MA 137: Calculus I for the Life Sciences - University of...

MA 137: Calculus I for the Life Sciences

David Murrugarra

Department of Mathematics,University of Kentucky

http://www.ms.uky.edu/~ma137/

Spring 2017

David Murrugarra (University of Kentucky) MA 137: Section 6.1 Spring 2017 1 / 15

mailto:[email protected]

http://www.math.uky.edu/

http://www.uky.edu/

http://www.ms.uky.edu/~ma137/


The Area Problem: Partitions

The idea we have used so far is to “to partition" or subdivide the given interval [a, b] into smaller subintervals on each of which

the variable x , and thus the function f (x), does not change much.

Definition (Partitions)A partition of an interval [a,b] is a set of points {x0, x1, x2, . . . , xn−1, xn},listed increasingly, on the x-axis with a = x0 and xn = b. That is:

a = x0 < x1 < x2 < · · · < xn−1 < xn = b.

These points subdivide the interval [a,b] into n subintervals

[a, x1], [x1, x2], [x2, x3], . . . , [xn−1,b].

The k th subinterval is thus of the form [xk−1, xk ] and it has length

4xk = xk − xk−1.



The Area Problem: Partitions

Definition (Norm of a Partition)The length of the longest subinterval is called the norm of P and isdenoted by ||P||,

||P|| = max1≤i≤n

{4xi}

We assume that our partition P is such that ||P|| → 0 as n→∞.In other words, we assume that the length of the longest (and, hence,of all) subinterval(s) tend(s) to zero whenever the number ofsubintervals in P becomes very large.



The Definite Integral

DefinitionLet f (x) be a function defined on an interval [a,b]. Partition the interval[a,b] in n subintervals of lengths 4x1, . . . ,4xn, respectively. Fork = 1, . . . ,n pick a representative point ck in the corresponding k th

subinterval. The definite integral of f from a to b is defined as

limn→∞

n∑k=1

f (ck ) · 4xk = lim||P||→0

n∑k=1

f (ck ) · 4xk

and it is denoted by ∫ b

af (x) dx .




∫ b

af (x) dx = lim

||P||→0

n∑k=1

f (ck ) · 4xk

The sumn∑

k=1

f (ck ) · 4xk is called a Riemann sum in honor of the

German mathematician Bernhard Riemann (1826-1866), whodeveloped the above ideas in full generality.The symbol

∫is called the integral sign. It is an elongated capital

S, of the kind used in the 1600s and 1700s. The letter S standsfor the summation performed in computing a Riemann sum.The numbers a and b are called the lower and upper limits ofintegration, respectively.The function f (x) is called the integrand and the symbol dx iscalled the differential of x . You can think of the dx as representingwhat happens to the term 4x in the limit, as the size 4x of thesubintervals gets closer and closer to 0.




The role of x in a definite integral is the one of a dummy variable. Infact ∫ b

ax2 dx and

∫ b

at2 dt

have the same meaning. They represent the same number. We recallthat a limit does not necessarily exist. However,

Theorem

If f is continuous on [a,b] then∫ b

af (x) dx exists.




As we observed earlier, it is computationally easier to partition theinterval [a,b] into n subintervals of equal length. Therefore eachsubinterval has length 4x = b−a

n (we drop the index k as it is no longernecessary). In this case, there is a simple formula for the points of thepartition, namely:

x0 = a + 0 · 4x , x1 = a + 1 · 4x , . . . , xn = a + n · 4x = b.

or, more concisely,

xk = a + kb − a

nfor k = 0,1,2, . . . ,n.



Right Versus Left Endpoint Estimates

Observe that xk , the right endpoint of the k th subinterval, is also theleft endpoint of the (k + 1)th subinterval. Thus the Riemann sumestimate for the definite integral of a function f defined over an interval[a,b] can be written in either of the following two forms:

n−1∑k=0

f(

a + kb − a

n

)· b − a

nand

n∑k=1

f(

a + kb − a

n

)· b − a

n

depending on whether we use left or right endpoints, respectively.

Figure: Left endpoint Riemann sum estimate. Figure: Right endpoint Riemann sum estimate.




Example (Neuhauser, Example # 1, p. 277/8)

We will try to find the area of the region below the parabola f (x) = x2

and above the x-axis between 0 and a.

To do this, we divide the interval [0,a] into n subintervals of equallength and approximate the area of interest by a sum of the areas ofrectangles, the widths of whose bases are equal to the lengths of thesubintervals and whose heights are the values of the function at theleft endpoints of these subintervals.



Definite Integrals and Areas

We stress the fact that if the function f takes on only positivevalues then the definite integral is nothing but the area of theregion below the graph of f , lying above the x-axis, and boundedby the vertical lines x = a and x = b.

Distance traveled by an object:

If the positive valued function under consideration is the velocityv(t) of an object at time t , then the area underneath the graph ofthe velocity function and lying above the t-axis represents thetotal distance traveled by the object from t = a to t = b.

Figure: The area of the region below the graph of f .



What if the Function Takes on Negative Values?

If f happens to take on both positive and negative values, then the Riemann sum is the sum of the areas of rectangles that lie

above the x-axis and the negatives of the areas of rectangles that lie below the x-axis. Passing to the limit, we obtain that, in

general, a definite integral can be interpreted as a difference of areas:

Figure: The definite integral is the difference of the areas in yellow minus the area in red.

∫ b

af (x) dx = [area of the region(s) lying above the x-axis]−

[area of the region(s) lying below the x-axis]



Example (Online Homework, HW 23, # 11)Determine which of the following regions has an area equal to thegiven limit without evaluating the limit:

limn→∞

n∑i=1

2n

(5 +

2in

)10

(a) The area of the region under the graph of y = x5 on the interval[5, 7].

(b) The area of the region under the graph of y = x10 on the interval[5, 7].

(c) The area of the region under the graph of y = x5 on the interval[2, 5].

(d) The area of the region under the graph of y = x10 on the interval[2, 5].




Example (Online Homework, HW 23, # 12)The value of the limit

limn→∞

n∑i=1

4n

√5 +

4in

is equal to the area below the graph of a function f (x) on an interval[A,B].

Find f , A, and B.

f (x) =

A = (use A = 0)

B =




Example (Online Homework, HW 23, # 7)Evaluate the following integral by interpreting it in terms of areas:∫ 3

0

(12

x − 1)

dx




Example (Online Homework, HW 23, # 10)

Let g(x) =∫ x

0f (t)dt , where f is the function whose graph is shown

below.

(a) Evaluate g(x) forx = 0,1,2,3,4,5, and 6.

(b) Estimate g(7).(c) At what value of x does g

attain its maximum?(d) At what value of x does g

attain its minimum?Figure: Graph of f .



MA 137: Calculus I for the Life Sciences - University of...

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