## Monday, May 15, 2017

### Unizor - Definite Integrals - Improper Integrals Examples 1

Notes to a video lecture on http://www.unizor.com

Improper Definite Integrals

Example 1.1

011/√x dx =
lima→0 a11/√x dx

Indefinite integral of f(x)=1/√x is F(x)=2√x.
Indeed, let's take a derivative of F(x)=2√x:
Dx F(x) = 2·(1/2)·x(1/2)−1 =
= x−1/2 = 1/√x

Using Newton-Leibniz formula,
a11/√x dx = F(1) − F(a) =
= 2√1 − 2√a

As a→0, this expression converges to 2

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Example 1.2

0e−x dx =
limb→∞ 0be−x dx

Indefinite integral of f(x)=e−x is F(x)=−e−x.
Evaluating integral:
0be−x dx = F(b) − F(0) =
= (−e−b) − (−e−0) = 1 −e−b

As b→∞, this expression converges to 1

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Example 1.3

01/(1+x²) dx =
limb→∞ 0b1/(1+x²) dx

Indefinite integral of f(x)=1/(1+x²) is F(x)=arctan(x).
Evaluating integral:
0b1/(1+x²) dx = F(b) − F(0) =
= arctan(b) − arctan(0) =
= arctan(b)

As b→∞, this expression converges to π/2

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Example 1.4

1(1/x²)·e1/x dx =
limb→∞ 1b(1/x²)·e1/x dx

Indefinite integral of
f(x)=(1/x²)·e1/x can be found by noticing that −1/x² is a derivative of 1/x and, therefore, a derivative of e1/x is e1/x·(−1/x²).
Therefore, indefinite integral of f(x)=(1/x²)·e1/x is F(x)=−e1/x.
Evaluating integral:
1b(1/x²)·e1/x dx = F(b)−F(1) =
= (−e1/b) − (−e1/1) = e − e1/b

As b→∞, this expression converges to e −1