1
Subject : Cal 221/Math 224 (Integral Calculus)
Instructor : Engr. Nurhassan S. Sappayani, Ed. D.
Time/Day : 8:30‒9:30/10:30‒11:30/1:00–2:00/3:00–4:00 (MWF)
7:30‒9:00/9:00‒10:30 (TTh)
Course : BSCS-2C/BSCS-2D/ BSCpE-2A/BSCpE-2B/BSCpE-2C/BSCpE-2D
MODULE 2
INDEFINITE INTEGRAL
Indefinite Integral is the set of functions F(x) + C, where C is any real number, such
that F(x) is the integral of f(x).
At the end of this section, the student should be able to:
1. learn Theorem 2 of Indefinite Integrals; and
2. Apply the theorem 2 in solving indefinite Integrals.
Section 2 ‒ Theorem 2
In previous Module 7, we found the derivative of exponential and logarithmic
functions. Theorem 2 presents the corresponding integral forms for those derivatives.
Theorem 2 2.1 ∫ 𝒆𝒙 𝒅𝒙 = 𝒆𝒙 + 𝑪
𝟏
2.2 ∫ 𝒅𝒙 = 𝒍𝒏 |𝒙| + 𝑪
𝒙
Notice that (2.2) now gives a way of integrating x –1 . The absolute value symbols are
necessary since the logarithm of a negative number is not defined.
3
3 Factoring , gives us
Example 1 Evaluate ∫ (5𝑒 𝑥 − 𝑥) dx 1
𝑥
3 ( 𝑥 ).
𝑥 3 𝑥 1
Solution ∫ (5𝑒 − 𝑥) dx = 5 ∫ 𝑒 𝑑𝑥 – 3∫ 𝑥 𝑑𝑥
= 5ex – 3 ln|x| + C ans.
2
Example 2 Evaluate ∫(6𝑥 − 3𝑒 𝑥 )𝑑𝑥 Using the Power Rule
6𝑥 2 from Theorem 1.1
Solution ∫(6𝑥 − 3𝑒 𝑥 )𝑑𝑥 = − 3𝑒 𝑥 + 𝐶 and Theorem 2.2
2
= 3x2 – 3ex + C ans.
Recall in the definitions of anti-derivative and indefinite integral, if F is an anti-
derivative of f, then F’(x) = f(x) and
∫ 𝑓 (𝑥 )𝑑𝑥 = 𝐹 (𝑥 ) + 𝐶
Another way to express these two ideas together is
∫ 𝐹 ′ (𝑥 )𝑑𝑥 = 𝐹 (𝑥 ) + 𝐶
This particular notation really emphasizes the idea that differentiation and
integration are “inverse processes.”
We are now ready to move to some examples on integration where the constant of
integration can be evaluated. In other words, we will be finding a specific anti-derivative for
a function.
Example 3 Find f(x) if f”(x) = 2x and f(1) = 5.
Solution To find f(x), we use the fact that
f(x) = ∫ 𝑓 ′(𝑥 )𝑑𝑥
Using the Power Rule
f(x) = ∫ 2𝑥 𝑑𝑥 from Theorem 1.1
2𝑥 2
f(x) = +C
2
f(x) = x2 + C
Since we are also given that f(1) = 5, this gives
f(1) = x2 + C @x=1
5 = (1) + C
2
5 = 1+C
5 – 1 = C or C = 4
So, f(x) = x2 + 4 ans.
Example 4 Find f(x) if f”(x) = 3x2 – 6x + 1 and f(0) = ‒10.
Solution To find f(x), we use theorem 1.4 from Module 1.
f(x) = ∫ 𝑓 ′(𝑥 )𝑑𝑥
f(x) = ∫(3𝑥 2 – 6𝑥 + 1) 𝑑𝑥
3𝑥 3 6𝑥 2
f(x) = − +𝑥+C
3 2
f(x) = x3 ‒ 3x2 + x + C
3
Since we are also given that f(0) = ‒10, this gives
f(0) = x3 ‒ 3x2 + x + C @x=0
‒10 = (0)3 ‒ 3(0)2 + 0 + C
‒10 = 0‒0+0+C
–10 = C or C = ‒10
So, f(x) = x3 ‒ 3x2 + x ‒ 10 ans.
Exercises #2
In exercises 1 through 8, evaluate each indefinite integral. (See Example 1, 2, 3, & 4).
Show your solution.
9 18𝑥 5 1 #5 answer, reduce to
1. ∫ (9𝑒 𝑥 − 𝑥) dx 5. ∫ ( − ) 𝑑𝑥 lowest term. (5 pts.)
6 𝑥
8 4
2. ∫ (4 𝑒 𝑥 + 𝑥) dx 3 pts. each 6. ∫(8𝑥 3 − 6𝑥 2 + 2𝑥) dx, f(‒1) = 0 8 points
3. ∫(10𝑥 + 6𝑒 𝑥 )𝑑𝑥 7. ∫(6𝑥 2 − 12𝑥 + 2) 𝑑𝑥, f(2) = ‒7 8 points
7 24𝑥 2
4. ∫ ( − ) 𝑑𝑥 5 points
𝑥 4
Notice: Answer sheet/s should be put in the expandable envelop (color light pink or any light color). Write legibly
and avoid many erasures on your solution/s. Don’t forget to write your complete name (family name first),
your course, year and section and the name of your instructor (in your expandable envelope and in your
answer sheets).
The following materials to be needed in solving the exercises:
1. Graphing paper
2. Ball pen (black or blue)
Source: Claudia Taylor (Brevard Community College) & Lawrence Gilligan (University of Cincinnati) (1985).
Applied Calculus. Brooks/Cole Publishing Company. Pacific Grove, California
