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 x0 x0 x0 x0 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