₦airaland Forum

thankyouJesus:
[b]Good morning to you all, how was the night?
Yesterday, we were considering differentiation, we were able to conclude that:
1. Differentiation means rate of change.
2. If y = axⁿ, then dy/dx = nax^n-1.
3. The differentiation of a constant is zero.
Today, we will be considering another aspect of differentiation.
DIFFERENTIATION OF POLYNOMIALS.
To differentiate a polynomial, we differentiate each term in turn.
General form:
if y = ax^b + cx^d + ex^f + . . .,
then dy/dx = bax^b-1 + dcx^d-1 + fex^f-1 + . . ..
. . . . . . . . . . . . . . . . . . . . . . . .
Example 1: if y = x⁴ + 5x³ - 4x² + 7x - 2.
Solution:
dy/dx = (4)(1)x^4-1 + (5)(3)x^3-1 - (4)(2)x^2-1 + (7)(1)x^1-1 - 0.
dy/dx = 4x³ + 15x² - 8x + 7.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 2: if y = 2x⁵ + 4x⁴ - x³ + 3x² - 5x + 7, find an expression for dy/dx and the value of dx/dx at x = 2.
Solution:
So, first of all,
dy/dx =
.
.
.
.
.
.
.
Very good, because, if
y = 2x⁵ + 4x⁴ - x³ + 3x² - 5x + 7.
Then,
y' = 10x⁴ + 16x³ - 3x² + 6x - 5.
Substituting x = 2, we have,
y' = 10(2)⁴ + 16(2)³ - 3(2)² + 6(2) - 5.
Therefore,
y' = 283.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 3:
if y = 2x² - 5x + 3, then
dy/dx = 4x - 5.
This double statement can be written as a single statement by putting 2x² - 5x + 3 in place of y in dy/dx. i.e
d/dx (2x² - 5x + 3) = 4x - 5.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 4:
In the same way,
d/dx (4x³ - 7x² + 2x + 3) =
Solution:
12x² - 14x + 2.
Note: Either of the two methods is acceptable: it is just a case of which is the more convenient in any situation.
. . . . . . . . . . . . . . . . . . . . . . . .
Summary:
1. To differentiate a polynomial, we differentiate each term in turn.
2. if y = ax^b + cx^d + ex^f + . . .,
then dy/dx = bax^b-1 + dcx^d-1 + fex^f-1 + . . ..
. . . . . . . . . . . . . . . . . . . . . . . .
Assignment:
1. No 39, 1994.
If y = 3t³ + 2t² - 7t + 3, find dy/dt at t = -1.
(a) -1 (b) 1 (c) -2 (d) 2.
2. No 19, 2002.
If y = x² - 1/x, find dy/dx
(a) 2x + 1/x² (b) 2x + x² (c) 2x - 1/x² (d) 2x - x²
3. No 36, 2008.
Find the derivative of y = x⁷ - x⁵
(a) x (x^{2 - 1) (b) 3x (x[sup]2} - 1) (c) 3x² - 1 (d) 7x⁶ - 5x⁴.
. . . . . . . . . . . . . . . . . . . . . . . .
Things to note before the next class:
1. There is difference between
(a) sinx - cosx
(b) sinx + cosx
(c) sinx cosx,
the difference is, (a) &(b) are polynomials because they are connected by either + or - while (c) is connected by multiplication. Another method is used to solve c.
2. If y = sinx, dy/dx = cosx.
3. If y = cosx, dy/dx = -sinx.
Have a splendid day.

Any question?

24th March, 2015.
07:07:33[/b]

Mathemagician1:
What happened to this thread? Let's help out while we can shall we. I volunteer to take section IV - Calculus tutorial.

SECTION IV: CALCULUS
I. Differentiation:
(a) limit of a function;
(b) differentiation of explicit algebraic and simple trigonometric functions - sine, cosine and tangent.

