{6}(x+12)^{17}$.Substitute $f(x)=1-x$: $g'(x)=(3-x)^{15}(x-4)^{6}(x+12)^{17}$.Critical points: $x=3$, $x=4$, $x=-12$.Check sign changes for $g'(x)$:At $x=-12$: sign changes from negative to positive → local minimum, so $p=-12$.At $x=3$: sign changes from positive to negative → local maximum, so $q=3$.At $x=4$: even power $(x-4)^6$ → no sign change → not an extremum.Answer:$|p+q| = |-12+3| = |-9| = 9$.