The Fundamental Theorem of Calculus After it is known that for a function f ( x) continuous on [a, b], the definite integral is defined as



b

a

f ( x) dx  lim  f  xi* x  lim  f (ck )xk for any partition P of [ a, b] , it can be proven that: n

n 

n

P 0

i 1

k 1

The Fundamental Theorem of Calculus If f ( x) is continuous on [a, b] and A( x) is any antiderivative of f ( x) , then x d x 1. If g ( x)   f (t ) dt , then g( x)  f ( x) or f (t ) dt  f ( x) if a < x < b a dx a 2.



b

f ( x) dx  A(b)  A(a)

a

Note I:

The second form of part 1 requires an application of the chain rule in the following case: d u ( x) f (t ) dt  f  u ( x)  u( x) . dx a

x3

a

Note II:

sin  t 2  dt  A  x3   A(a) by part 2 above where A(t ) is any antiderivative of sin  t 2  .

Taking the derivative of both sides, we have

Working on the right side, we have

d x3 d d sin  t 2  dt  A  x3   A(a) .  dx a dx dx

d A(a)  0 since A(a) is a constant. dx





2 d A  x3   sin  x3  3x 2 since the derivative of the antiderivative A is the dx original function and the 3x 2 term comes from application of the chain rule. d x3 Thus sin  t 2  dt  sin  x 6  3x 2 is an illustration of Note I above.  a dx

We also have

Examples: 1. Part 2: 2. Part 2:



5

2

 x  3x

5



2



5

2 3 dx  x  x

2

 2

 (2) 2  52 245  53    (2)3    2 2  2 

 2 x  sin( x)  dx  x  cos( x) 2

2

   2  99 2  (5 )2  cos(5 )      cos    1 2   2  4    2 5

 

3. Part 1:

d 3 x2 cos0.5 (t ) dt  cos0.5  3x 2  6 x  2 dx

4. Part 1:

4 4 d x3 sin  t 4  dt  sin  x3  3x 2  sin  x 2  2 x  x 3x sin  x12   2sin  x8  2  x dx













Examples continued………… 5. If g ( x)   f (t ) dt on [3, 2] , then g ( x) is “related” to the area(s) bounded by the graph of f ( x) , x

2

the x-axis, and the lines x = –2 and x . We use a diagram consisting of a straight line and a semi-circle of radius 1 to represent a continuous f ( x) and illustrate the application of areas in evaluating definite integrals. The graph below shows f ( x) . NOTE: In attempting the problems below, keep in mind that because g ( x)   f (t ) dt , x

2

we must also have that g( x)  f ( x) .

f ( x)

Problems related to example 5

(justify your answers):

(a)

Find g (0) .

(b)

Find g (3) .

(c)

Find g (2) .

(d)

Find relative extrema for g ( x) .

(e)

Where is g ( x) concave downward?

(f)

At what value(s) of x does g ( x) have an inflection point?

(g)

(i) (ii)

At what value(s) of x is g( x) undefined? At what value(s) of x is g ( x) undefined?

Problem: Using the information above found in example 5, sketch a graph of g ( x) . The solutions for example 5: f (x)

A3

A1 A2

SOLUTIONS: We use the modified diagram above. (a)

To calculate g (0)   f (t ) dt , we note that A1 = A2 and that this integral is A1 – A2 = 0.

(b)

To calculate g (3)   f (t ) dt , we note that

0

2

3

2



3

2

2

f (t ) dt   f (t ) dt  A3  (1  0.5)  1.5 3

because A3 is the combination of a rectangle with area 1 and a triangle with area 0.5.

(c)

To calculate g (2)   f (t ) dt we separate into intervals as follows: 2

2



2

2

f (t ) dt   f (t ) dt   f (t ) dt  0   f (t ) dt . 0

2

2

2

0

0

We must deal with the areas between the curve and the x-axis between 0 and 2. We need the area (and to make it negative) between f ( x) and the x-axis. The plan is to calculate the area of the entire rectangle and then subtract the area of the semi-circle. We have 2 – (1/2)(1)2 because the semi-circle is half the area of a circle with radius 1. Thus the final answer is –(2 – (1/2)(1)2) = –2 + (/2).

(d)

Extrema occur where the sign of g( x) changes. We know that g( x)  f ( x) from the fundamental theorem. Examination of the graph of f ( x)  g( x) shows a sign change from + to – at x = –1, indicating a relative (maybe absolute?) maximum at x = –1.

(e)

g ( x) is concave downward where g ( x)  f ( x) is negative. To find where g ( x)  f ( x)  0 , we look at the slopes of f ( x) which are negative for –3 < x < 0 and again for 1 < x < 2.

(f)

g ( x) has inflection points where g ( x) changes sign. Since g( x)  f ( x) , we know that g ( x)  f ( x) , which values can be found from the slopes of f ( x) . The slopes of f ( x) change from – to + at x = 0 and from + to – at x = 1. There are no other changes.

(g)

(i) g( x) is defined everywhere since g( x)  f ( x) and f ( x) is continuous on [–3, 2]. (ii) g ( x) is defined wherever f ( x) exists since g ( x)  f ( x)  g ( x)  f ( x) . f ( x) is given by the slopes of f, but the slopes are not the same to the left and right of 0. f ( x)  f (0) f ( x)  f (0) More formally, lim  1  lim  x0 x0 x0 x0 f ( x) does not exist at x = 0; therefore g ( x) does not exist at x = 0.

Problem: Using the information above found in example 5, sketch a graph of g ( x) . Concave down on (–3, 0) and (1, 2) g ( x)

g (–2) = g (0) = 0