2. Application of differentiation:
(a) rate of change
(b) maxima and minima

3. Integration:
(a) integration of explicit algebraic and simple trigonometric functions.
(a) area under the curve.

I call on our able moderators and all maths gurus on this forum to lend a helping hand in demystifying mathematics for the younger generation.

thankyouJesus:
[b]Good evening all, how is your day fairing?
In our last class, we were considering differentiation of polynomials. We were able to achieve that
1. To differentiate a polynomial, we differentiate each term in turn.
2. If y = ax^b + cx^d + . . ., then, dy/dx = bax^b-1 + dcx^d-1 + . . ..
Now, we will be considering a different aspect:
HIGHER DERIVATIVES.
If y = 2x⁴ - 5x³ + 3x² - 2x + 4, then, by the previous method:
dy/dx = 8x³ - 15x² + 6x - 2.
This expression for dy/dx is itself a polynomial in powers of x and can be differentiated in the same way as before, i.e. we can find the derivative of dy/dx.
d/dx [dy/dx] is written d²y/dx² and is the second derivative of y wrt x (spoken as "dee two y by dee x squared".
So, in this example, we have:
y = 2x⁴ -5 x³ + 3x² - 2x + 4.
dy/dx or y' = 8x³ - 15x² + 6x - 2.
d²y/dx² or y" = 24x² - 30x + 6.
We could, if necessary, find the third derivative of y in the same way:
d³y/dx³ or y"' = . . . . . . . . . .
.
.
.
.
.
.
Very good,
y"' = 48x - 30
. . . . . . . . . . . . . . . . . . . . . . . . Summary:
if y = ax^b + cx^d + ex^f + gx^h + . . .
Then,
dy/dx = bax^b-1 + dcx^d-1 + fex^f-1 + hgx^h-1 + . . .
Also,
y" = (b-1)bax^b-1-1 + (d-1)dcx^d-1-1 + (f-1)fex^f-1-1 + (h-1)hgx^h-1-1 + . . .
. . . . . . . . . . . . . . . . . . . . . . . .
Things to note before next class:
1. Y = sinx
dy/dy = cosx
2. Y = cosx
dy/dx = -sinx
3. Y = tanx
dy/dx = sec²x.
. . . . . . . . . . . . . . . . . . . . . . . .
Assignment:
Pick up JAMB mathematics past question and solve anything "solvable" in differentiation.

Any question?

.
Have a blissful night rest.
24th March, 2015.
19:22:36.[/b]

abelincon:
plz.i need ur help to solve dis mathematical
problems for.i will be grateful
1. In a party there are 3boys and 4 girls.in how
may different ways can they dance.(a) 24 (b) 12
(c) 29 (d) 18

2. If elements of a series are
1,3,2,4,3,7,5,11,8,18,13,... then the next element is
(a) 19 (b) 21 (c) 29 (d) 24


3. if a sweet costs #2.00,a chocolate costs #3.00,a cake costs #5.00 and a mango costs #10.00,then buy each item,at least one and as many items as many items as possible from #50.00. the number of sweet you buy is what (a) 16 (b) 12 (c) 18 (d) 14

thankz

Mathemagician1:
(D) MULTIPLICATION AND DIVISION OF POLYNOMIALS CONT...

Example 7: See attached.

We've already covered division of polynomials under factor and remainder theorem.

THE END

Mathemagician1:
(D) MULTIPLICATION AND DIVISION OF POLYNOMIALS CONT...

Example 6: see attached.

thankyouJesus:
[b]Good afternoon all, in response to Tolaxin demand, I am going to fast forward (quickly rush differentiation) to integration, but I need someone to stop me (quote me and type stop).
Here we go at light speed!
Check the questions below and note the difference
(a) y = sin
(b) y = sinx
(c) y = sin²x
(d) y = sin(2x + 4)
(e) y = sinx + cosx
(f) y = sinx cosx
(g) y = sinx cosx + sinx cosx.
The difference
(a) y = sin
This is a wrong question because no function is given, i.e
y = sin(nothing)
(b) y = sinx
This can be differentiated to be
y' = cosx
(c) y = sin²x
This question can also be written as
y = sinx sinx,
since the two trigonomentaries are connected by "multiplication(product)", a different approach is adopted (product rule)
(d) y = sin(2x + 4)
.
.
.
.
.
.
.
No, you are wrong
y = sin X (2x + 4)
is wrong because of (a) above, therefore, you treat both
y = sin(2x + 4)
as single entity because they are "chained" together, to solve this, a different approach is adopted (chain rule).
Note:
y = sin(2x + 4)
is not equal (the same as) to
y = sin2x + sin4
(e) y = sinx + cosx,
this is polynomial, because the right hand expression (sinx + cosx) are connected together by +, to solve this, you differentiate each term in turn
(f) y = sinx cosx
since, sinx cosx are connected together by "multiplication(product)", product rule is use to solve.
(g) y = sinx cosx + sinx cosx
this is a mixture of product rule and polynomial, to solve this, product rule is used in turn.
P:S: Master this piece before the next class.
25th March, 2015.
12:59:13[/b]

