Welcome to Part 16 of our 21 part Calculus Revisited series. Today we got down, dirty, and improper.
Improper Integrals
(I)
∞
∫ f(x) dx =
a
lim ( ∫(f(x) dx, a, t) as t → ∞
If f(r) is undefined:
(II)
a
∫ f(x) dx =
r
lim ( ∫f(x) dx, t, a) as t → r+ (t approaches r from the right side)
(III)
r
∫ f(x) dx =
a
lim ( ∫(f(x) dx, a, t) as t → r- (t approaches r from the left side)
Problems
1. Calculate:
4
∫ 2/(x - 2) dx
2
Note 2/(x - 2) is defined at x = 2. This is an improper integral.
∫( 2/(x - 2) dx, 2, 4)
= lim ∫( 2/(x - 2) dx, t, 4) as t → 2+
= lim (2 ln(4 - 2) - 2 ln(t - 2)) as t → 2+
Note: ln t has no limit as t → 0
Therefore the integral has no finite answer. Yes, that can happen.
2. Calculate:
3
∫ dx / √(3 -x)
0
Note that 1/√(3 - x) is undefined at x = 3. Another improper integral.
∫( 1 / √(3 - x) dx, 0, 3)
= lim ∫( 1 / √(3 - x) dx, 0, t) as t → 3-
= lim (- 2 * √(3 - t) + 2 * √(3 - 0) ) as t → 3-
= lim ( -2 * √(3 - t) + 2√3) as t → 3-
= 2√3 ≈ 3.46410
3. Calculate:
∞
∫ e^-x/2 dx
0
So:
∫ (e^(-x/2) dx, 0, ∞)
= lim (e^(-x/2) dx, 0, t) as t → ∞
= lim (-1/2* e^(-t/2) + 1/2 * e^0 ) as t → ∞
= 0 + 1/2
= 1/2
Next time we are going to work with sequences.
Until then, have a great day!
Eddie
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