Calculus I, Section 5.3, #2
The Fundamental Theorem of Calculus
Let g(x) = Rx
0f(t) dtwhere fis the function whose
graph is shown at right.1
t
y
0 1 4 6
1f
(a) Evaluate g(x)for x= 0,1,2,3,4,5, and 6.
g(0) = R0
0f(t) dt= 0 from the property R0
0f(x) dx= 0.
g(1) = R1
0f(t) dt. From the graph of f, we see that fis nonnegative on the interval [0,1], so the
integral is equal to the area between the graph and the x-axis. This region is a triangle, so the
area = 1
2·1·1 = 1
2. Thus, g(1) = 1
2.
g(2) = R2
0f(t) dt. From the graph of f, we see that fis nonnegative on the interval [0,1], and negative
on [1,2]. We know that from 0 to 1, the integral is 1
2. On the interval [1,2] the area is again 1
2, but the
function is negative, so the integral is −1
2. Thus g(2) = 1
2−1
2= 0.
g(3) = R3
0f(t) dt. Continuing the same reasoning–integral is positive if function is above the x-axis,
integral is negative if function is below the y-axis–we can see g(3) = 1
2−1
2−1
2=−1
2.
g(4) = R4
0f(t) dt=1
2−1
2−1
2+1
2= 0.
g(5) = R5
0f(t) dt=1
2−1
2−1
2+1
2+3
2=3
2.
g(6) = R6
0f(t) dt=1
2−1
2−1
2+1
2+3
2+5
2=8
2= 4.
(b) Estimate g(7).
g(7) = Z7
0
f(t) dt
=Z6
0
f(t) dt+Z7
6
f(t) dt
and since the area under the graph from 6 to 7 is ≈2.2,
= 4 + 2.2
= 6.2
(c) Where does ghave a maximum value? Where does it have a minimum value?
ghas a maximum value of 6.2 that occurs at x= 7. ghas a minimum value of −1
2that occurs at
x= 3.
1Stewart, Calculus, Early Transcendentals, p. 399, #2.