thankyouJesus:
[b]Good evening all, I want to believe "you" have mastered all (the previous class).
We got no time to waste.
FUNCTIONS OF A FUNCTION.
If
y = sinx,
y is a function of the angle x, since the value of y depends on the value given to x.
If
y = sin(2x - 3),
y is a function of the angle (2x -3) which is itself a funtion of x.
Therefore, y is a function of (a function of x) and is said to be a funtion of a function of x.
DIFFERENTIATION OF A FUNCTION OF A FUNCTION.
To differentiate a function of a function, we must first introduce the chain rule.
With the example above,
y = sin(2x -3),
we put
u= 2x - 3
i.e
y = sinu
where
u = 2x - 3.
dy/dx = dy/du "times" du/dx.
This is the chain rule and is particular useful when determining the derivatives of functions of a function.
In summary:
Chain rule is used when you cannot split a function into two, examples
(a) sin(2x + 3),
you cant split this into
sin(nothing) X (2x + 3)
(b) cos(4x² + 2x - 6),
you cant split this into
cos(nothing) X (4x² + 2x - 6)
(c) There are exceptions to this rule,
(2x + 6)²,
the above can also be express as
(2x + 6) (2x + 6),
since, they are connected by "multiplication-product", product rule can be used and also chain rule.
Generalising, if
(ax^R + bx^R + . . .)^N,
where R is real numbers,
N is natural numbers greater than 1. Chain rule and product rule can both be used.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 1. Find y' if
y = sin(2x + 3).
Solution:
y = sin(2x + 3)
let u = 2x + 3
du/dx = 2,
since u = 2x + 3
y = sinu
dy/du = cosu
using chain rule,
dy/dx = dy/du "times" du/dx
dy/dx = cosu "times" 2
dy/dx = 2 cosu
dy/dx = 2 cos(2x + 3).
Short cut:
if y = sin(ax^R + bx^R + . . .)
y' = d/dx ((ax^R + bx^R) + . . .) d/dx sin(ax^R + bx^R).
In what you can understand,
if y = sin (2x + 3)
dy/dx = 2 cos(2x + 3),
because if I differentiate (2x + 3), I will get 2 and also if I differentiate "sin", I will have cos.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 2: find y' if
y = cos(7x -
Solution:
y' = -7 sin(7x -
. . . . . . . . . . . . . . . . . . . . . . . .
Exceptions:
if y = (ax^R + bx^R + . . .)ⁿ
dy/dy = n(ax^R + bx^R + . . .)^n-1d/dx (ax^R + bx^R + . . .).
Example
y = (2x + 3)⁶
y' = 6(2x + 3)⁵(2)
y' = 12(2x + 3)⁵.
. . . . . . . . . . . . . . . . . . . . . . . .
No summary and assignment (remember light speed).
25th March, 2015.
22:03:06[/b]

thankyouJesus:
[b]Good evening all gurus in the house, I remain my humble self thankyouJesus, allow me to work on existing protocols.
In ado to our previous classes, we are still considering differentiation.
Today, we will be considering:
1. EXPONENTIAL DIFFERENTIATION.
General formula, if
y = e^u,
where u is a function of x, then
y' = d/dx (u) e^u.
In other words, differentiate the power of the exponential and use it to multiply the original question.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 1. If y = e^2x, find y'.
Solution:
y = e^2x,
dy/dx = 2e^2x.
2 is from the differention of 2x.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 2: if y = e^sinx, find y'.
Solution:
y = e^sinx
y' = cosx e^sinx.
. . . . . . . . . . . . . . . . . . . . . . . . 2. LOGARITHMIC DIFFERENTIATION.
General formular, if
y = 1n (u) or
y = log_e (u),
then,
dy/dx = d/dx (u) / (u).
In other words, if y is equal to "1n" u (where u is a function of x) or log u to base e (exponential), the differentiation is equal to differention of the function divided by the original function.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 1: if y = 1n(2x + 7), find dy/dx.
Solution:
y = 1n(2x + 7)
y' = 2 / (2x + 7).
The 2 is as a result of d/dx (2x + 7).
. . . . . . . . . . . . . . . . . . . . . . . .
Example 2: if y = log_e (sinx).
Solution:
y = log_e (sinx)
y' = cosx / sinx
y' = cotx
. . . . . . . . . . . . . . . . . . . . . . . .
Summary:
Differentiation is very simple, all you need is constant practise.
. . . . . . . . . . . . . . . . . . . . . . . .
Assignment:
y = e^{2x sinx} 1n [(6x+ 7) (tanx)].
. . . . . . . . . . . . . . . . . . . . . . . .

