WEEK #10:
Integrals
Integration by Parts, Definition and Application of
Goals:
• Integration by Parts
• Riemann Sums
• Definite Integrals
• Fundamental Theorem of Calculus
Textbook reading for Week #10: Read Sections 4.3, 4.4, 4.5
2
Integration by Parts
From Section 4.3
So far in studying integrals we have used
• direct anti-differentiation, for relatively simple functions, and
• integration by substitution, for some more complex integrals.
However, there are
Z many integrals that can’t be handled with these techniques.
Try to evaluate
xe4x dx.
Week 10 – Integration by Parts, Definition and Application of Integrals
3
This particular integral can be evaluated with a different integration technique,
integration by parts. This rule is related to the product rule for derivatives.
Expand
d
(uv) =
dx
Integrate both sides with respect to x and simplify.
Express
Z
dv
dx relative to the other terms.
u
dx
4
Integration by Parts
For short, we can remember this formula as
Z
Z
udv = uv − vdu
Integration by parts:
• Choose a part of the integral to be u, and the remaining part to be dv
• Differentiate u to get du
• Integrate dv to get v
Z
Z
• Replace u dv with uv − vdu
• Hope the new integral is easier to evaluate
Week 10 – Integration by Parts, Definition and Application of Integrals
Use integration by parts to evaluate
Z
5
xe4x dx.
6
Here is a problem from the homework that will test both the techniques and understanding. (Variant of Section 4.3 #35)
Example: Suppose the mass M of a toad grows according to the differential
equation
dM
= t2e−3t with M (0) = 0
dt
Question: When is this toad growing fastest?
A. When M (t) is largest.
dM
B. When
is largest.
dt
C. At t = 0.
dM
D. At some point where
= 0.
dt
d 2 −3t
te
= 0.
E. At some point where
dt
Week 10 – Integration by Parts, Definition and Application of Integrals
dM
= t2e−3t with M (0) = 0
dt
Find the value of t when M is growing most quickly.
7
8
dM
= t2e−3t with M (0) = 0
dt
Find the mass of the toad at t = 1.
Week 10 – Integration by Parts, Definition and Application of Integrals
9
Integrals as Sums
From Section 4.4
So far we’ve needed integration to undo the effect of a derivative. For example,
dy
= t2 + 4e2t
dt
means
Z
y = t2 + 4e2t dt
=
However, there is another interpretation of the integral that can be equally important and useful: the integral as a sum or area.
10
Consider an object whose velocity is given by
v = 1 + t2
Use a ∆x and ∆t argument to estimate the distance traveled between t = 0
and t = 0.2 seconds.
Week 10 – Integration by Parts, Definition and Application of Integrals
Estimate the distance traveled between t = 0.2 and t = 0.4.
11
12
Here is a graph of v(t). How would the distances you just calculated appear
on this graph?
Velocity
2
1
0
0.2
0.4
0.6
Time
0.8
1
If you were to continue in the same manner, show on the graph how you would
estimate the total distance covered between t = 0 and t = 1.0.
Week 10 – Integration by Parts, Definition and Application of Integrals
13
Sketch on the following graphs two other ways we could use other rectangles
to approximate the total distance traveled between t = 0 and t = 1.0.1
2
Velocity
Velocity
2
1
0
0.2
0.4
0.6
Time
0.8
1
1
0
Give a name to each of these approximations.
1
More sophisticated approximations are introduced in the assignment.
0.2
0.4
0.6
Time
0.8
1
14
On the graph below, what area represents the exact distance traveled between
t = 0 and t = 1.0?
Velocity
2
1
0
0.2
0.4
0.6
Time
0.8
1
Week 10 – Integration by Parts, Definition and Application of Integrals
15
Riemann Sum and Summation Notation
The original rectangles we used have the advantage that they are easy to use in
area calculations.
Using the function f (t) = 1 + t2 to simplify calculations, write out long-hand
the estimated distance using the left-hand estimate. Go from t = 0 to t = 1,
using ∆t = 0.2.
16
Construct the same for approximation using the right-hand estimate.
Week 10 – Integration by Parts, Definition and Application of Integrals
17
Define ∆t = 0.2, and t0 = 0, t1 = 0.2, . . . , ti = 0 + i∆t, . . . , t5 = 1.0. Write out
both distance estimates using summation notation.
These sums, approximating an area using terms of the form (∆t)f (t), are
called Riemann Sums.
18
3t
between t = 0
Example: Express the area under the graph of a(t) =
3+t
and t = 10 as a Riemann Sum, using 5 intervals. Use a left-hand estimate.
Week 10 – Integration by Parts, Definition and Application of Integrals
3t
a(t) =
3+t
Estimate the total area.
19
20
Sketch the interpretation of this total area on the graph below.
3
a(t)
2
1
0
2
4
t
6
8
10
Week 10 – Integration by Parts, Definition and Application of Integrals
21
How could we make our rectangular approximation a better approximation of
the exact area under the graph?
Let n be the number of intervals used between t = 0 and t = 10. Express ∆t,
ti, and the total sum using n.
22
How could we express the exact area under the graph using this sum?
We define this exact area calculation as the definite integral:
Week 10 – Integration by Parts, Definition and Application of Integrals
23
Write down the integral that represents the amount of oxygen absorbed by a
5t
lung from t = 0 to t = 5 seconds, if it is being absorbed at a rate of r(t) =
2+t
mol/second.
24
Write down the integral that represents the change in the amount of salt in a
lake from t = 30 to t = 60, if
• the rate of salt flowing in is rin = 4 kg/day
• the rate of salt flowing out is rout = 3 − e−0.02t kg /day
Week 10 – Integration by Parts, Definition and Application of Integrals
25
The Definite Integral and Anti-Derivatives
We will now tie together our two recent ideas: anti-derivatives of functions, and
the specific areas under curves.
• The indefinite integral, a function relating to anti-derivatives:
dy
= f (t)
dt
is equivalent to
Z
y = f (t) dt
• The definite integral, a number representing the change in a function y between two time values, given the rate, r(t):
Z b
r(t) dt
Change in y between t = a and t = b =
a
where r(t) is the rate of change or accumulation
26
Since we’re using the same symbol for both, they must clearly be related. The
relationship is defined by the
Fundamental Theorem of Calculus
Fundamental Theorem of Calculus
Suppose f is continuous. If you can find an antiderivative F (x) of f (x); that is, a
function F (x) such that F 0(x) = f (x), then
Z b
f (x)dx = F (b) − F (a)
a
In other words, if we can find an antiderivative F (x), then calculating the value of
the integral requires a simple evaluation of the anti-derivative F (x) at two
points. This is much easier than computing an area using finite Riemann sums,
and also provides an exact value of the integral instead of an estimate.
Week 10 – Integration by Parts, Definition and Application of Integrals
27
Consider an object which whose velocity is given by
v = 1 + t2 m/s
Set up the definite integral that defines the distance traveled over 1 second.
Approximate the value of this integral using a Riemann sum with 5 intervals,
using the left-hand approximation.
28
Compute the exact value of the integral using the Fundamental Theorem of
Calculus.
Week 10 – Integration by Parts, Definition and Application of Integrals
29
Use the Fundamental Theorem to evaluate the following three definite integrals:
Z 3
e−0.1t dt
• I1 =
0
• I2 =
• I3 =
Z
Z
5
e−0.1t dt
3
5
e−0.1t dt
0
30
Using a sketch of the graph, explain why the sum of the first two integrals
equals the third.