Integration

If we were to look at a function on a graph, the integral of *f(x) *would describe the area underneath the function. To represent the integral of *f(x)*, we would write \(\int{f(x) dx}\), with *dx *telling us we are integrating with respect to x. (This is called the differential).

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Jetzt kostenlos anmeldenIf we were to look at a function on a graph, the integral of *f(x) *would describe the area underneath the function. To represent the integral of *f(x)*, we would write \(\int{f(x) dx}\), with *dx *telling us we are integrating with respect to x. (This is called the differential).

Conceptually, we think of integration as being the inverse of differentiation. This means that to find an integral, we can think of having to 'undo' the process of differentiation. When we have 'undone' the integration, we call this result the antiderivative.

Conceptually, integration can be thought of as the "opposite" of differentiation. Both are mathematical operations that are covered in the mathematical field of calculus.

**Integration** is the sum of infinitely small pieces to find the total; this might be the area under a curve, the length of a curve, or other physical quantities like displacement given velocity.

For example, when you integrate a speed function, you get a distance function because you're summing up all the infinitesimal distances travelled over a time period.

There are two primary types of integration: **definite and indefinite**. A **definite integral **has actual **limits** and gives a **numerical value**, which could represent a physical quantity like an area. An **indefinite integral **doesn't have set limits and **results in a function**, often represented by a graph.

**A definite integral is one with limits**, so we could view this as the area under a function between two points, say point a and point b. For a function *f(x)*, we would write this as \(\int^{b}_a f(x) dx\). This can be visualized as

The way to visualize this is to split the area under the function into n equal strips between a and b. This means we have the width of each strip, \(\delta x = \frac{b-a}{n}\). We then take the height of each strip as \(f(x_i^*)\), with the point \(x^*_i\) at some point in strip i. This is shown below.

The area of the strips at this point is given as \(\sum^n_{i=1} f(x_i^*) \delta x\). To find the value of the integral, we need to use an infinite number of strips to cover the inside of the curve fully. This means as we take the limit, we get \(\lim_{n \rightarrow \infty} \sum^n_{i=1} f(x^*_i) \Delta x = \int^b_a f(x)dx\).

\(\lim_{n \rightarrow \infty} \sum^n_{i=1} f(x^*_i) \Delta x = \int^b_a f(x)dx\). In practice, this becomes easier as we find the antiderivative (without the integration constant) and then evaluate it at the two limits, taking the bottom limit away from the top limit.

Find \(\int ^2_0 2x \space dx\)

The first step is to find the antiderivative of 2x. This means we need to think of a function that differentiates to 2x. Thinking about this, we get to x^{2}. Now we know the antiderivative; we need to evaluate this at the limits.

\(\int^2_0 2x dx = [x^2]_{x=0}^{x=2} = (2)^2 - (0)^2 = 4\)

The purpose of an indefinite integration is to **find the antiderivative**. The antiderivative is given as a function and doesn't tell us directly the area under the function. If we want to check whether we have the correct antiderivative, we can differentiate the antiderivative, and we should arrive back at the original function. If our original function is *f(x)*, we often denote *F(x) *as the antiderivative of* f(x)*.

When we find an indefinite integral, it is important that **we add a constant of integration**, meaning that if we were to find \(\int{f(x) dx}\), we would give our answer as *F(x) + C*. This* + C* reflects that this antiderivative function could have any constant and still be differentiated to the original function.

Find \(\int 3x^2 dx\)

\(x^3\) differentiates to \(3x^2\), so that is our antiderivative. However, in full, our answer is \(x^3 + C\), as we must include this constant of integration.There are several fundamental rules for integration in mathematics. These rules form the basis for more complex techniques and are central to calculus. These are:

**Power Rule**: This states that \( \int x^n dx = \frac{x^(n+1)}{n+1} + C\), where n ≠ -1, and C represents the constant of integration.**Constant Rule**: The integral of a constant times a function is the constant times the integral of the function. That is, \(\int c \cdot f(x) dx = c \cdot \int f(x) dx\).**Sum Rule**: The integral of the sum of two functions is the sum of the integrals of the functions. This rule is stated as \(\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx\)**Difference Rule**: The integral of the difference between two functions is the difference between the integrals of the functions. This rule is stated as \(\int [f(x) - g(x)] dx = \int f(x) dx - \int g(x) dx\)**Exponential Rule**: \(\int e^x dx = e^x + C\).**Substitution Rule (u-substitution)**: This method involves making a substitution to simplify the integral. If we have \(\int f(g(x)) \cdot g'(x) dx\), we can substitute \(u = g(x), \, du = g'(x) dx\), and then perform the simpler integral \(\int f(u) \, du\).**Integration by Parts**: This is a method used for integrating the product of two functions. It's the integration equivalent of the product rule for differentiation. The formula is \(\int u\, v \, dx = u \int v \, dx - \int [u' \cdot (\int v \, dx)] dx\)**Trigonometric Rules**: There are several trigonometric rules:*Sine rule*: \(\int \sin(x) \, dx = -\cos(x) + C\)*Cosine rule*: \(\int \cos(x) \, dx = \sin(x) + C\)*Secant Squared Rule:*\(\int \sec^2(x) \, dx = \tan(x) + C\)*Cosecant Squared Rule*:\(\int \csc^2(x) \, dx = -\cot(x) + C\)

*Secant · Tangent Rule*: \(\int \sec(x)\tan(x) \, dx = \sec(x) + C\)*Cosecant · Cotangent Rule*: \(\int \csc(x)\cot(x) \, dx = -\csc(x) + C\)

For definite derivatives, we have additional rules that can help solving the integral:

- For constants a and b, and functions f, g, then \(\int(af(x) + bg(x)) dx = a\int f(x)dx + b\int g(x) dx\)
- \(\int^b_a f(x) dx = -\int^{a}_b f(x) dx\)
- \(\int^{a}_a f(x) dx= 0\)
- for \( c \space \epsilon \space [a, b], \int^b_a f(x)dx = \int^c_a f(x)dx + \int^b_c f(x)dx\)

Not all antiderivatives can be found easily by inspection. Here, we can use an integration method instead to allow us to find the antiderivative.

By the product rule (as seen in differentiation), for two functions u(x) and v(x) then \((u(x)v(x))' = u(x)v'(x) + u'(x)v(x)\)

If we integrate both sides with respect to x, we get: \(\int{(u(x) v(x))'dx} = \int{u(x) v'(x)dx} + \int{u'(x)v(x)dx}\)

which then simplifies to -

\(u(x)v(x) = \int{u(x)v'(x)dx} + \int{u'(x)v(x)dx}\)

We rearrange this to -

\(\int{u(x)v'(x)dx} = u(x)v(x) - \int{u'(x)v(x)dx}\)

This is now our formula for integration by parts, and we will demonstrate this through an example.

Use integration by parts to find \(\int{x \cos x \space dx}\).

We are going to let \(u(x) = x\) and \(v'(x) = \cos(x)\). We now look to find \(u'(x)\) and differentiating, we find and integrate to find. This means that \(v(x) \quad u'(x) = 1 \quad v(x) = \sin x \quad \int{x \cos x = x \sin x} - \int{1 \cdot sin x}\)

We can now evaluate this last integral to give \(\int{x \cos x} = x \sin x + \cos x + C\).

Note did we have included the integration constant here. We could have included this earlier; however, we can combine them all into one here.

There is also an option to use substitution to simplify an integral. Here we change the variable we integrate with respect to. In the case of a definite integral, the limits also need to be changed using the substitution. We must also change the integrand. This is best demonstrated by an example. Trying to choose the right substitution takes time. However, it becomes easier.

Use substitution to find \(\int{2xe^{x^2}dx}\)

Let us take \(u = x^2\), which means that \(\frac{du}{dx} = 2x\). We can rearrange this to get \(dx=\frac{du}{2x}\).

Now substituting this in, we get

\(\int{2xe^{x^2}dx} = \int{2xe^u \cdot \frac{du}{2x}} = \int{e^u \space du} = e^u + C = e^{x^2} + C\).

We can also be given a function parametrically and be expected to integrate this. Suppose we are given that \(y = f(t)\) and \(x = g(t)\), then the integral of the curve defined by these functions is given as \(\int{y \frac{dx}{dt} dt}\). We can think of this as the dt's cancelling to give \(\int y \space dx\), which is what we'd expect in a normal integral.