Any question?

. . . . . . . . . . . . . . . . . . . . . . . .
Stay bless.
30th March, 2015.
19:13[/b]

thankyouJesus:
[b]Good evening all, welcome to differentiation class, where everything goes. If you are just joining us, you are on a long thing.
Without wasting time unlike Jega who made us watched season film, let's quickly get down to business.
. . . . . . . . . . . . . . . . . . . . . . . .
1. PARAMETRIC DIFFERENTIATION.
General formula:
if y = a function different from x (let's say u), and
x = a function different from itself (lets say u).
Then,
dy/dx = (dy/du) (du/dx)
. . . . . . . . . . . . . . . . . . . . . . . .
Example 1: if y = 2t + 4 and x = 3t - 7, find dy/dx.
Solution:
y = 2t + 4
dy/dt = 2
x = 3t - 7
dx/dt = 3
dt/dx = 1/3
dy/dx = (dy/dt) (dt/dx)
dy/dx = (2) (1/3)
dy/dx = 2/3.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 2: if y = sina and x = 2a, find y'.
Solution:
y = sina
dy/da = cosa
x = 2a
dx/da = 2
da/dx = 1/2
dy/dx = cosa / 2
. . . . . . . . . . . . . . . . . . . . . . . .
2. IMPLICIT FUNCTIONS.
Consider this:
y = x.
The co-efficient and power of y are both 1.
i.e
1y¹ = 1x ¹.
Differentiating both sides,
(1)(1)y^1-1dy/dx = (1)(1)x^1-1,
y⁰dy/dx = 1,
dy/dx = 1.
Note: Whenever you differentiate y, always introduce dy/dx.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 1: if y = 2xy + y² + 7, find dy/dx.
Solution:
y = 2xy + y² + 7.
(1)(1)y^1-1dy/dx = 2x(dy/dx) + 2y + 2y(dy/dx) + 0.
dy/dx = 2x(dy/dx) + 2y + 2y(dy/dx),
collect like terms,
dy/dx - 2x(dy/dx) - 2y(dy/dx) = 2y,
dy/dx (1 - 2x - 2y) = 2y,
dy/dx = 2y / (1 - 2x - 2y)
. . . . . . . . . . . . . . . . . . . . . . . .

Any question?

. . . . . . . . . . . . . . . . . . . . . . . .
31st March, 2015.
18:53:05[/b]

