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The integral, along with the derivative, are the two fundamental building blocks of calculus. Put simply, an integral is an area under a curve; This area can be one of two types: definite or indefinite. Definite integrals give a result (a number that represents the area) as opposed to indefinite integrals, which are represented by formulas. Indefinite Integrals (also called antiderivatives) do not have limits/bounds of integration, while definite integrals do have bounds. Watch the video for a quick introduction on to definite integrals, or read on below for more definitions, how-to articles and videos.
- Additive Interval Property
- Divergent Integrals
- Double Integrals
- Isotropic / Anisotropic Definition, Examples
- Fundamental Theorem of Calculus
- Fresnel Integrals
- Gauge Integral
- Improper Integrals
- Integral Bounds / Limits of Integration
- Integral Function
- Integral Kernel
- Integral Operator
- Iterated Integrals
- Lebesgue Integration
- Line Integral
- Mean Value Theorem for Integrals
- Multiple Integrals: Definition, Examples
- Numerical Quadrature (Numerical Integration)
- Order of Integration
- Ordinary Integral
- Product Integral
- Riemann Integral
- Singular Integral: Simple Definition
- Sum Rule
- Triple Integral (Volume Integral)
General How-To Integrals
- Area Between Curve and Y-Axis
- Area Function.
- Indefinite Integrals of power functions
- Constant Rule of Integration
- Finding definite integrals
- Integration of Even and Odd Functions
- Integration Using Long Division: Definition, Examples
- Integration by parts
- Integration by Separation
- Log Rule for Integration
- Integral of a Natural Log
- Integrate with U Substitution
- How to Integrate Y With Respect to X
- Method of Partial Fractions
- Rationalizing Substitutions
- Tabular Integration (The Tabular Method)
- Trig Substitution
Integral Calculus Advanced Problem Solving
- Find Total Distance Traveled (opens in new window)
- How to find the volume of an egg(opens in new window)
- How to prove the volume of a cone(opens in new window)
- How to find the area between two curves
An integral kernel is a given (known) function of two variables that appears in an integral equation; This unknown function appears with an integral symbol.
The kernel is symmetric if If K(x, y) = K(y, x).
Notation for the Integral Kernel
The kernel is denoted by K(x, y):
As well as K(x, y), you might also see slightly different notation depending on what variables are used in the equation. For example:
- A(x, y),
- Ta(x, y), or
- K(x, x′).
What notation is used sometimes depends on exactly what the kernel is representing. Some specific representations include (Wolf, 2013):
- A translation operation 𝕋a: Ta(x, y),
- Inversions: I0(x, y),
- The operator of differentiation: ∇(x, y).
Avramidi (2015) describes an integral operator on the Hilbert space L2 ([a, b]) as follows:
Where the function K(x, x′) is the integral kernel. Note that the author also uses “K” on the left hand side of the equation to denote the operator, a distinction that “…shouldn’t cause any confusion because the meaning of the symbol is usually clear from the context”.
Integral Kernel, or Symbol?
Although the term “integral kernel” is widely used, many authors prefer the alternate term symbol instead, to avoid confusion with many other meanings for the word kernel in mathematics. For example, in geometry, a kernel is the set of points inside a polygon from where the entire boundary of the polygon is visible; In statistics, a kernel is a weighting function used to estimate probability density functions for random variables in kernel density estimation.
Integral Kernel: References
Avramidi, I. (2015). Heat Kernel Method and its Applications 1st ed. Birkhäuser
Paulsen, V. & Raghupathi, M. (2016). An Introduction to the Theory of Reproducing Kernel Hilbert Spaces.
Wolf, K. (2013). Integral Transforms in Science and Engineering. Springer Science & Business Media.
Generally speaking, an integral operator is an operator that results in integration or finding the area under a curve. It is defined by the integral symbol: ∫.
The integral operator is sometimes called a standard integral operator  to separate it from special cases used in complex analysis, operator theory and other areas of mathematical analysis.
Special Cases of Integral Operator
The first operators appeared at the beginning of the 20th century, at the beginning of the theory of complex-variable functions. Many operators have been developed over the years and are defined very narrowly for special circumstances. They include:
- Alexander integral operator: Defined for a class of analytic functions on the unit disk D :
- Fredholm operator: Arises in the Fredholm equation, an integral equation where the term containing the kernel function has constants as limits of integration.
- The Volterra integral equation is similar to the Fredholm equation, except that it has variable integral limits.
- A variety of pseudo-differential operators are used to study elliptic differential equations. These operators, as well as Fourier integral operators, make it possible to handle differential operators with variable coefficients in about the same way as differential operators with constant coefficients using Fourier transforms .
Integral Operator: References
 Anderson, A. SOME CLOSED RANGE INTEGRAL OPERATORS ON SPACES
OF ANALYTIC FUNCTIONS. Retrieved April 23, 2021 from: http://www2.hawaii.edu/~austina/documents/research/aatgpaper2.5.1.pdf
 Gao, C. (1992). On the Starlikeness of the Alexander Integral Operator. Proc. Japan Acad. 68. Ser. A.
