So, V = ∫(x = -1 to 3) 2π (4 - x) [(x + 3) - (x^2 - x)] dx
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4) This diverges via limit comparison with the harmonic series (limit = 1).
5) Write out the terms:
(1/3 - 1) + (1/4 - 1/2) + (1/5 - 1/3) + ... + (1/101 - 1/99)
= -1 - 1/2 + 1/100 + 1/101, since all other terms cancel in pairs
6) Rewrite as y' - xy = x^(-3) e^x.
Multiply both sides by the integrating factor e^(∫ -x dx) = e^(-x^2/2):
e^(-x^2/2) y' - xe^(-x^2/2) y = x^(-3) e^x e^(-x^2/2)
==> (d/dx) [e^(-x^2/2) y] = x^(-3) e^((2x-x^2) / 2)
Integrate both sides:
e^(-x^2/2) y = C + ∫(t = 0 to x) t^(-3) e^((2t-t^2) / 2) dt
Most likely, the integral on the right can't be simplified.
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7) Assuming that the n-th term is (1 - 2/n)^(3n^2) * x^n, use the root test:
r = lim(n→∞) |(1 - 2/n)^(3n^2) * x^n|^(1/n)
..= |x| * lim(n→∞) (1 - 2/n)^(3n)
..= |x| * lim(n→∞) {[1 + 1/(-n/2)]^(-n/2)}^(-6)
..= |x| / e^6.
So, the series converges for |x| < e^6.
(The series at the endpoints can be shown to diverge by noting that the n-th term does not converg to 0 in either case.)
Good luck!