Tuesday, May 8, 2012

Calculus Revisited #16: Improper Integrals

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

This blog is property of Edward Shore. © 2012


1 comment:

  1. Thanks for sharing this about Integral as-Common anti derivative that are used in rational functions are anti derivative of constant is ∫ k dx = kx + c b) and anti derivative of power function is ∫ xn dx = xn + n + 1 c)
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