Suppose we are given a curve defined by \(y = 2-t^2, \space x = t^3\), with t ranging from 0 to 1, and we want to find the area under this curve when \(t = 0, \space x= 0, \text{ and }t = 1, \space x=1\), so our integral is given as \(^1_0\int(2-t^2) \cdot 3t^2 dt\). We can evaluate this to get \(^1_0\int{(2-t^2) \cdot 3t^2 dt = ^1_0\int6t^2-3t^4dt} = [2t^3-\frac{3}{5} t^5]^{x=1}_{x = 0} = 2 - \frac{3}{5} = \frac{7}{5}\)

In the following section, we will go over worked examples of integrals, and use the integration rules shown above to drive them home.

From the derivative of a polynomial, you should know that \(\frac{d}{dx}x^n =nx^{n-1}\). For an integral, we can reverse this to get \(\int{x^n \space dx} = \frac{1}{n+1} x^{n+1}, \space n≠-1\). This rule will become second nature the more integrals that you do.

Integrate \(12x^5\) with respect to x.

\(\int{12x^5 \space dx} = 12 \int x^5 dx = 12 \cdot \frac{1}{6} x^6 = 2x^6 + C\)

The above formula for polynomials will not work for \(\frac{1}{x}\). So, let's look at this a different way:

\(\int{\frac{1}{x} dx}\). Let \(x = e^y\), then \(\frac{dx}{dy} = e^y\) so \(dx = e^y dy\).

Filling this in, we get: \(\int{\frac{1}{x} dx} = \int{\frac{1}{e^y} \cdot e^y dy} = \int dy = y = \ln|x| +C\)

Note we put x in a modulus function to ensure that the logarithm input is valid. We can extend this further. By making a suitable substitution, we can show: \(\int{\frac{f'(x)}{f(x)} dx} = \ln|f(x)| + C\).

Like everything we have seen so far, we can treat integration as the inverse of differentiation, and this continues with trigonometric functions. We may have to use substitutions to solve these, and we can also introduce trigonometric functions as a substitution.

Find \(\int{\tan(x) dx}\).

\(\int{\tan(x) dx} = \int{\frac{\sin x}{\cos x} dx}\)

Now let \(u = \cos(x)\) and then \(\frac{du}{dx} = -\sin x\). This means that \(\int \frac{\sin x}{\cos x} dx = -\int \frac{1}{u} du = -\ln|\cos x| + C= \ln|\cos x|^{-1} + C = \ln|\sec x| + C\)Use a trigonometric substitution to find \(\int \frac{1}{\sqrt{9-x^2}} dx\).

Let \(x = 3 \sin u\) so \(\frac{dx}{du} = 3 \cos u\), and \(dx = 3 \cos u \cdot du\).

Substituting this in, we get \(\int \frac{1}{\sqrt{9-x^2}} dx = \int \frac{3 \cos u}{\sqrt{9-9\sin^2 u}} du\).

As \(\sin^2 y + \cos^2 y = 1, \space 1- \sin^2 u = \cos^2 u\). So, \(\int \frac{3 \cos u}{\sqrt{9-9 \sin^2 u}} du = \int \frac{3 \cos u}{3 \sqrt{1- \sin^2 u}} du = \int \frac{3 \cos u}{3 \cos u} du = \int du = u + C = \arcsin (\frac{x}{3}) + C\)

Integration is the inverse of differentiation.

A definite integral is bounded by limits and is evaluated there.

An indefinite integral is an antiderivative and an integration constant.

Integration by parts is defined as \(\int u(x)v'(x) dx = u(x)v(x) - \int u'(x) v(x) dx\)

For a polynomial, \(\int x^n dx = \frac{1}{n+1} x^{n+1}, \space n≠-1\); \(\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C\)

Integration is the process of finding the antiderivative of a function.

An example of integration would be Integration of 3x^2dx=x^3+C

Integration by parts is when we have two functions multiplied together, and then we have an integral of this. The equation for this is given by

Int of u(x)v'(x)dx=u(x)v(x)- Int of u'(x)v(x)dx

What is a definite integral?

An integral calculated between two limits is called a definite integral.

Is it possible to calculate the definite integral for all functions?

No

The curve y=f(x) intersects the x-axis between x=a and x=b. It encloses an area of 57 above the x-axis and an area of 22 below the x-axis. What is the total area enclosed between the curve and the x-axis between x=a and x=b?

79

Find the total area of the finite region bounded by the curve y= x(x - 1)(x + 3) and the x-axis.

11.83

What happens if you differentiate a function and then integrate it?

You get back the original function.

What happens if you integrate a function and then differentiate it?

You get back the original function.

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