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4 The Definite Integral
4.1 Introduction to Area and 4.2 The Definite Integral
y = 1 − x^2
R
x
y (^) The idea of the definite integral arose from the problems of calculating lengths, areas, and vol- umes of curvilinear geometric figures, i.e. ob- jects with a curved boundary. Consider the graph of the function f ( x ) = 1 − x^2 between x = 0 and x = 1. There is no formula for finding the exact area underneath the graph of f and above the x - axis, from x = 0 to x = 1. Let us denote this region by R. We actually discussed about how to ap- proach this problem the first day in class: cover the region R with familiar shapes whose areas can be found easily. Well, I think it is unanimous that rectangles are the easiest one among all.
Let us demonstrate how to do these approximations with 4 rectangles of equal width. To this end, we first divide the interval [0, 1] into 4 equally sized subintervals: [ 0,
] ⋃ [
] ⋃ [
] ⋃ [
]
We can clearly tell that each rectangle has width 1/4. There are certainly many choices in terms of how to arrange these rectangles to cover the region R , but let us look at two particular choices.
- We inscribe rectangles in the region R. In this case, the height of each inscribed rectangle is given by the value of f at the right-endpoint of the subinterval.
x
y
- We circumscribe the region R by rectangles. In this case, the height of each circumscribed rectangle is given by the value of f at the left-endpoint of the subinterval.
x
y
The figures below are the analogous approximations with 16 rectangles.
x
y
x
y
We would expect our approximations to get better and better as the number of rectangles in- crease. This suggests a reasonable method to compute the area between the graph of f ( x ), the x -axis and the lines x = a and x = b :
Example 4.2. Use the Special Sum Formulas to re-evaluate the sums in Example 4.1.
(a)
∑^5
i = 1
(2 i + 3) =
(b)
∑^5
k = 1
( k^2 − 1) =
Example 4.3. Find the area of the region underneath the graph of f ( x ) = 1 − x^2 and above the x -axis between x = 0 and x = 1 by approximating it with inscribed rectangles and then taking a limit.
x
y
- Divide the interval [0, 1] into n subintervals of equal length.
- Label the endpoints of successive subintervals as x 0 , x 1 ,... , xn. These points are given by the formula:
- Label the inscribed rectangles as R 1 , R 2 ,... , Rn.
- Compute ∆ x , the length of each of these subintervals.
- Now listen to me: Look only at a particular rectangle Ri. The height of Ri is given by the value of f ( x ) at the right-endpoint of the corresponding subinterval. - This right endpoint is: - Consequently, the height of the rectangle Ri is - Finally, the area of the rectangle Ri is
- Write an expression using the sigma notation for the total area of the n inscribed rectangles and then evaluate this sum. The answer should be a function of n only.
- Take the limit of your answer from part (c) as n −→ ∞ to find the actual area.
Riemann sum
The type of sums that arises using rectangles approximation are called Riemann sums. The usual procedure is as follows:
- Break the interval [ a , b ] into n equally-spaced subintervals, each of length ∆ x =
- Pick a sample point xi from the i th subinterval and set f ( xi ) to be the height of the rectangle Ri on that subinterval.
- The area of all n rectangles, which we call a Riemann sum, is
The sample point can be any point in the subinterval, including endpoints as well. Below we list three common choices. Suppose f ( x ) is a continuous function defined on the interval [ a , b ].
Let n be any positive integer and set ∆ x = b − a n
- The left-endpoint approximation to the area under the graph of f ( x ) between x = a and x = b is given by
Ln :=
∑^ n i = 1
f ( a + ( i − 1)∆ x ) ∆ x
- The right-endpoint approximation to the area under the graph of f ( x ) between x = a and x = b is given by
Rn :=
∑^ n i = 1
f ( a + i ∆ x )∆ x
- The midpoint approximation to the area under the graph of f ( x ) between x = a and x = b is given by
Mn :=
∑^ n i = 1
f
a +
i −
∆ x
∆ x
Example 4.5. Compute the left-endpoint, right-endpoint, and midpoint approximations to the area under the graph of f ( x ) = x^2 between x = 0 and x = 3 with n = 3.
Definite Integral
Suppose f is a function defined on the interval [ a , b ]. The definite integral of f from x = a to x = b is given as (^) ∫ b a
f ( x ) d x = (^) n lim→∞
∑^ n i = 1
f ( xi )∆ x ,
provided this limit exists. If it does exist, then f ( x ) is said to be integrable.
Remark 4.6. If f ( x ) is integrable, then the limit of the Riemann sum will be the same regardless of the sample points xi chosen on each subinterval.
For f ( x ) ≥ 0, a Riemann sum approximates the area under the graph of f ( x ) and above the x -axis. However, it is possible that the terms f ( xi )∆ x in a Riemann sum is negative, which occurs when f ( xi ) < 0. WAITTTTTTT, we know damn well that area cannot be negative, so does this mean we just break math..........??? The precise geometric meaning is that,
The definite integral gives the signed area of the region between the graph of f ( x ) and the x -axis. ∫ (^) b
a
f ( x ) d x =
Example 4.7. This geometrical interpretation of the integral as a signed area allows us to compute
certain definite integrals geometrically. Evaluate both
− 1
1 − x^2 d x and
− 2
x + 1 d x.
Example 4.9. Find
1
( x^2 − x ) d x by taking the limit of the right-endpoint approximations.
Solution: The width is ∆ x =
n
n , and the points xi has the formula
xi = 1 + i ∆ x = 1 + 3 i n , i = 0, 1,... , n.
