1. Learning Objectives for Section 13.4 The Definite Integral
1. The student will be able to approximate areas by using left and right sums.
2. The student will be able to compute the definite integral as a limit of sums.
3. The student will be able to apply the properties of the definite integral.
Barnett/Ziegler/Byleen Business Calculus 11e

2. Introduction
We have been studying the indefinite integral or antiderivative of a function.
We now introduce the definite integral. This integral will be the area bounded by f (x), the x axis, and the vertical lines x = a and x = b, with notation
Barnett/Ziegler/Byleen Business Calculus 11e

3. Estimating
One way to approximate the area under a curve is by filling the region with rectangles and calculating the sum of the areas of the rectangles.
Take the width of each rectangle to be x = 1. If we use the left endpoints, the heights of the four rectangles are f (1), f (2), f (3) and f (4), respectively.
f (x) = 0.5 x2 + 2
L4 = f (1) Δx + f (2) Δx + f (3) Δx + f (4) Δx
= = 23
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4. Estimating Area (continued)
We can repeat this using the right side of each rectangle to determine the height.
The width of each rectangle is again x = 1. The heights of each of the four rectangles are now f (2), f (3), f (4) and f (5), respectively.
The sum of the rectangles is then
R4 = = 35
f (x) = 0.5 x2 + 2
The average of L4 and R4 would be an even better approximation: Area  ( )/2 = 29.
Barnett/Ziegler/Byleen Business Calculus 11e

5. Estimating Area (continued)
The previous average of 29 is very close to the actual area of ….
Our accuracy can be improved if we increase the number or rectangles, and let x get smaller.
f (x) = 0.5 x2 + 2
The error in our process can be calculated if the function is monotone. That is, if the function is only increasing or only decreasing.
Let Ln and Rn be the approximate areas, using n rectangles of equal width, and the left or right endpoints, respectively.
Barnett/Ziegler/Byleen Business Calculus 11e

6. Estimating Area (continued)
If the function is increasing, convince yourself by looking at the picture that
Ln  Area  Rn
If f is decreasing, the inequalities go the other way.
f (x) = 0.5 x2 + 2
If you use Ln to estimate the area, then
Error = |Area – Ln|  |Rn – Ln|.
If you use Rn to estimate the area, then
Error = |Area – Rn|  |Rn – Ln|.
Either way you get the same error bound.
Barnett/Ziegler/Byleen Business Calculus 11e

7. Theorem 1 Theorem 1 It is not hard to show that
|Rn – Ln| = | f (b) – f (a)| x,
and that for n equal subintervals,
Theorem 1
For our previous example:
Barnett/Ziegler/Byleen Business Calculus 11e

8. Theorem 2
If f (x) is either increasing or decreasing on [a, b], then its left and right sums approach the same real number I as n  .
This number I is the area between the graph of f and the x axis from x = a to x = b.
Barnett/Ziegler/Byleen Business Calculus 11e

9. Approximating Area In the previous example, we had f (x) = 0.5 x 2 + 2
If we wanted a particular accuracy, say 0.05, we could use the error formula to calculate n, the number of rectangles needed:
Solving for n yields n = 960. We would need at least 960 rectangles to guarantee an accuracy of 0.05.
Barnett/Ziegler/Byleen Business Calculus 11e

10. Definite Integral as Limit of Sums
We now come to a general definition of the definite integral.
Let f be a function on interval [a, b]. Partition [a, b] into n subintervals at points
a = x0 < x1 < x2 < … < xn–1 < xn = b.
The width of the kth subinterval is xk = (xk – xk-1).
In each subinterval, choose an arbitrary point ck
xk-1  ck  xk.
Barnett/Ziegler/Byleen Business Calculus 11e

11. Definite Integral as Limit of Sums (continued)
Then define
Sn is called a Riemann sum. Notice that Ln and Rn are both special cases of a Riemann sum.
Barnett/Ziegler/Byleen Business Calculus 11e

12. A Visual Presentation of a Riemann SumΔx
f (c 2)
f (c 1)
. . .
a = x 0
x 1
x 2
x n -1
x n = b
c 1
c 2
c n
The area under the curve is approximated by the Riemann sum
Barnett/Ziegler/Byleen Business Calculus 11e

13. Area (Revisited)
Let’s revisit our original problem and calculate the Riemann sum using the midpoints for ck.
f (x) = 0.5 x2 + 2
The width of each rectangle is again x = 1. The heights of the four rect-angles are now f (1.5), f (2.5), f (3.5) and f (4.5), respectively.
The sum of the rectangles is then
S4 = = 28.5
This is quite close to the actual area of
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14. The Definite Integral
Theorem 3. Let f be a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I as max xk  0.
This limit I of the Riemann sums for f on [a, b] is called the definite integral of f from a to b, denoted
The integrand is f (x), the lower limit of integration is a, and the upper limit of integration is b.
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15. Negative Values
If f (x) is positive for some values of x on [a, b] and negative for others, then the definite integral symbol
y = f (x) B a
represents the cumulative sum of the signed areas between the graph of f (x) and the x axis, where areas above are positive and areas below negative. A b
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16. Examples
Calculate the definite integrals by referring to the figure with the indicated areas.
Area A = 3.5
Area B = 12 B a A b c
y = f (x)
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17. Definite Integral Properties
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18. Examples Assume we know that Then A) B)
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19. Examples (continued) C) D) E)
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20. Summary
We summed rectangles under a curve using both the left and right ends and the centers and found that as the number of rectangles increased, accuracy of the area under the curve increased.
We found error bounds for these sums.
We defined the definite integral as the limit of these sums and found that it represented the area between the function and the x axis.
We learned how to compute areas under the x axis.
Barnett/Ziegler/Byleen Business Calculus 11e