thankyouJesus:
cc dupre
[b]Last aspect of differentiation for this class.
Differentiation Application (Maximum and Minimum values).
I hate stories when it comes to Mathematics, without wasting much time.
If y = x³/3 - x² - 2x + 5.
1. Stationary points are given by dy/dx = 0.
2. The type of each stationary point is determined by substituting the roots of the equation dy/dx = 0 in the expression for d²y/dx².
If y" is negative, then y is a maximum.
If y" is positive, then y is a minimum.
If y" is zero, then y may be a point of inflexion.
We shall need both the first and second derivatives, so make sure you are ready.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 1: if y = x³/3 - x² - 2x + 5, then y' = . . . . . . . . .and y" = . . . . . . . .
Solution:
y = x³/3 - x² - 2x + 5.
.
.
.
.
.
.
.
y' = x² -x - 2.
y" = 2x - 1.
(a) Stationary points occur at y' = 0.
Therefore, x² - x - 2 = 0.
Therefore,
(x - 2) (x + 1) = 0.
Therefore,
x = 2 and x = -1.
i.e. stationary points occur at x = 2 and x = -1.
(b) To determine the type of each stationary point, substitute x = 2 and then x = -1 in the expression for y".
At x = 2, y" = 4 - 1 = 3, i.e positive, therefore, x = 2 gives y_min.
At x = -1, y" = -2 - 1 = -3, i.e negative, therefore x = -1 gives y_max.
To find the min and max values, substituting in y = f(x) gives x = 2,
y_min = 5/3
and x = -1,
y_max = 37/6.
. . . . . . . . . . . . . . . . . . . . . . . . Summary:
This is all about differentiation concerning JAMB syllabus, differentiation is a wide topic.
. . . . . . . . . . . . . . . . . . . . . . . .
Assignment:
1. No 27, 1991
at what value of x is the function y = x² - 2x - 3 minimum.
Ans: 1.
2. No 35, 1999
If the maximum value of y = 1 + hx - 3x² is 13, find h.
Ans: 12.
3. No 36, 1999
find the value of x for which the function y = x³ - x has a minimum value.
Ans: root 3 / 3.
4. No 22, 2002
find the maximum value of y in the equation y = 1 - 2x - 3x².
Ans: 4/3
5. No 37, 2003
determine the maximum value of y = 3x² - x³.
Ans: 4.
6. No 50, 2006
find the value of x for which the function 3x³ = 9x².
Ans: 0.
7. No 5, 2007
find the value of x for which the function f(x) = 2x³ - x² - 4x + 4.
Ans: 1.
8. No 38, 2008
find the minimum value of the function y = x (1 + x).
Ans: -1/2
9. No 38, 2009
what value of x will make the function x (4 - x) a maximum.
Ans: 2.
10. No 42, 2010
at what value of x does the function y = -3 - 2x + x² attain a minimum value.
Ans:1.
. . . . . . . . . . . . . . . . . . . . . . . .

Any question?

. . . . . . . . . . . . . . . . . . . . . . . .
The next topic will be Integration by God's grace (after/before Easter. . . . . .not sure *camp things*).
. . . . . . . . . . . . . . . . . . . . . . . .
02nd April, 2015.
10:14:02[/b]

Mathemagician1:
(E) ROOTS OF POLYNOMIALS

From example 1 and 2 of this topic we've learnt how to find the roots of second  degree polynomials and our roots have always been either positive or negative roots. But it's not always so when finding roots of polynomials. There are three types of roots, there are Positive roots, Negative roots and Non-real roots.
‎
Example 3: Find the possible rational roots of P(x) = 2x³ + 3x² - 8x + 3‎
‎
3rd degree polynomial
Factors = 3
Root = 3‎
Possible root = [Using Descarte's rule of signs] +ve root = 2, 0; -ve roots = 1; Non-real roots = 0, 2.‎
Solution = See attached

CLASSWORK 
‎
(3) Find the possible rational roots of ‎F(x) = 2x³ - x² - 2x + 1‎
 ‎
Example 4: Find the possible rational roots of G(x) = x³ - 3x² + x - 3

3rd degree polynomial‎
Factors = 3‎
Root = 3
Possible root = [Using Descarte's rule of signs] +ve root = 3, 1; -ve roots = 0; Non-real roots = 0, 2.‎
Solution = See attached

CLASSWORK 

‎(4) Find the possible rational roots of ‎P(x) = 2x³ - x² + 2x - 1‎

(5) Find the possible rational roots of ‎P(x) = x³ - 3x² - 2x + 6

THE END

Mathemagician1:
(G) GRAPHS OF POLYNOMIALS 

We often say that a picture is worth a thousand words. In fact, pictures and graphs are an effective way to present information. In this tutorial, we'll learn to draw graphs of polynomials of up to the third degree. 

Have you ever wondered why graphs look so different? Straight, curve, slant, in circles, shapeless etc. Equations of different degrees possess different graph given their peculiarities.

A linear graph of the form y = mx + c is always going to be a straight line graph, however, a quadratic graph of the form y = x² is certainly going to take a U-like shape, while a graph of the form y = x³ will look like a curvy lady sitting on a chair, of course in a curvy way.

Please see attached for illustration.

Let's see if all graphs take the shape as illustrated in the introduction. 

Example 1: Graph the equation y = 3x + 6

As you can see the equation is in the form y = mx + c so it's graph is definitely going to be a straight line.

To be cont...

Mathemagician1:
(2) If a < 0, the parabola opens downward.

