Revision:OCR Core 2 - Integration

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1 5. Integration
2 Also See

5. Integration

Finding a function from its derivative

One might have seen how the graph of the derivative of a function is related to the function itself.

Consider the following:

$f(x) = x^{2}$

One can say that:

$f^{'}(x) = 2x$

This is a straight line graph which passes through the origin. As one goes from $- \infty$ to 0 (on the x-axis), one has negative values on the $f^{'}(x)$ axis, and therefore one can say that there is a negative gradient on the axis of the original graph. Also one can say that there is a turning point at 0 on the x-axis, as the values on the $f^{'}(x)$ axis (of the plot of the derivative) become positive, and therefore the gradient has changed.

One can also say that as the graph of the derivative shows a constant gradient, the second derivative of the original function is constant, and therefore the function has a constantly changing gradient (by constant one refers to the manner of change, not the change over a given measurement).

Could one draw the graph of a function having the knowledge of the graph of the derivative of the function? One must think carefully about this.

Consider:

$f(x) = x^{n}$ ,

and:

$g(x) = x^{n} + k$

(Where "k" denotes a real constant).

One can calculate the derivatives of both of these functions:

$f^{'}(x) = nx^{n - 1}$

$g^{'}(x) = nx^{n - 1}$

One can see that if one was to take either of these derivatives and plot them as a graph, one would get precisely the same graph, even though the graphs of the functions are different.

The problem comes when a constant value is added to a function, when this is differentiated this will give nothing, and therefore one cannot differentiate (not in the Mathematical sense) between the different functions from their derivative.

One can predict the shape of a function from it's derivative, however, one cannot place this function upon the y-axis in such a way that one can calibrate the y-axis.

The process of getting from $f^{'}(x)$ to has a name, integration, and is the other area of calculus.

Consider that if one was given:

$f^{'}(x) = nx^{n - 1}$ ,

one might be asked to find .

Initially one might expect (through simple recognition) that the answer would be:

$f(x) = x^{n}$ .

This is not fully correct.

One has seen that through the addition of a constant value to a function, there is no change to the derivative, hence one must use a different answer:

$f(x) = x^{n} + k$

(Where "k" is a real constant).

One would refer to this as the indefinite integral of the function "f".

One should always remember to use the arbitrary constant "k" (or another letter).

It is desirable to have a rule for integration (much like differentiation), and thankfully it is a rather simple process if one remembers that integration is the opposite of differentiation.

Consider:

$f(x) = x^{n}$

If this were a derivative, what would the original function be.

One can say that when one differentiates this type of function, one reduces the power by 1, so the power will be when this is integrated.

Consider now:

$g(x) = x^{n + 1}$

$g^{'}(x) = (n + 1)x^{n}$

One wishes for this derivative to be $x^{n}$ , hence one must multiply by $\displaystyle \frac{1}{n + 1}$ .

Hence, if one integrates the original function "f", one gets:

$\displaystyle \frac{1}{n + 1} x^{n + 1}$ .

Obviously, at this stage, one can say that if n = -1, there is no meaning in this rule. This is the exception to the rule, and one should not consider this further (until Core 3).

Example

1. A graph has a derivative:

$\frac{dy}{dx} = x^{3} + x^{2} + x + 1$ ,

and passes through the origin. What is the original function?

This is a simple matter of integration. If one takes the integral of the differential, one has the general form for the function.

Integration is similar to differentiation in that one can split a function up such that one can integrate each part. Integration gives:

$\displaystyle y = \frac{x^{4}}{4} + \frac{x^{3}}{3} + \frac{x^{2}}{2} + x + k$

Now, one must find the value of "k".

Consider that when y = 0, x = 0 (due to the passing through the origin), hence:

Hence:

$y = \displaystyle y = \frac{x^{4}}{4} + \frac{x^{3}}{3} + \frac{x^{2}}{2} + x$ .

Notation for indefinite integrals

It is convenient to have notation for integration.

$\displaystyle \int y \ \mathrm{d}x$

This is the notation for the indefinite integral. This is equivalent to "integrate "y" with respect to "x"".

Hence:

$\displaystyle \int \left( x^{n} \right) \ \mathrm{d}x = \frac{1}{n + 1} x^{n + 1} + k$

One can omit the brackets if the term is simple, however it is often useful to use them as it can remove confusion.

Calculating areas

One of the most well-known uses of integration is the calculation of the area under a curve (bounded by lines which extend from the x-axis [in the context of integration with respect to "x"]).

