Integration of Algebraic functions
Tangents and Normals
If you differentiate the equation of a curve, you will obtain a formula for the gradient of the curve. Before you learnt differentiation, you would have calculated the gradient of a curve by drawing a tangent and measuring the gradient of this. This is due to the fact that the gradient of a curve at a point is equal to the gradient of the tangent at that point.
The equation of the tangent to a point on a curve can thus be obtained by differentiation.
Example:
Calculate the equation of the tangent to the curve y = x3 at the point (2, 8). dy/dx = 3x2
Gradient of tangent when x = 2 is 3 × 22 = 12.
From the coordinate geometry topic, the equation of the tangent is thus:
y - 8 = 12(x - 2) since the gradient of the tangent is 12 and we know that it passes through (2, 8)
thus y = 12x - 16
You may as well be asked to obtain the gradient of the normal to the curve. The normal to the curve is the line perpendicular (at right angles) to the tangent to the curve at that point.
Bear in mind that if two lines are perpendicular, the product of their gradients is -1.
Thus if the gradient of the tangent at the point (2, 8) of the curve y = x3 is 12, the gradient of the normal is -1/12, since -1/12 × 12 = -1 .
The equation of the normal at the point (2, 8) is thus:
y - 8 = -1/12 (x - 2)
therefore the equation of the normal at (2,8) is 12y + x = 98 .
Integration by Substitution
It is possible to change a difficult integral to a simpler integral through the method of substitution.
For instance, assuming that we are integrating a difficult integral with respect to x, We may be able to allow x = sin t, say, to make the integral simpler. As long as we alter "dx" to "cos t dt" (due to the fact that if x = sin t then dx/dt = cost) we can at this point integrate with respect to t and we will obtain the same answer as if we had done the original integral.
With the use of substitutions, we can show that:
The second one is particularly significant. If you want to integrate a fraction, where the top is the differential of the bottom, the answer is merely ln of the bottom plus a constant.
Example
Find the integral of:
(a) -sin x cos²x
(b) 3x²
x³ + 1
(a) With the use of the first of the two above formulae above, imagine f(x) = cos x and n = 2. Thus [f(x)]² = cos²x and f "(x) = -sin x. Hence, since n = 2, the answer is merely (cos³x)/ 3 + c
(b) Since the top is the differential of the bottom, we can make use of the second of the two formulae above to obtain the answer of ln(x³ + 1) + c.
Using a Substitution
Sometimes you will be told to integrate a function by using a substitution. Unless the substitution is a simple one, you will likely be told what substitution to make use of in the exam.
Example
Integration by Parts
From the product rule, we can get the following formula, which is highly essential in integration:
It is used when integrating the product of two expressions (a and b in the bottom formula). When making use of this formula to integrate, we say we are "integrating by parts".
Occasionally, you will have to integrate by parts two times (or perhaps even more times) before you obtain an answer.
Example
Find ∫xe-x dx
Integrating by parts (with v = x and du/dx = e-x), we get:
-xe-x - ∫-e-x dx (since ∫e-x dx = -e-x)
= -xe-x - e-x + constant
We can as well occasionally make use of integration by parts when we want to integrate a function that cannot be divided into the product of two things. The trick we make use of in such situations is to multiply by 1 and take du/dx = 1.
Example
Find ∫ ln x dx
To integrate this, we make use of a trick, rewrite the integrand (the expression we are integrating) as 1.lnx . We then let v = ln x and du/dx = 1 .
Thus ∫ ln x dx = x ln x - ∫ x (1/x) dx
= x lnx - ∫ dx
= x lnx - x + constant
Integration Techniques
Significant Formulae
Rewriting the Integrand
Dividing
If you are asked to integrate a fraction, try multiplying or dividing the top and bottom of the fraction by a number
Example
If we divide everything on the numerator and everything on the denominator by x2, we obtain:
= ∫ (3x – 4x-1 – 5x-2) dx
= 3x2 – 4lnx + 5x-1 + c
2
Divide into Partial Fractions
A few times it will help if you split a fraction up before trying to integrate. This can be done with the use of the method of partial fractions.
Example
(At this point we divide the fraction into partial fractions)
= -3lnx + 4ln(x - 1) + (x - 1)-1 + c
The use of Trigonometric Formulae
When integrating trigonometric expressions, it will frequently help to rewrite the integral with the use of trigonometric formulae.
Example
∫ cos2x dx
cos2x = 2cos2x - 1
cos2x = ½ (cos2x + 1)
∫ cos2x dx = ½ ∫ (cos2x + 1) dx
= ½ ( ½ sin2x + x) + c
= ¼ sin2x + ½ x + c
Differentiation From First Principles
Differentiating a linear function
A straight line has a constant gradient, or in other words, the rate of change of y with respect to x is a constant.
