Aim: How do we evaluate definite integrals analytically?
Often the function inside the definite integral doesn’t represent lines or semi-circles. In this lesson, we
would explore evaluating definite integral analytically.
1
EX: Evaluate
 x( x  2)dx
0
1
 x3
4
2
(
x

2
x
)
dx

 x  
0
3
0 3
1
2
Note: the upper bound goes in first, then the lower bound.
EX1: Evaluate each definite integral
1
1
dx
a)  x 
x
0
4
b)  2e x dx
2
4
1
x
0
1
x
 1
1
5

dx    2 x1/2     2    0  
2
x

2
0  2
2
 2e dx  2e
x
x
2
  2e 2  2e
1
2
One of the great applications of the definite integral is that we can find area of a region that is
abnormally shaped.
1
and the x-axis on the interval [1, e] .
x
Even though the region is not a shape in which we know the area formula. However, the area is equal to
the value of the definite integral, which we can find analytically.
EX2: Find the area bounded by the graph of g ( x)  2 
e
1
e
A   (2  )dx   2 x  ln x 1  (2e  ln e)  (2  ln1)  2e  1
x
1
Note: find the area of the bounded region  evaluating the definite integral.
This method of evaluating the definite integral is derived from a theorem.
Fundamental Theorem of Calculus: If a function f is continuous on [a, b] and f is the antiderivative of
b
f ' on [a, b], then  f '( x)dx  f (b)  f (a)
a
This is the heart of calculus. It relates the derivative function and the regular function. We already
know its geometric application in finding area. In the next two lessons, we would explore its other
application, which is calculating total change when given a rate.
For example: Let f be a function that is defined
for all x  0 and the derivative function
1
is f '( x )  . What is f (2) if we know f (e)  1 ?
x
e
1
 x dx  f (e)  f (2)
2
ln x 2  1  f (2)
e
According to the FTC:
ln e  ln 2  1  f (2)
 ln 2   f (2)
f (2)  ln 2
EX3: If f '( x ) 
2
and f ( e )  5 , what is f (e) ?
x
According to the FTC:
e
2
 x dx  f (e)  f ( e )
e
 2 ln x  e 
e
f ( e)  5
2 ln e  2 ln e  f (e)  5
2  1  f ( e)  5
f (e)  6
EX4: Let f be a function that is defined everywhere and the derivative function is f '( x)  2 x . What is
f (2) if we know f (1)  1 ?
2
 2 xdx  f (2)  f (1)
1
2
 x 2   f (2)  1
1
4  1  f (2)  1
4  f (2)
The power of derivative is that given the function, we can find its rate at any point, provided it exists.
The power of the Fundamental Theorem of Calculus lets us calculate total change in a quantity when
given the rate of change of the quantity.
These two areas are the heart of calculus.
How do we know if something is a rate?
First case: it can be that obvious: the question states that the rate …
Second case: from its unit. It’s always presented in something per something
Third case: velocity and acceleration are rates.
EX5: Village C has a population of 5000 in year 3 and is growing at the rate modeled by r (t )  100t
people per year till year 7.
a) What is the population change from year
t  3 to t  7 ?
b) What is the population at year t  7 ?
7
7
 (100t )dt  50t
|  50(49  9)  2000 people
2 7
3
 (100t )dt  p(7)  p(0)
3
7
3
p (7)   (100t )dt  5000  7000 people
3
EX6: During a spring runoff, a reservoir receives water at the rate of f (t )  4e20.3t thousands of gallons
per day, where t is the number of days since the runoff began. The reservoir has 500 thousand gallons of
water in it when the runoff began.
a) Write the definite integral that represents the
number of gallons of water entered the reservoir in
the first 30 days.
b) Write the definite integral that represents the
number of gallons of water that were in the
reservoir after 30 days.
30
30
2  0.3t
 4e dt
 4e
0
2  0.3t
dt  500
0
EX7: A cup of coffee at 90 is put into a 20 room when t  0 . The coffee’s temperature at time t
minutes is changing at the rate of r (t )  7e0.1t C per minute. Write the expression that represents the
coffee’s temperature when t  10 .
10
 7e
0.1t
dt  F (10)  F (0)
0
10
F (10)   7e 0.1t dt  90
0
EX8: A metal rod is made of material of varying density. The density is f ( x)  1  sin(
x
) grams/cm at
4
the point x cm from the left end. Write the expression that represents the weight of the leftmost 2 cm of
the rod.
2
 (1  sin
x
4
0
) dx
Of all these cases, it’s obvious that rate is given. The subtle case in which the FTC is applied deals with
velocity and position, acceleration and velocity.
EX9: The position function of an object in feet is given by x(t )  t 3  6t 2  9t  2, where t is measured in
seconds.
a) Find the change in position over the interval [0,
5]
5
t4

35
3
2
(
t

6
t

9
t

2)
dt

  2t  4.5t  2t  
0
4
0 4
5
3
2
b) If the object was at x  1 initially, what is its
position after 5 seconds?
35
x
 1  9.75
4
c) Find the total distance the object traveled on the interval [0, 5]
5
55
3
2
0 | t  6t  9t  2 | dt  4
EX10: For 0  t  12 , a particle moves along the x-axis. The velocity of the particle at time t is given by

v(t )  cos t . The particle is at position x = 2 at time t = 0.
6
a) Find the position of the particle at time t = 4.
b) Write the expression that represents the total
distance traveled by the particle from 0  t  6 .
2

0 6 cos tdt  x(2)  x(0)
 2
| cos t | dt
6 0
2
2




x(4)  2   cos tdt  2   sin t   2  (sin 2)
6
6
6
0
0
dy
 9.81t meters per second. An
dt
object is dropped from a height of 200 meters. What’s its position after 2 seconds?
EX11: an object in free fall along the y-axis has a velocity given by
2
 9.81tdt  y(2)  y(0)
0
2
 9.81t 2 
y (2)  200   9.81tdt  200  
  200  2(9.81)  180.38
 2 0
0
2
Note that position and velocity differs by one derivative. The FTC can be applied whenever the two
functions differs by one derivative
3
ft / sec2 . The initial
t
velocity is 2 feet per second. What is the velocity of the rocket at t  80 second?
EX12: A rocket launched upward has an acceleration modeled by a(t ) 
81
 3t
1/2
dt  v(81)  v(0)
0
81
v(81)   3t
0
81
1/2
dt  2  3t1/2 (2)   2  6(9)  2  56 ft / sec
0