Published on 15 Sep 2018

School

Department

Course

Professor

Miami University

MTH 151

Calculus I

Winter 2018

Term Test 3

Prof: Olga Brezhneva

Exam Guide

CONTENTS

Exam Guide Covers the Following Topics:

Unit 3: Chapter 5: Integrals

Areas and Distances

Definite Integral

The Fundamental Theorem of Calculus

Indefinite Integrals and the Net Change Theorem

Substitution

Areas and Distances

Typically seen questions

●Given the graph, what is the distance travelled…

○Between 1 and 4 pm?

○In total? (from t = 0 to t = 5)

●So we know that the graph shows the relationship between velocity (miles per hour)

and time (hours)

○Because D (distance) is equal to the product of R (rate) and T (time), we can find

and solve for distance using the formula D = RT

○Using the givens, between 1 and 4 pm is 3 hours, and the speed is constant at 40

mph

■So D = (40)(3) = 120 miles

●Looking at the graph, you may realize that 120 is equal to the area of the rectangle

formed using the time as a base and velocity as a height

## Document Summary

Given the graph, what is the distance travelled . In total? (from t = 0 to t = 5) So we know that the graph shows the relationship between velocity (miles per hour) and time (hours) Using the givens, between 1 and 4 pm is 3 hours, and the speed is constant at 40 mph. So d = (40)(3) = 120 miles. This is a fact that within a velocity time graph, the area under the curve is the distance travelled. Knowing this information, we can then solve for the distance by finding the area under the curve between t = 0 and t = 5. We can either break this into two triangles and a rectangle, or simply use the equation of area for a trapezoid. ( )(b1 + b2)(h) in which b1 is equal to base 1, b2 is base 2, and h is the height. So say we have the function y = x^2.