Integral

Integration is about finding the surface s, given a, b and y = f(x). The formula for the integral from a  to b, graphed above, is:
    Formula:   [math]\displaystyle{ \int\limits_{a}^{b} f(x)\,dx }[/math]

What is the integral (animation)

In calculus, an integral is the space under a graph of an equation (sometimes said as "the area under a curve"). An integral is the reverse of a derivative, and integral calculus is the opposite of differential calculus. A derivative is the steepness (or "slope"), as the rate of change, of a curve. The word "integral" can also be used as an adjective meaning "related to integers".

The symbol for integration, in calculus, is: [math]\displaystyle{ \textstyle \int_{\,}^{\,} }[/math] as a tall letter "S".

Integrals and derivatives are part of a branch of mathematics called calculus. The link between these two is very important, and is called the fundamental theorem of calculus. The theorem says that an integral can be reversed by a derivative, similar to how an addition can be reversed by a subtraction.

Integration helps when trying to multiply units into a problem. For example, if a problem with rate, [math]\displaystyle{ \left(\tfrac{\text{distance}}{\text{time}}\right) }[/math], needs an answer with just distance, one solution is to integrate with respect to time. This means multiplying in time to cancel the time in [math]\displaystyle{ \left(\tfrac{\text{distance}}{\text{time}}\right)\times\text{time} }[/math]. This is done by adding small slices of the rate graph together. The slices are close to zero in width, but adding them together indefinitely makes them add up to a whole. This is called a Riemann sum.

Adding these slices together gives the equation that the first equation is the derivative of. Integrals are like a way to add many tiny things together by hand. It is like summation, which is adding [math]\displaystyle{ 1+2+3+4....+n }[/math]. The difference with integration is that we also have to add all the decimals and fractions in between.

Another time integration is helpful is when finding the volume of a solid. It can add two-dimensional (without width) slices of the solid together indefinitely—until there is a width. This means the object now has three dimensions: the original two and a width. This gives the volume of the three-dimensional object described.

Riemann integration

If [math]\displaystyle{ f }[/math] is a real-valued function defined on an interval [math]\displaystyle{ [a,b] }[/math] (where [math]\displaystyle{ a\lt b }[/math]), then the Riemann integral of [math]\displaystyle{ f }[/math] over [math]\displaystyle{ [a,b] }[/math] is a certain real number. It may not always be the case that we can find such a real number. If we can, the we say that [math]\displaystyle{ f }[/math] is Riemann integrable over [math]\displaystyle{ [a,b] }[/math].

To define the Riemann integral, we will give the following conditions for an infinite sequence:

[math]\displaystyle{ \mathbf{(i)} }[/math] We construct the [math]\displaystyle{ n }[/math]th term of the sequence as follows:

(1) Split the interval [math]\displaystyle{ [a,b] }[/math] into [math]\displaystyle{ n }[/math] sub-intervals [math]\displaystyle{ [p_{0},p_{1}], [p_{1},p_{2}]...[p_{n-1},p_{n}] }[/math], where [math]\displaystyle{ p_{0}=a }[/math], [math]\displaystyle{ p_{n}=b }[/math], and [math]\displaystyle{ p_{0}\lt p_{1}\lt p_{2}...\lt p_{n-1}\lt p_{n} }[/math].

(2) For each integer [math]\displaystyle{ k }[/math] from [math]\displaystyle{ 1 }[/math] to [math]\displaystyle{ n }[/math], choose an arbitrary point [math]\displaystyle{ x_{k} }[/math] in the interval [math]\displaystyle{ [p_{k-1},p_{k}] }[/math].

(3) Find the sum [math]\displaystyle{ \sum_{k=1}^{n}((x_{k}-x_{k-1})f(x_{k})) }[/math]. This is the [math]\displaystyle{ n }[/math]th term.

[math]\displaystyle{ \mathbf{(ii)} }[/math] If we are given any positive real number [math]\displaystyle{ \delta }[/math], we can find a positive integer [math]\displaystyle{ N }[/math] such that for all positive integers [math]\displaystyle{ n }[/math] greater than [math]\displaystyle{ N }[/math], the length of every sub-interval in step (1) is less than [math]\displaystyle{ \delta }[/math].