Example
Consider the straight line y = 3x + 2 shown below
A graph of the straight line y = 3x + 2.
We can calculate the gradient of this line as follows. We take two points and calculate the change in y divided by the change in x.
When x changes from −1 to 0, y changes from −1 to 2, and so
Irrespective of which pair of points we select the value of the gradient is always 3.
Values of the function y = 3x + 2 are illustrated below:
Look at the table of values and observe that for every unit increase in x we always obtain an increase of 3 units in y. In other words, y increases as a rate of 3 units, for every unit increase in x. We say that “the rate of change of y with respect to x is 3”.
Observe that the gradient of the straight line is the same as the rate of change of y with respect to x.
NOTE: For a straight line: the rate of change of y with regard to x is the same as the gradient of the line.
Differentiation from first principles of some simple curves
For any curve, it is clear that if we select two points and join them, this yields a straight line.
For different pairs of points we will obtain different lines, with very different gradients. We show below:
Linking different pairs of points on a curve yields lines with different gradients
Example : Assuming we look at y = x2.
Observe that as x increases by one unit, from −3 to −2, the value of y decreases from 9 to 4. It has reduced by 5 units. But when x increases from −2 to −1, y decreases from 4 to 1. It has reduced by 3. Therefore, even for a simple function like y = x2 we observe that y is not altering constantly with x. The rate of change of y with respect to x is not a constant.
Calculating the rate of change at a point
We now explain how to estimate the rate of change at any point on a curve y = f(x). This is defined to be the gradient of the tangent drawn at that point as illustrated below:
The rate of change at a point P is defined to be the gradient of the tangent at P.
NOTE: The gradient of a curve y = f(x) at a given point is defined to be the gradient of the tangent at that point.
We make use of this definition to calculate the gradient at any particular point.
Consider the graph below which illustrates a fixed point P on a curve. We as well show a sequence of points Q1, Q2, . . . getting closer and closer to P. We observe that the lines from P to each of the Q’s get nearer and nearer to becoming a tangent at P as the Q’s get nearer to P.
The lines through P and Q approach the tangent at P when Q is very close to P. Therefore, if we calculate the gradient of one of these lines, and allow the point Q approach the point P along the curve, then the gradient of the line ought to approach the gradient of the tangent at P, and thus the gradient of the curve.
Example :
We ought to carry out the calculation for the curve y = x2 at the point, P, where x = 3.
The graph below illustrates the graph of y = x2 with the point P marked. We choose a nearby point Q and link P and Q with a straight line. We will select Q so that it is quite close to P. Point R is vertically below Q, at the same height as point P, so that △PQR is right-angled.
The graph of y = x2. P is the point (3, 9). Q is a nearby point.
Assuming we select point Q so that PR = 0.1. The x coordinate of Q is then 3.1 and its y coordinate is 3.12. Knowing these values we can calculate the change in y divided by the change in x and thus the gradient of the line PQ.
We can take the gradient of PQ as an approximation to the gradient of the tangent at P, and thus the rate of change of y with respect to x at the point P.
The gradient of PQ will be a better approximation if we take Q closer to P. The table below illustrates the effect of reducing PR successively, and recalculating the gradient.
The gradient of the line PQ, QR/PR appears to approach 6 as Q approaches P.
Notice that as Q gets closer to P the gradient of PQ seems to be getting nearer and nearer to 6.
We will now repeat the calculation for a general point P which has coordinates (x, y).
The graph of y = x2. P is the point (x, y). Q is a nearby point.
Point Q is selected to be close to P on the curve. The x coordinate of Q is x + dx where dx is the symbol we make use of for a small change, or small increment in x. The corresponding change in y is written as dy. Therefore, the coordinates of Q are (x + dx, y + dy).
Due to the fact that we are considering the graph of y = x2, we know that y + dy = (x + dx)2.
As we let dx become zero we are left with just 2x, and this is the formula for the gradient of the tangent at P. We have a brief way of expressing the fact that we are allowing dx approach zero. We write
‘lim’ stands for ‘limit ’and we say that the limit, as x tends to zero, of 2x+dx is 2x. Observe that when x has the value 3, 2x has the value 6, and thus this general result agrees with the earlier result when we calculated the gradient at the point P(3, 9).
We can do this calculation in the same way for many curves. We have a special symbol for the phrase
We write this as dy/dx and say this as “dee y by dee x”. This is as well known as the derivative of y with regard to x.
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