INTEGRATION: ANTIDERIVATIVES AND INDEFINITE INTEGRALS MR.VELAZQUEZ AP CALCULUS
INTRO TO INTEGRATION
Mathematicians spent a lot of time working with the topic of derivatives, describing how functions change at any given instant.
They then sought a way to describe how those changes accumulate over time, leading them to discover the calculation for area under a curve. This is known as integration, the second main branch of calculus.
Finally, Liebniz and Newton discovered the connection between differentiation and integration, known as the Fundamental Theorem of Calculus, an incredible contribution to the understanding of mathematics.
ANTIDERIVATIVES
Before we look at the various uses and techniques of calculating area, we will first examine the concept of the antiderivative.
A function πΉ is an antiderivative of π on an interval πΌ if πΉβ² π₯ = π(π₯) for all π₯ in πΌ.
Note that we call this function an antiderivative, and not the antiderivative.
Itβs easy to see why this is true. Consider the function below:
π π₯ = 3π₯2
πΉ1 π₯ = π₯3 πΉ2 π₯ = π₯3 β 5 πΉ3 π₯ = π₯3 + 27
In this example, πΉ1, πΉ2 and πΉ3 are all valid antiderivatives of π(π₯).
ANTIDERIVATIVES
If πΉ is an antiderivative of π on a given interval πΌ, then πΊ is also an antiderivative of π if and only if πΊ is of the form πΊ π₯ = πΉ π₯ + πΆ for all π₯ in πΌ, where πΆ is a constant.
Function Antiderivative
π π₯ = 2π₯ πΊ π₯ = π₯2 + πΆ
β π₯ =1
π₯2π» π₯ = β
1
π₯+ πΆ
π π₯ = cos π₯ π½ π₯ = sin π₯ + πΆ
NOTATION FOR ANTIDERIVATIVES
ΰΆ±π(π₯) ππ₯ = πΉ π₯ + πΆ
Indefinite Integral Antiderivative
IntegrandVariable of Integration
An antiderivative of π(π₯)
Constant
Itβs important to note that both of these termsβindefinite
integral and antiderivativeβrefer to the same thing.
BASIC INTEGRATION RULES
ΰΆ±0ππ₯ = πΆ
ΰΆ±π ππ₯ = ππ₯ + πΆ
ΰΆ±π π(π₯) ππ₯ = πΰΆ±π π₯ ππ₯
ΰΆ± π π₯ Β± π π₯ ππ₯ = ΰΆ±π π₯ ππ₯ Β±ΰΆ±π π₯ ππ₯
ΰΆ±π₯π ππ₯ =π₯π+1
π + 1+ πΆ, π β β1
Integral of a Zero
Integral of a Constant
Constant Multiple Rule
Sum & Difference
Power Rule
BASIC INTEGRATION RULES
ΰΆ±cos π₯ ππ₯ = sin π₯ + πΆ
ΰΆ±sin π₯ ππ₯ = βcos π₯ + πΆ
ΰΆ±sec2 π₯ ππ₯ = tan π₯ + πΆ
Trigonometric Integrals
ΰΆ±sec π₯ tan π₯ ππ₯ = sec π₯ + πΆ
ΰΆ±csc π₯ cot π₯ ππ₯ = βcπ π π₯ + πΆ
ΰΆ±csc2 π₯ ππ₯ = cot π₯ + πΆ
EXAMPLES
Find the indefinite integrals shown below, using basic integration rules.
ΰΆ±4π₯ ππ₯ ΰΆ±(cos π₯ + 3 sec2 π₯)ππ₯
EXAMPLES
Find the indefinite integrals shown below, using basic integration rules.
ΰΆ± 3π₯4 β 5π₯2 + π₯ ππ₯
EXAMPLES
Find the indefinite integrals shown below, using basic integration rules.
ΰΆ±π₯ + 1
π₯ππ₯
Hint: Rewrite the
integrand as two
fractions.
EXAMPLES
Find the indefinite integrals shown below, using basic integration rules.
ΰΆ±sin π₯
cos2 π₯ππ₯
Hint: Rewrite the
integrand as a product.
USING INITIAL CONDITIONS
We have seen that when we take an indefinite integral, we are actually getting the set of all possible antiderivatives of a function.
But suppose we want to find one particular antiderivative solution. For this, we would need an initial conditionβin most cases this is simply the value of πΉ π₯ for one specific value of π₯.
πΉ π₯ = ΰΆ± 2π₯ β 1 ππ₯ = π₯2 β π₯ + πΆ
Suppose we know that the antiderivative passes through the point (2, 4). We can use this to find the particular solution.
GENERAL SOLUTION
πΉ 2 = 4 β 2 2 β 2 + πΆ = 4 β 4 β 2 + πΆ = 4 β πͺ = π
π π = ππ β π + π PARTICULAR SOLUTION
EXAMPLE
Find the antiderivative of πΉβ² π₯ =1
π₯2for π₯ > 0 which satisfies the condition
πΉ 1 = 0.
CLASSWORK & HOMEWORK
MATH JOURNAL: Donβt forget!
CLASSWORK: INDEFINITE INTEGRALS: Find the following indefinite
integrals. For the last two, find the particular solution using the initial
condition given.
ΰΆ± 12 + π₯ ππ₯
ΰΆ± π₯ +1
2 π₯ππ₯
ΰΆ±π₯2 + 2π₯ β 3
π₯4ππ₯
πΉ π₯ = ΰΆ± 10π₯ β 12π₯3 ππ₯ , πΉ 3 = 2
πΊ π₯ = ΰΆ±2 sin π₯ ππ₯ , πΊ 0 = 1
1.
2.
3.
4.
5.
Homework:
Pg. 255-256, #1-63 (odd)