I'll advice that you learn how to convert quadratic equation in the form y = ax² + bx + c to standard form a(x - h)² + k cos with this you can sketch instead of plotting the points

Geofavor:
how can a number be less than zero?


hmm, more clarity on this would be nice

Mathemagician1:
I am sorry but I cannot continue with this tutorial or any other on Nairaland after the establishment of Rule 9.‎
‎
As a detribalised Nigerian, I've been on Nairaland for over 3 years contributing to the success of this forum especially in the education section, never engaged in politics or fancy the fun of it. My sole aim is to help demystify mathematics among young people irrespective of their tribe or religion. I am neither Pro-Biafran or anti-Biafran but I cannot deny the pain and suffering of my people and the agitation for a better tomorrow.‎
‎‎
Seun's comment against Biafra in the guise of loving Nigeria more than everyone else isn't only ridiculous but anti-Igbo and insensitive. Gagging the voice of Igbos through the establishment of Rule 9 is undemocratic and against my principles and conviction. 
‎
I cannot continue contributing to his forum while my kinsmen are denied their right to free speech. ‎I'm so so sorry sir but you can always write me on WhatsApp and I'll help if I can.‎

thankyouJesus:
[b]Ékárò nì bì yì o, Otutu o ma.
This is integration class, if you miss the differentiation class, quickly refer to previous pages.
INTEGRATION.
Integration is the reverse process of differentiation. When we differentiate we start with an expression and proceed to find its derivative. When we integrate, we start with the derivative and then find the expression from which it has been derived.
E.g d/dx (x⁴) = 4x³.
Therefore, the integral of 4x³ with respect to x we know to be x⁴. This is written:
[size=40]$[/size]4x³ dx = x⁴.
Generalizing:
[size=40]$[/size]xⁿ dx = [(xⁿ⁺¹)/(n+1)] + c.
This is true except when n = -1, for then we should be dividing by 0------(refer to note 5 below).
C is called the constant of integration and must be always be included.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 1: evaluate [size=40]$[/size]4x³ dx.
Solution:
[size=40]$[/size]4x³ dx = [(4x³⁺¹)/(3+1)] + c
=[(4x⁴)/(4)] + c
= x⁴ + c.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 2: determine y = [size=40]$[/size]4x³ dx, given that y = 3 when x = 2.
Solution:
As before;
y = x⁴ + c
but y = 3 when x = 2, so that;
3 = 2⁴ + c
3 = 16 + c
c = -13.
So, in this case;
y = x⁴ - 13.
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INTEGRATION OF POLYNOMIAL EXPRESSIONS.
In the previous class we differentiated a polynomial expression by dealing with the separate terms, one by one. It is not surprising, therefore, that we do much the same with the integration of polynomial expressions.
Polynomial expressions are integrated term by term with the individual constants of integration consolidated into one symbol c for the whole expression.
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Example : Evaluate [size=40]$[/size](4x³ + 3x² - 2x + 7) dx.
Solution:
[size=40]$[/size](4x³ + 3x² - 2x + 7) dx = [size=40]$[/size](4x³ + 3x² - 2x + 7(1)) dx
= [size=40]$[/size](4x³ + 3x² - 2x + 7x⁰) dx
= x⁴ + x³ - x² + 7x + c.
. . . . . . . . . . . . . . . . . . . . . . . . Summary:
1. Integration is the reverse process of differentiation. When we differentiate we start with an expression and proceed to find its derivative. When we integrate, we start with the derivative and then find the expression from which it has been derived.
2. [size=40]$[/size]xⁿ dx = [(xⁿ⁺¹)/(n+1)] + c.
This is true except when n = -1, for then we should be dividing by 0.
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Assignment:
Note the following;
1. d/dx (sinx) = cosx
[size=40]$[/size]sinx dx = -cosx + c
2. d/dx (cosx) = -sinx
[size=40]$[/size] cosx dx = sinx + c
3. d/dx (tanx) = sec²x
[size=40]$[/size]sec²x dx = tanx + c
4. d/dx (e^x) = e^x
[size=40]$[/size]e^x dx = e^x + c
5. d/dx (1n x) = 1/x
[size=40]$[/size]1/x dx = 1nx + c
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Any question?

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Have a blessed day.[/b]

Geofavor:
i have a question
how did you get your values for X in the examples?
Like when you said if x = -4, y = -3, why did you decide to make x equal to -4 there, why not -2 or anyother number?

Geofavor:
how can a number be less than zero?