If it is the case that every sequence that satisfies conditions [math]\displaystyle{ \mathbf{(i)} }[/math] and [math]\displaystyle{ \mathbf{(ii)} }[/math] is convergent, and converges to the same limit, only then we say that [math]\displaystyle{ f }[/math] is Riemann integrable over [math]\displaystyle{ [a,b] }[/math]. To put it in other words, if we can find real number [math]\displaystyle{ L }[/math] such that every sequence that satisfies conditions [math]\displaystyle{ \mathbf{(i)} }[/math] and [math]\displaystyle{ \mathbf{(ii)} }[/math] is convergent and they all converge to [math]\displaystyle{ L }[/math], then [math]\displaystyle{ f }[/math] is Riemann integrable over [math]\displaystyle{ [a,b] }[/math], and [math]\displaystyle{ L }[/math] is called the Riemann integral of [math]\displaystyle{ f }[/math] over [math]\displaystyle{ [a,b] }[/math].

If we can find a divergent sequence that satisfies both conditions [math]\displaystyle{ \mathbf{(i)} }[/math] and [math]\displaystyle{ \mathbf{(ii)} }[/math], then [math]\displaystyle{ f }[/math] is not Riemann integrable over [math]\displaystyle{ [a,b] }[/math]. If we can find two convergent sequences that satisfy both conditions [math]\displaystyle{ \mathbf{(i)} }[/math] and [math]\displaystyle{ \mathbf{(ii)} }[/math], but they converge to different limits, then [math]\displaystyle{ f }[/math] is not Riemann integrable over [math]\displaystyle{ [a,b] }[/math].

Sum of functions

The integral of a sum of functions is the sum of each function's integral. that is,

[math]\displaystyle{ \int\limits_{a}^{b} [f(x) + g(x)]\, dx = \int\limits_{a}^{b} f(x)\, dx + \int\limits_{a}^{b} g(x)\, dx }[/math].

The proof of this is straightforward: The definition of an integral is a limit of sums. Thus

[math]\displaystyle{ \int\limits_{a}^{b} [f(x) + g(x)]\, dx = \lim_{n \to \infty} \sum_{i=1}^n \left(f(x_i^*) + g(x_i^*)\right) }[/math]

[math]\displaystyle{ = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) + \sum_{i=1}^n g(x_i^*) }[/math]

[math]\displaystyle{ = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) + \lim_{n \to \infty} \sum_{i=1}^n g(x_i^*) }[/math]

[math]\displaystyle{ = \int\limits_{a}^{b} f(x)\, dx + \int\limits_{a}^{b} g(x)\, dx }[/math]

Note that both integrals have the same limits.

Constants in integration

When a constant is in an integral with a function, the constant can be taken out. Further, when a constant c is not accompanied by a function, its value is c * x. That is,

[math]\displaystyle{ \int\limits_{a}^{b} cf(x)\, dx = c \int\limits_{a}^{b} f(x)\, dx }[/math] and

This can only be done with a constant.

[math]\displaystyle{ \int\limits_{a}^{b} c\, dx = c(b-a) }[/math]

Proof is again by the definition of an integral.

Other

If a, b and c are in order (i.e. after each other on the x-axis), the integral of f(x) from point a to point b plus the integral of f(x) from point b to c equals the integral from point a to c. That is,

[math]\displaystyle{ \int\limits_{a}^{b} f(x)\, dx + \int\limits_{b}^{c} f(x)\, dx = \int\limits_{a}^{c} f(x)\, dx }[/math]

if they are in order. (This also holds when a, b, c are not in order if we define

[math]\displaystyle{ \textstyle \int\limits_{a}^{b} f(x) \,dx= -\int\limits_{b}^{a} f(x)\, dx }[/math].)

[math]\displaystyle{ \int\limits_{a}^{a} f(x)\, dx = 0 }[/math]

This follows the fundamental theorem of calculus (FTC): [math]\displaystyle{ F(a)-F(a) = 0 }[/math].

[math]\displaystyle{ \int\limits_{a}^{b} f(x)\, dx = -\int\limits_{b}^{a} f(x)\, dx }[/math]

Again, following the FTC: [math]\displaystyle{ F(b)-F(a) = -[F(a)-F(b)] }[/math].

Integral Media

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  A definite integral of a function can be represented as the signed area of the region bounded by its graph and the horizontal axis; in the above graph as an example, the integral of f(x) between a and b is the yellow (−) area subtracted from the blue (+) area

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  Approximations to integral of √x from 0 to 1, with 5 yellow right endpoint partitions and 10 green left endpoint partitions

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  Demonstrates a step towards the Lebesgue integration of a function using a simple function.

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  The improper integral\int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x}} = \pihas unbounded intervals for both domain and range.

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  Double integral computes volume under a surface z=f(x,y)

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  A line integral sums together elements along a curve.

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  The definition of surface integral relies on splitting the surface into small surface elements.

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  Numerical quadrature methods: rectangle method, trapezoidal rule, Romberg's method (at varying piece count), Gaussian quadrature

Related pages

• Contour integral
• Multiple integral
• Numerical integration
• Riemann sum
• Surface integral