One can demonstrate this using a simple, straight line example.

Take the graph of .

One can calculate the area between "a", and "b" using the idea of a trapezium.

At the x-coordinate "a", the coordinates are (a, a), and at the x-coordinate "b", the coordinates are (b, b).

Hence, the area is:

$\frac{1}{2} \times (b - a) \times (b + a) = \frac{1}{2} (b^{2} - a^{2}$

Now consider the integral of the function:

$\displaystyle \int x \ \mathrm{d}x = \frac{1}{2} x^{2} + k$

Ignoring the constant, there is a link. If one allows the integral (disregarding the constant) to be denoted:

(for the case where x = x).

One can see that the area is equal to:

One always uses the simplest integral for this procedure.

Proof of the area procedure

Consider the graph of a general function:

If one was to want to find the area under this graph between they x-coordinates of a, and b, one can use a method of limitation.

Consider that one can construct a series of infinite trapezia which will describe the area under the curve, as the number of the trapezia increases (up to infinity) the approximation of the area becomes more accurate.

Let there be a point "x", which is a point upon the x-axis, between a, and b.

The area under the curve between a, and "x" is a function of "x", as if one changes the position of "x", the area changes.

Let there be a change in "x", of $\delta x$ , and therefore a change in the area, A, of $\delta A$ , and a change in y of $\delta y$ .

Consider now that one can think of an extension of the trapezium. If one extends a line across, one can make a rectangle, whose area is a bound of $\delta A$ .

One can say that $\delta A$ is between $(y + \delta y) \delta x$ , and $y \delta x$ .

Also, one might consider that one is referring to a change over a range, as this range ( $\delta x$ ) tends to nothing, the change ( $\delta A$ ) becomes more accurate, until it becomes the real change $\frac{dA}{dx}$ .

As one limits $\delta x$ to 0, $\delta y$ tends to 0 also. Consider now that if the value of $\delta y$ tends to 0, the value of $y + \delta y$ tends to y. It follows that the bounds for the area are both the same, and therefore:

$\frac{dA}{dx} = y$ .

Or:

$\frac{dA}{dx} = f(x)$

Hence, the integral of is A.

One now has to consider the problem that arises due to the arbitrary constant.

If one allows there to be a function of x:

such that ,

one can consider further.

If "x" is equal to a, there is no area between them, and therefore A = 0; hence:

Hence:

So now one can consider the original idea, x = b:

(This agrees with the idea that was initially believed).

Integration notation

There is a simple change to the notation for the indefinite integral in order to make it suited to the definite integral (the integral one uses which has limitations, for the calculation of area [in the previous example]).

$\displaystyle \int _{a}^{b} y \ \mathrm{d}x$

This means "the area between the x-axis and the curve "y", bounded by the lines extending from the x-axis at "b", and "a"".

The two values (a, and b) are often called limits, or bounds of integration. The function that is being integrated is called the integrand.

Hence one does:

$\displaystyle \int _{a}^{b} f(x) \ \mathrm{d}x = \left[ I(x) \right] _{a}^{b} = I(b) - I(a)$

(Where denotes the simplest integral of [where k = 0]).

Some properties of definite integrals

One must be vigilant of several things when integrating. Sometimes one will get a value which is negative. This is simply meaning that the area that has been calculated lies below the x-axis, and therefore if one is asked for an area, one should give the figure without the negative sign.

Another issue that can cause problems is if the area is partially above, and partially below the x-axis (part of the area will cancel the other out). One must ensure that one does the integration in two stages in these cases. Consider that:

$\displaystyle \int _{a}^{c} f(x) \ \mathrm{d}x = \int _{a}^{b} f(x) \ \mathrm{d}x + \int _{b}^{c} f(x) \ \mathrm{d}x$

Proof is:

$\displaystyle \int _{a}^{c} f(x) \ \mathrm{d}x = \left[ I(x) \right] _{a}^{c} = I(c) - I(a)$ ,

and:

$\displaystyle \int _{a}^{b} f(x) \ \mathrm{d}x + \int _{b}^{c} f(x) \ \mathrm{d}x = \left[ I(x) \right] _{a}^{b} + \left[ I(x) \right] _{b}^{c}$

$\displaystyle = I(b) - I(a) + I(c) - I(b) = I(c) - I(a)$

(Hence the two are equivalent).

If one has a question in which part of the curve lies above the x-axis, and part lies below (and one wishes to calculate the area of a region under the curve which is partially below, and partially above), one should use two (or more) steps to calculate the area.

