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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