15 Sep 2018

School

Department

Course

Professor

Miami University

MTH 251

Calculus II

Winter 2018

Term Test 2

Professor: Konstantinos Beros

Exam Guide

Table of content:

-Section 7.8: Improper Integrals (Continue)

-Section 8.1: Arc Length

-Section 8.2: Surface Area of Solids of Revolution

-Exam Review Problems

Section 7.8: Improper Integrals (Continue)

Definition of improper Integrals of Type 1(The interval is infinite):

i. ()

exists for every t, then we can define ()=

lim

()

, provided that this limit exists.

ii. If the integral ()

exists for every tb, then we define ()=

lim

()

provided this limit exists.

The improper integrals, ()

and ()

are called convergent if their respective

limits exist, and divergent if their limits do not exist.

iii. If ()

=()

both exist then we define ()

=

()=()

Example

+ 4

=lim

+ 4

To solve

use u substitution where u=, du=2zdz, and

=, so

+ 4

=1

21

+ 4

So, plugging this back in,

lim

+ 4

=lim

1

21

+ 4

lim

1

2(1

2tan2)0