Note that x 0 = 1 +
n = 0 and xn = 1 +
3 n n = 1 + 3 = 4. The right-endpoint approximation is
Rn =
∑^ n i = 1
f ( xi )∆ x
∑^ n i = 1
f
3 i n
n
∑^ n i = 1
[(
3 i n
3 i n
)] (
n
) [
Since f ( x ) = x^2 − x.
]
∑^ n i = 1
[(
3 i n
3 i n
)] (
n
) [
Factor out
3 i n
]
∑^ n i = 1
3 i n
3 i n
n
∑^ n i = 1
3 i n
9 i n^2
∑^ n i = 1
[
9 i n^2
27 i^2 n^3
] [
Multiply 9 i n^2 into the first parenthesis.
]
n^2
( (^) n ∑ i = 1
i
n^3
( (^) n ∑ i = 1
i^2
) [
Factor out the constants
n^2 and
n^3
]
n^2
n ( n + 1) 2
n^3
n ( n + 1)(2 n + 1) 6
) [
Use the Special Sum Formulas.
]
9( n + 1) 2 n
27( n + 1)(2 n + 1) 6 n^2
Thus ∫ (^4)
1
( x^2 − x ) d x = (^) n lim→∞ Rn = (^) n lim→∞
[ (^) 9( n + 1) 2 n
27( n + 1)(2 n + 1) 6 n^2
]
n^ lim→∞
9 n + 9 2 n
n^ lim→∞
27(2 n^2 + 3 n + 1) 6 n^2
n^ lim→∞
9 n 2 n
n^ lim→∞
54 n^2 6 n^2
Terminology and properties of definite integrals
Below we list some properties of definite integrals.
∫ (^) a
a
f ( x ) d x = 0
∫ (^) b
a
f ( x ) d x = −
∫ (^) a
b
f ( x ) d x
- If f is integrable on an interval containing the points a , b , c , then ∫ (^) b
a
f ( x ) d x =
∫ (^) c
a
f ( x ) d x +
∫ (^) b
c
f ( x ) d x
no matter what the order of a , b , c is.
- If f is continuous on [ a , b ], then f is integrable on [ a , b ].
Example 4.10. Not all functions are integrable! Consider the following function on the interval [0, 1], defined by
f ( x ) =
x if 0 < x ≤ 1, 0 if x = 0.
Example 4.12. Find the following derivatives.
(a) d d x
(∫ (^) x
0
t^2 d t
(b) d d x
(∫ (^) x
1
sin t 1 + t d t
(c) d d x
x
p u du
(d) d d x
x^3 2
t t^4 + 1
d t
(e) d d x
cos x
t^5 d t
Example 4.13. Find a formula for
g ( x ) =
∫ (^) x
1
t^4 d t.
Hint: Note that FTC1 gives us g ′( x ). What is g (1)?
Solution: Applying FTC1 to g ( x ) gives
g ′( x ) = d d x
(∫ (^) x
1
t^4 d t
= x^4.
This says that g ( x ) is the general antiderivative of x^4 , i.e.
g ( x ) =
g ′( x ) d x =
x^4 d x = x^5 5
+ C.
For x = 1, g (1) =
1
t^4 d t = 0.
This means that 0 = g (1) =
+ C =⇒ C = −
Hence g ( x ) = x^5 5
We just demonstrated that it is possible to compute definite integrals using antiderivatives.
(d)
∫ (^) π /
0
2 cos(2 x ) d x
The following is merely a restatement of the FTC2. It has the benefit of being phrased in a way that is useful in applications in the physical and natural sciences.
Net Change Theorem
The integral of a rate of change is the net change: ∫ (^) b
a
F ′( t ) d t = F ( b ) − F ( a ).
Example 4.15. If an object moves along a straight line with position function s ( t ) and velocity v ( t ) = s ′( t ), then the integral of the velocity is the change in position, i.e. ∫ (^) b
a
v ( t ) d t =
∫ (^) b
a
s ′( t ) d t = s ( b ) − s ( a ).
Suppose a particle moves along a straight line with velocity given by v ( t ) = t^3 − 5 t^2 + 6 t. If s (0) = 2, where is the particle at t = 3?
The substitution rule
Essentially, FTC2 tells us that the key to evaluating definite integrals is finding an antiderivative of the integrand. So far we have only looked at fairly simple functions, what about scary, complicated functions? The integration technique of substitution is “undoing” the Chain Rule in disguise.
Substitution Rule
Suppose g ( x ) is a differentiable function whose range is an interval I and f ( x ) is continuous on I. If F is an antiderivative of f , then ∫ f ( g ( x )) g ′( x ) d x =
f ( u ) du = F ( u ) + C = F ( g ( x )) + C.
Example 4.16. Find the following indefinite integrals.
(a)
∫ (^) p 3 x + 1 d x
(b)
∫ (^) cos θ sin^3 θ
dθ
When using the Substitution Rule to evaluate a definite integral, we can
- either use substitution to find the indefinite integral, i.e. the antiderivative, then evaluate this antiderivative at the given endpoints, or
- we can use the substitution u = g ( x ) to change the limits of integration, i.e. ∫ (^) b
a
f ( g ( x )) g ′( x ) d =
∫ (^) g ( b )
g ( a )
f ( u ) du.
Example 4.18. Evaluate the following definite integrals.
(a)
0
x^2 (2 + x^3 )^5 d x
(b)
∫ p π
0
θ sin( θ^2 ) dθ
(c)
1
x^2 + 1 p x^3 + 3 x
d x
