Evaluate the definite integral
Answers
Answer:
Fundamental Theorem of Calculus, Part II
Suppose
f(
x
)
is a continuous function on
[
a
,
b
]
and also suppose that
F
(
x
)
is any anti-derivative for
f
(
x
)
. Then,
∫
b
a
f
(
x
)
d
x
=
F
(
x
)
|
b
a
=
F
(
b
)
−
F
(
a
)
To see the proof of this see the Proof of Various Integral Properties section of the Extras chapter.
Recall that when we talk about an anti-derivative for a function we are really talking about the indefinite integral for the function. So, to evaluate a definite integral the first thing that we’re going to do is evaluate the indefinite integral for the function. This should explain the similarity in the notations for the indefinite and definite integrals.
Also notice that we require the function to be continuous in the interval of integration. This was also a requirement in the definition of the definite integral. We didn’t make a big deal about this in the last section. In this section however, we will need to keep this condition in mind as we do our evaluations.
Next let’s address the fact that we can use any anti-derivative of
f
(
x
)
in the evaluation. Let’s take a final look at the following integral.
∫
2
0
x
2
+
1
d
x
Both of the following are anti-derivatives of the integrand.
F
(
x
)
=
1
3
x
3
+
x
and
F
(
x
)
=
1
3
x
3
+
x
−
18
31
Using the Fundamental Theorem of Calculus to evaluate this integral with the first anti-derivatives gives,
∫
2
0
x
2
+
1
d
x
=
(
1
3
x
3
+
x
)
∣
∣
∣
2
0
=
1
3
(
2
)
3
+
2
−
(
1
3
(
0
)
3
+
0
)
=
14
3
Much easier than using the definition wasn’t it? Let’s now use the second anti-derivative to evaluate this definite integral.
∫
2
0
x
2
+
1
d
x
=
(
1
3
x
3
+
x
−
18
31
)
∣
∣
∣
2
0
=
1
3
(
2
)
3
+
2
−
18
31
−
(
1
3
(
0
)
3
+
0
−
18
31
)
=
14
3
−
18
31
+
18
31
=
14
3
Evaluate the definite integral
Answer
Represented as ∫f(x) dx
It has an upper limit and lower limit
We can deferential the terms from the given integral
Like sin X= cos X
Cos X=-sin X etc
The terms will change according to the question