There are Positive numbers and negative numbers. 0 stands in the middle. Positive numbers are greater than 0 eg. 1,2,3,4... while negative numbers are less than 0 eg -1,-2,-3,-4...

hmm, more clarity on this would be nice

Geofavor:
Mathemagician, can you, at least, answer my questions? I'm not on whatsapp.

thankyouJesus:
[b]Good evening ladies and gentlemen. Please, bear with us, we are sorry for the "Orubebe" (disturbing this thread with mathematics), if not, we will "Oba Akiolu" you.
FUNCTIONS OF A LINEAR FUNCTION OF X.
It is often neccessary to integrate any one of the expressions shown in our list of standard integrals (the note in the previous class) when the variable x is replaced by a linear expression in x. That is, of the form ax + b.
Note:
1. [size=40]$[/size](2x + 6)^n>1 dx the 2x +6 is a linear expression because the degree power of x is 1 and algebraic substitution is adopted.
2. [size=40]$[/size]cos(5x + 6) dx, the 5x + 6 is linear, therefore algebraic substitution is adopted.
3. [size=40]$[/size]e^{(8x + 7)} dx, 8x + 7 is linear, algebraic substition method.
4. [size=40]$[/size](2x² + 3)³ dx, the degree of x is 2, this is not linear but quadractic expression, to solve this, the brackets is expanded, the result will yield polynomials, integrate in turns.
5a. [size=40]$[/size](x sinx) dx, they are both linear expressions but they are connected together by multiplication, algebraic method would't work (we will consider a question like this in a later class).
5b. [size=40]$[/size]{(2x + 6)(x² + 6x)} dx, algebraic method can solve this, till later class..
In summary, algebraic substitution method is used to solve linear expression, but there are some special cases where it work also, it is all about substituting.
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Example 1: evaluate [size=40]$[/size](2x + 7)⁴ dx.
Solution:
[size=40]$[/size](2x + 7)⁴ dx
let u = 2x + 7,
differentiate u wrt x
du/dx = 2
making dx the subject of the relation
dx = du/2
substituting all the above into the question
=[size=40]$[/size]u⁴ du/2
1/2[u⁴⁺¹/4+1] + c
=1/2[u⁵/5] + c
=1/10[u⁵] + c
=u⁵/10 + c,
substituting back
=(2x + 7)⁵/10 + c
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Example 2: evaluate [size=40]$[/size]cos(2x + 7) dx.
Solution:
[size=40]$[/size]cos(2x + 7) dx
let u = 2x + 7
dx = du/2

[size=40]$[/size]cos(2x + 7) dx
=[size=40]$[/size]cosu du/2.
[color=grey]Note: cos is a standard integral (note of the previous class), integral of cos is sin.[/color]
= 1/2[sinu] + c
= sinu/2 + c
= [sin(2x + 7)]/2 + c.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 3: evaluate [size=40]$[/size]e^{(2x + 4)} dx.
Solution:
[size=40]$[/size]e^{(2x + 4)} dx
let u = 2x + 4
dx = du/2
[size=40]$[/size]e^{(2x + 4)} dx
= [size=40]$[/size]e^u du/2
recall that e is also standard integral
= e^u/2 + c
= e^{(2x + 4)}/2 + c
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Summary:
Integration is easy if only you can understand differentiation.
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Assignment:
Practise the above, if you can the master the above, concerning JAMB/POST UTME, integration will be a walk over, only if you can master this.
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Any question?

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To all those we are stepping on their toes, we are deeply very sorry, dont "Orubebe" us so we would't "Oba Akinolu" you.
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Good Night.
AFD[/b]