 Hormander, L. Fourier Integral Operators, I. Retrieved April 23, 2021 from: https://projecteuclid.org/journalArticle/Download?urlid=10.1007%2FBF02392052
Finding the area between two curves in integral calculus is a simple task if you are familiar with the rules of integration (see indefinite integral rules). The easiest way to solve this problem is to find the area under each curve by integration and then subtract one area from the other to find the difference between them. You may be presented with two main problem types. The first is when the limits of integration are given, and the second is where the limits of integration are not given.
Area Between Two Curves: Limits of Integration Given
Example problem 1: Find the area between the curves y = x and y = x2 between x = 0 and x = 1.
Step 1: Find the definite integral for each equation over the range x = 0 and x = 1, using the usual integration rules to integrate each term. (see: calculating definite integrals).
Step 2: Subtract the difference between the areas under the curves. You’ll need to visualize the curves (sketch or graph the curves if you need to); you’ll want to subtract the bottom curve from the top one. The curve on top here is f(x) = x, so:
1⁄2 – 1⁄3 = 1⁄6.
Limits of Integration NOT Given
Example problem: Find the area between the curves y = x and y = x2.
Step 1: Graph the equations. In most cases, the limits of integration will be clear, especially if you’re using a TI-calculator with an Intersection feature (just find the intersections of the two graphs). If you can find the intersection by graphing, skip to Step 3.
Step 2: Find the common solutions of these two equations if you cannot find the intersection by graphing (treat them as simultaneous equations).
Substituting y = x for x in y = x2 gives an equation y = y2, which has only two solutions, 0 and 1.
Putting the values back into y = x to give the corresponding values of x: x = 0 when y = 0, and x = 1 when y = 1. The two points of intersection are (0,0) and (1,1).
Step 3: Complete the steps in Example Problem 1 (limits of integration given) to complete the calculation.
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Integration by separation takes a complicated-looking fraction and breaks it down into smaller parts that are easier to integrate.
For example, the following fraction is challenging (impossible?) to integrate using the usual rules of integration:
However, you can rewrite it as a series of fractions, using algebra:
Simplifying, this becomes:
These fractions can be individually integrated, using the power rule and the common integral ∫1⁄xdx = ln |x|:
Trig substitution helps you to integrate some types of challenging functions:
- Radicals of polynomial functions, like √(4 – x2),
- Rational powers of the form n/2, e.g. (x2 + 1)(3/2).
Although trig substitution is fairly straightforward, you should use it when more common integration methods (like u substitution) have failed.
The technique is very similar to u substitution: you substitute a new term (one made from integer powers of trig functions) in place of the one you have, in order to make the integration easier. At the end, you simply substitute the original function back in.
Why Is Trig Substitution a Last Resort?
Although it’s straightforward, trig substitution requires you to have a lot of background knowledge. Unlike a table of integrals, you can’t just look up an integral for a particular expression. It’s a must that you are able to recognize the trigonometric identities. Let’s look at an example to see why this is so important.
Example question: Integrate
To solve this, you need to consider all of the trig identities to see which would be a good fit. If you aren’t familiar with them, this could be a stumbling block before you’ve even started. In order to solve this particular integral, you need to recognize that it looks very similar to the trig identity
1 + tan2 x = sec2 x.
Here are the solution steps:
Step 1: : Rewrite the expression using a trig substitution (and derivative). The goal here is to get the expression into something you can simplify with a substitution:
Here, I substituted in tan2θ for x.
As the substitution for x has been made, I also had to change the “dx” to represent the derivative of tan2θ (instead of plain old derivative of “x”). So the new “dx” became sec2 θ dθ.
Step 2: : Simplify by using a trig identity. In this example, we’ve been heading towards changing 1 + tan2 x to sec2 x. There’s no magic here—if you chose the correct trig function in Step 1, you should already know which trig identity you’re going to use here:
Step 3: : Simplify using algebra (if possible). For this example, notice that we can cancel out the sec2 in the numerator and denominator,
∫ 1 dθ.
Step 4: Integrate. The integral of a constant function is just the constant * x (or constant * θ) + C, so:
∫ 1 dθ = θ + C
Step 5: Substitute your original term back in. In Step 1, I substituted tan-1 θ for x, so putting that back in gives the solution:
= tan -1 x + C
Useful Background Information
As you may be able to tell from the above example, trig substitution requires you to have some strong background skills in algebra, derivatives, and trigonometric identities.
“…any teacher of Calculus will tell you that the reason that students are not successful in Calculus is not because of the Calculus, it’s because their algebra and trigonometry skills are weak” ~ Jones (2010)
Also extremely helpful:
- Integrals of Trig Functions,
- U Substitution for Trigonometric Functions,
The following table shows how to express one of the common six trig functions as a pair of other trig functions. These may also come in handy:
Trig Substitution: References
Banner, A. (2007). The Calculus Lifesaver: All the Tools You Need to Excel at Calculus (Princeton Lifesaver Study Guides) Illustrated Edition. Princeton University Press.
Jones, J. (2010). Skills Needed for Success in Calculus 1.
Kouba, D. (2017). Finding Integrals Using the Method of Trigonometric Substitution. Retrieved November 9, 2020 from: https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/trigsubdirectory/TrigSub.html
Stephanie Glen. "Integrals / Integral Calculus" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/integrals/
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