MATH1051 Lecture Notes - Lecture 2: Formula D
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5.1 – 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
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○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
○(½)(b1 + b2)(h) in which b1 is equal to base 1, b2 is base 2, and h is the height
○(½)(3 + 5)(40) = 160 miles
Area Under the Curve Between Interval [a, b] (Theoretical)
●Estimating area using rectangles
●So say we have the function y = x^2
○We want to find the area under the curve between [0, 3]