Calculus- Early Transcendentals, 2021a

6.2. The Fundamental Theorem of Calculus 239
Solution.
1.
2.
3.
4.
∫ b
a
∫ a
b
∫ a
a
∫ c
a
∫ b
∫ b
2 f (x) − 3g(x)dx = 2 f (x)dx− 3 g(x)dx = 2(7) − 3(3)=5.
a
a
∫ b
2g(x)dx = −2 g(x)dx = −2(3)=−6.
a
f (x) · g(x)dx = 0.
∫ b ∫ b
f (x)dx+ f (x)dx = f (x)dx = 7.
c
a
♣
We next evaluate a definite integral using three different techniques.
Example 6.17: Three Different Techniques
Evaluate
∫ 2
0
x + 1 dx by
1. Using FTC II (the shortcut)
2. Using the definition of a definite integral (the limit sum definition)
3. Interpreting the problem in terms of areas (graphically)
Solution. 1. The shortcut (FTC II) is the method of choice as it is the fastest. Integrating and using the
‘top minus bottom’ rule we have:
∫ 2
0
x + 1 dx = x2
2 + x ∣ ∣∣∣
2
=
0
[ 2
2
] [ 0
2
]
2 + 2 −
2 + 0 = 4.
2. We now use the definition of a definite integral. We divide the interval [0,2] into n subintervals of
equal width Δx, and from each interval choose a point x ∗ i . Using the formulas
Δx = b − a
n
and
x i = a + iΔx,
we have
Δx = 2 and x i = 0 + iΔx = 2i
n
n .
Then taking x ∗ i ’s as right endpoints for convenience (so that x∗ i = x i ), we have:
∫ 2
0
n
x + 1 dx = lim n→∞ ∑ f (x ∗ i )Δx
i=1