Improper Integrals |

**Intro**

The fundamental theorem of Calculus states that we can evaluate a definite integral by

*F*(*b*) – *F*(*a*) -- the difference between the antiderivative values at the limits of

integration **IF** the integrand function is** continuous on the closed interval [ **

This means that the

the integrand

**Improper Integrals** are those which do not satisfy one or both of these conditions.

Remember that the definite integral represents the area under the curve over the

interval of integration. If one or both of the limits of integration is infinity,

the question becomes can we find a number value for the area under the curve.

If the integrand function has a discontinuity somewhere on the interval of

integration, again, we have to find ways to evaluate the area under the curve.

.

**Note:** If the integral can be evaluated, we say it **converges** or it is **convergent**.

If the integral becomes infinite or can't be evaluated, we say it **diverges** or it is **divergent**.

.

**Note:** We may have to use l'Hopital's Rule or other limit techniques to find the final limit.

.

There are 2 types of improper integrals

**Type 1/ Integrals with Infinite Limits of Integration**

In this type, one (or both) of the **limits of integration is infinity**.

We **replace** the **infinity** **with** a **parameter** (usually* ***t***, ***s*** *or ** r**),

we integrate over a well defined interval

Let's look at pictures and the integral set up to understand the approach.

i) | ii) |

iii) |

** *f *(*x*) **must be continuous** over the interval of integration.

Set *a* = 0 where possible for efficiency.

.

**Examples:**

**1/** Assign an area if possible to the region between *y = e ^{ x}* and the

**solution:**

We need to integrate from to *x = *1, -- so it's the **lower limit **that **is undefined**.

The area is:

As *t* approaches , *e ^{ t} *approaches 0. So the integral converges to

.

**2/** Sketch the graph of and evaluate if possible the integral from to .

**solution:**

Area =
This integral converges to give an area of o square units. |

.

**3/** Evaluate if possible the integral of *y = x ^{ }*– 3/4 , from

We want to evaluate

This integral diverges. We cannot assign a value to it. |

.

**Notes:**

1) Always replace the infinite limit(s) with a **parameter(s)**, then, at the end, limit it to infinity.

2) Draw the graph if you can since in many cases, it's obvious if the area converges or diverges.

**Type 2/ Integrals with Discontinuous Integrands:**

Like the first type, there are 3 possible situations:

- 1) discontinuous at the right end point of the interval

- 2) discontinuous at the left end point of the interval

- 3) discontinuous somewhere on the interval

Here are the images and the parameter set up:

i) If *f *(*x*) is discontinuous at ** b** , then:

ii) If *f *(*x*) is discontinuous at **a** , then:

iii) If *f *(*x*) is discontinuous at **c** where a < c < b, then:

If either of the 2 integrals diverges, the whole thing diverges.

**Example:** (type iii: discontinuity on the interval).

This integral has a vertical asymptote at *x =* 1, so *c* = 1.

It becomes

We antidifferentiate to get

The 1st integral becomes . The integral diverges.

.

**Watch out for ones with both kinds of discontinuities!!**

For example:

has an **infinite limit of integration** and an **asymptote** at *x =* 1 on the interval.

We break this guy up into 3 integrals.

The setup is:

from t to 0 |
from 0 to s |
from s to 3 |

or | or | or |

**Practice**

1/ Determine if these improper integrals converge or diverge:

a) | b) |

c) | d) |

e) | f) |

g) | h) |

**Solutions**

a) | b) |

The integral diverges. | |

The integral converges to 1. |
The volume of the solid created byrevolving about the from The This solid has If it were a paint can, it could never hold enough paint to cover the outside of the can. |

c)
This integral converges to |
d) now substitute to get: This integral converges to . |

e)
The integral diverges. |
f) There is an asymptote at
When we antidifferentiate the 1st integral we get:
The integral diverges. |

.

g) We'll split it at x = 0 but first let's simplify the integrand with a substitution.If |

The antiderivative is u = – x² so we get:
Since , we get: – ½ [ (1 – 0) + (0 – 1) ] = 0 The integral converges to 0. |

h) Let's deal with the integrand's discontinuities first. This means we need only deal with the infinite limit of integration. So we get: Now, let's do partial fraction decomposition on the integrand.
Setting
Applying the rule of logs that says a difference of 2 logs is the log of the quotient, and now we evaluate the limit. It becomes |

(*all content **© MathRoom Learning Service; 2004 - *).