Given f(x)=1−2x+∫0xe(x−t)f(t)dt.
Differentiating and simplifying:
e−xf′(x)−e−xf(x)=−2e−x+(1−2x)e−x(−1)+e−xf(x)
This reduces to:
f′(x)−2f(x)=2x−3
Solving the linear ODE dxdy−2y=2x−3:
y⋅e−2x=∫e−2x(2x−3)dx
On solving, we get f(x)=1−x.
Now, g(x)=∫0x(3−t)15(t−4)6(t+12)17dt
g′(x)=(3−x)15(x−4)6(x+12)17
=−(x−3)15(x−4)6(x+12)17
Sign analysis of g′(x):
g′(x)changes from +→−at x=3→ local maxima (q=3)
g′(x)changes from −→+at x=−12→ local minima (p=−12)
∣p+q∣=∣−12+3∣=9