One other rule that one should be aware of is:

$\displaystyle \int _{a}^{b} f(x) \ \mathrm{d}x = - \int _{b}^{a} f(x) \ \mathrm{d}x$ .

This can be explained easily:

$\displaystyle \int _{a}^{b} f(x) \ \mathrm{d}x = \left[ I(x) \right] _{a}^{b} = I(b) - I(a)$

Also:

$\displaystyle \int _{b}^{a} f(x) \ \mathrm{d}x = \left[ I(x) \right] _{b}^{a} = I(a) - I(b)$

Hence:

$\displaystyle - \int _{b}^{a} f(x) \ \mathrm{d}x = -( I(a) - I(b)) = I(b) - I(a) = \int _{a}^{b} f(x) \ \mathrm{d}x$ .

The area between a curve and the y-axis

It is possible, and often very useful to calculate the area between a curve and the y-axis, between two bounds upon the y-axis.

There is nothing special about this calculation, all one needs to do is make an expression of "x" in terms of "y".

Example

1. Calculate the area between the curve and the y-axis, bounded by the lines y = 0, and y = 1, of:

$y = x^{2}$ .

One will simply do:

$\displaystyle x = \sqrt{y} = y^{ \frac{1}{2}}$ .

Now:

$\displaystyle \int _{0}^{1} \left( \sqrt{y} \right) \ \mathrm{d}y = \left[ \frac{2}{3} y^{ \frac{3}{2}} \right] _{0}^{1} = \frac{2}{3}$

Infinite and improper integrals

Some integrals cannot be calculated in the afore-mentioned manner. Consider the graph of:

$y = \frac{1}{x}$ .

If one wanted to calculate the area under the curve, and one limit was 0, one can see that there would be a problem.

The way one would go about solving a problem like this is to do the following:

$\displaystyle \int _{a}^{b} f(x) \ \mathrm{d}x = \left[ I(x) \right] _{a}^{b} = I(b) - I(a)$

Consider now that one wishes to find the area between the curve and the x-axis, bounded by x = b, and x = 0. One can allow $a \to 0$ , and predict the result.

Those integrals involving an infinite bound (or limit) are called infinite integrals, and those involving a value for which the integrand is not defined are referred to as improper integrals.

Example

1. Find the value of the following infinite integral:

$\displaystyle \int _{1}^{ \infty } \left( \frac{1}{x^{5}} \right) \ \mathrm{d}x$ .

So, do the following:

$\displaystyle \int _{1}^{s} \left( \frac{1}{x^{5}} \right) \ \mathrm{d}x = \left[ \frac{-1}{4} x^{-4} \right] _{1}^{s} = \frac{-1}{4s^{4}} - \frac{-1}{4 \times (1^{4})} = \frac{-1}{4s^{4}} + \frac{1}{4}$

Now:

$\displaystyle \int _{1}^{ \infty } \left( \frac{1}{x^{5}} \right) \ \mathrm{d}x = \lim _{s \to \infty} \int _{1}^{s} \left( \frac{1}{x^{5}} \right) \ \mathrm{d}x = \frac{1}{4}$ .

Calculating other areas

Sometimes one will need to calculate the area between two curves. This is a relatively simple process.

Consider:

One can calculate the area between these curves in a simple manner (allow the limits of integration to be "b", and "a").

$\displaystyle \int _{a}^{b} f(x) \ \mathrm{d}x - \int _{a}^{b} g(x) \ \mathrm{d}x$ .

This will give the area between the two curves. Often one will be asked to found the enclosed area. This means that one must find the intersections, and then use these figures as the bounds.

One can also (due to the rules of integration) do this in a single integral:

$\displaystyle \int _{a}^{b} \left( f(x) - g(x) \right) \ \mathrm{d}x$

One merely has to ensure that the integrand is correct, and then the area can be calculated.

One should be aware that one can be asked to calculate areas other than the area underneath the graph. Often one might need to calculate areas of regions on the Cartesian plane. Part of the region will be some calculation of the area under the curve (or similar), but it may involve some clever thinking. One should always try to simplify the calculations into integrals, and known shapes.

Also See

Read these other OCR Core 2 notes:

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Article Updates

Revision:OCR Core 2 - Integration

Contents

5. Integration

Finding a function from its derivative

Notation for indefinite integrals

Calculating areas

Proof of the area procedure

Integration notation

Some properties of definite integrals

The area between a curve and the y-axis

Infinite and improper integrals

Calculating other areas

Also See