thankyouJesus:
[b]Good afternoon ladies and gentlemen, how is your day going?
INTEGRALS OF THE FORM [size=40]$[/size]f(x)f'(x) dx.
Consider the integral [size=40]$[/size]sinx cosx dx. This is not one of our standard integrals, so how shall we tackle it? This is an example of a type of integral which is very easy to deal with but which depends largely on how keen your wits are. You will notice that if we differentiate one of the two, we obtain the other expression.
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Note 1: Whenever you have integral of two products, the first thing is, can I differentiate to cancel out the other? If yes, algebraic substitution method is adopted:
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Example 1: evaluate [size=40]$[/size]sinx cosx dx.
Solution:
[size=40]$[/size]sinx cosx dx.
This is product of two expressions (sinx and cosx), since the differentiation of sin is cos and the differentiation of cos is -sin, therefore, I can pick any one of the two.
[size=40]$[/size]sinx cosx dx
let u = sinx
du/dx = cosx
dx = du/cosx
substituting back into the question
[size=40]$[/size]u cosx du/cosx
the cosx will strike out the other cosx, we are left with
[size=40]$[/size]u du
{u¹⁺¹/1+1} + c
u²/2 + c
sin²x/2 + c
. . . . . . . . . . . . . . . . . . . . . . . .
Example 2: evaluate [size=40]$[/size](2x + 3)(x² + 3x)² dx.
Solution:
[size=40]$[/size](2x + 3)(x² + 3x)² dx,
mere looking at the two shows that, if you differentiate the other without the power, you will get the other, so,
let u = x² + 3x
du/dx = 2x + 3
dx = du/(2x + 3)
substituting back
[size=40]$[/size](2x + 3)u² du/(2x + 3)
[size=40]$[/size]u²du
u³/3 + c
(2x + 3x)³/3 + c.
. . . . . . . . . . . . . . . . . . . . . . . .
Note 2: Can I manipulate to cancel out the other?
Example 1: evaluate [size=40]$[/size](x + 2)(x² + 4x)² dx.
Solution:
[size=40]$[/size](x + 2)(x² + 4x)² dx
mere looking shows that the other can strike out the other except that
.
.
.
.
.
.
.
Very good, multiply by 2.
[size=40]$[/size](x + 2)(x² + 4x)² dx
let u = x² + 4x
dx = du/(2x + 4)
[size=40]$[/size](x + 2)u² du/(2x + 4)
[size=40]$[/size](x + 2)u² du/{2(x + 2)}
[size=40]$[/size]u² du/2
[1/2(u³/3)] + c
u³/6 + c
(x² + 4x)³/6 + c
. . . . . . . . . . . . . . . . . . . . . . . .
Note 3: Is there a way I can go round?
Example 1: evaluate [size=40]$[/size]e^sinxcosx dx.
Solution:
[size=40]$[/size]e^sinxcosx dx
mere looking shows that sin will strike out cos and vice versa, but one is in power, i.e if you pick cos, you will have -sin, there is no "mathematical magic" that can strike out e^sin with sin, not just possible.
We will pick what as u?
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.
Very good
u = sinx
dx = du/cosx
[size=40]$[/size]e^ucosx du/cosx
[size=40]$[/size]e^udu
hope you can still remember e is a standard integral?
e^u + c
e^sinx + c
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Summary:
Whenever you come across product of two functions:
1. The first thing is? Can one strike out the other after differentiation? If yes, . . . . . .you know what to do.
2. If 1 above does not work, can I manipulate to strike out the other? If yes. . . . . . .you know what to do.
3. If 1 and 2 above fail me, what next? Can I go round?

A big Care must be taken

.
Note: Not all products require algebraic substitution method e.g [size=40]$[/size]x e^sinx dx, [size=40]$[/size]x sinx dx, in this case, another method is adopted, but concerning JAMB/POST, integration for products end here.
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Any question?

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2 more sub topics, that will lead us to any of trig, variation, geometry, polynomials e.t.c
AFD.[/b]

thankyouJesus:
Good evening ladies and gentlemen of the high table, all protocols are noted and well duly obsesved.
INTEGRATION AS A SUMMATION.
These are integration with boundaries, i.e upper bound and lower bound.

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Example 1: Evaluate [size=40]$[/size]²₁(2x + 4) dx.
Solution:
[size=40]$[/size]²₁(2x + 4) dx
integrating without the boundaries yield:
[size=40][[/size]x² + 4x[size=40]][/size]²₁
substituting the upper and lower boundaries through the introduction of substraction sign
=[2(2)² + 4(2)] - [2(1)² + 4(1)]
=[2(4) + 8] - [2 + 4]
=[8 + 8] - [6]
=16 - 6
=10

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Example 2: Evaluate [size=40]$[/size]⁹⁰₀sinx dx
Solution:
[size=40]$[/size]⁹⁰₀sinx dx
[size=40][[/size]-cosx[size=40]][/size]⁹⁰₀
=(-cos90) - (-cos0)
= (-0) - (-1)
= 1

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Any question?

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One more to go