INTEGRAL CALCULUS | CONCEPT OVERVIEW The topic of INTEGRAL CALCULUS can be referenced on page 28 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: The inverse of differentiation is formally known as the process of INTEGRATION. An INTEGRAL is classified as being DEFINITE or INDEFINITE based on whether it has DEFINED LIMITS. If the integral has limits, it is a DEFINITE INTEGRAL. If the integral does not have limits, then it is an INDEFINITE INTEGRAL. In the case of INDEFINITE INTEGRALS, we will always add a CONSTANT OF INTEGRATION at the end of our INTEGRATION, which represents the undefined range of the function. To solve for this CONSTANT OF INTEGRATION, we typically use INITIAL or BOUNDARY CONDITIONS. The GENERAL FORM of a DEFINITE INTEGRAL can be referenced under the main subject of MATHEMATICS on page 28 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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For a curve represented by the function 𝑦 = 𝑓(𝑥), the area under the curve from 𝑥 = 𝑎 to 𝑥 = 𝑏 can represented as a DEFINITE INTEGRAL, defined as: !
lim
!→!
!
𝑓 𝑥! ∆𝑥! = !!!
𝑓 𝑥 𝑑𝑥 !
Also, ∆𝑥! → 0 for all “𝑖” A table of derivatives and integrals can be referenced on Page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. These general integral equations can be used along with the following methods of integration: A. Integration by Parts B. Integration by Substitution C. Separation by Rational Functions Into Partial Fractions
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INTEGRAL CALCULUS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. !
The area under the curve 𝑦 = between the limits 𝑦 = 2 and 𝑦 = 10 is most close to: !
(HINT: Use the Integrations calculator hack) A. 1.61 B. 2.39 C. 3.71 D. 3.97
SOLUTION: The GENERAL FORM of a DEFINITE INTEGRAL can be referenced under the main subject of MATHEMATICS on page 28 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The area under the curve 𝑓(𝑥) between two limits, 𝑥! and 𝑥! , is equal to “𝐴”, represented generically as:
𝐴=
!!
𝑓 𝑥 𝑑𝑥
!!
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In this problem, we are given the formula:
𝑓(𝑥) =
1 𝑥
We are also given the limits in which we are working, which are defined as: 1 2 1 𝑥= 10 𝑥=
Taking this information and plugging it in to our general formula, we have: !/!
𝐴=
1 𝑑𝑥 !/!" 𝑥
Evaluating the integral between the stated limits, we get:
𝐴 = ln 𝑥
1/2 1/10
This can be expanded to read:
𝐴 = 𝑙𝑛
1 1 − 𝑙𝑛 2 10
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The area under the curve between the limits 𝑦 = 2 and 𝑦 = 10 is: 𝐴 = 1.61 The correct answer choice is A. 𝟏. 𝟔𝟏