The document discusses integration, which is the reverse process of differentiation. It provides formulas for integrating polynomial expressions like xn and axn and explains how to find the constant of integration C. Examples are given for integrating expressions and finding the area under a curve by integrating between limits of integration.
Solving second order ordinary differential equations (boundary value problems) using the Least Squares Technique. Contains one numerical examples from Shah, Eldho, Desai
Solving second order ordinary differential equations (boundary value problems) using the Least Squares Technique. Contains one numerical examples from Shah, Eldho, Desai
01. Differentiation-Theory & solved example Module-3.pdfRajuSingh806014
Total No. of questions in Differentiation are-
In Chapter Examples 31
Solved Examples 32
The rate of change of one quantity with respect to some another quantity has a great importance. For example the rate of change of displacement of a particle with respect to time is called its velocity and the rate of change of velocity is
called its acceleration.
The following results can easily be established using the above definition of the derivative–
d
(i) dx (constant) = 0
The rate of change of a quantity 'y' with respect to another quantity 'x' is called the derivative or differential coefficient of y with respect to x.
Let y = f(x) be a continuous function of a variable quantity x, where x is independent and y is
(ii)
(iii)
(iv)
(v)
d
dx (ax) = a
d (xn) = nxn–1
dx
d ex =ex
dx
d (ax) = ax log a
dependent variable quantity. Let x be an arbitrary small change in the value of x and y be the
dx
d
(vi) dx
e
(logex) = 1/x
corresponding change in y then lim
y
if it exists, d 1
x0 x
is called the derivative or differential coefficient of y with respect to x and it is denoted by
(vii) dx
(logax) =
x log a
dy . y', y
dx 1
or Dy.
d
(viii) dx (sin x) = cos x
So, dy dx
dy
dx
lim
x0
lim
x0
y
x
f (x x) f (x)
x
(ix) (ix)
(x) (x)
d
dx (cos x) = – sin x
d (tan x) = sec2x
dx
The process of finding derivative of a function is called differentiation.
If we again differentiate (dy/dx) with respect to x
(xi)
d (cot x) = – cosec2x
dx
d
then the new derivative so obtained is called second derivative of y with respect to x and it is
Fd2 y
(xii) dx
d
(xiii) dx
(secx)= secx tan x
(cosec x) = – cosec x cot x
denoted by
HGdx2 Jor y" or y2 or D2y. Similarly,
d 1
we can find successive derivatives of y which
(xiv) dx
(sin–1 x) = , –1< x < 1
1 x2
may be denoted by
d –1 1
d3 y d4 y
dn y
(xv) dx (cos x) = –
,–1 < x < 1
dx3 ,
dx4 , ........, dxn , ......
d
(xvi) dx
(tan–1 x) = 1
1 x2
Note : (i)
y is a ratio of two quantities y and
x
(xvii) (xvii)
d (cot–1 x) = – 1
where as dy
dx
dy
is not a ratio, it is a single
dx
d
(xviii) (xviii)
(sec–1 x) =
1 x2
1
|x| > 1
quantity i.e.
dx dy÷ dx
dx x x2 1
(ii)
dy is
dx
d (y) in which d/dx is simply a symbol
dx
(xix)
d (cosec–1 x) = – 1
dx
of operation and not 'd' divided by dx.
d
(xx) dx
(sinh x) = cosh x
d
(xxi) dx
d
(cosh x) = sinh x
Theorem V Derivative of the function of the function. If 'y' is a function of 't' and t' is a function of 'x' then
(xxii) dx
d
(tanh x) = sech2 x
dy =
dx
dy . dt
dt dx
(xxiii) dx
d
(xxiv) dx
d
(coth x) = – cosec h2 x (sech x) = – sech x tanh x
Theorem VI Derivative of parametric equations If x = (t) , y = (t) then
dy dy / dt
=
(xxv) dx
(cosech x) = – cosec hx coth x
dx dx / dt
(xxvi) (xxvi)
(xxvii) (xxvii)
d (sin h–1 x) =
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1. Lecture 6 - Integration
C2 Foundation Mathematics (Standard Track)
Dr Linda Stringer Dr Simon Craik
l.stringer@uea.ac.uk s.craik@uea.ac.uk
INTO City/UEA London
2. Lecture 6 skills
integrate a polynomial expression
integrate the gradient function (dy
dx ) of a curve
find C, the constant of integration
integrate to find the area under a curve
3. Differentiation
If y = xn, then
dy
dx
= nxn−1
If y = x3 + x2 + x + 1, then
dy
dx
= 3x2
+ 2x + 1
If y = 2x3 + 5x2 + 7x + 200, then
dy
dx
= 6x2
+ 10x + 7
If y = x3
3 + x2
2 + x + 52, then
dy
dx
= x2
+ x + 1
4. The reverse of differentiation
Question: If dy
dx = 2x, what is y?
Answer: y = x2 + C?
Question: If dy
dx = 3x2, what is y?
Answer: y = x3 + C?
Question: If dy
dx = x2, what is y?
Answer: y = x3
3 + C?
5. Integration is the reverse of differentiation
If dy
dx = xn, then
y =
xn+1
n + 1
+ C, n = 1
C is called the constant of integration
If dy
dx = x5, then
y =
x5+1
5 + 1
+ C =
x6
6
+ C
If dy
dx = x2, then
y =
x3
3
+ C
If dy
dx = 3x2, then
y = 3 ×
x3
3
+ C = x3
+ C
6. Integration
Integration is the reverse of differentiation.
If we are given dy
dx , we integrate to find y.
We also use integration to find the area under a curve.
The integral of a function f(x) with respect to x is written as
f(x) dx
7. Integrating a polynomial
Let x be a variable, and a, b, n, m be constants
xn
dx =
xn+1
n + 1
+ C
axn
dx = a xn
dx =
axn+1
n + 1
+ C
axn
+ bxm
dx =
axn+1
n + 1
+
bxm+1
m + 1
+ C
Note: these formulae work for all n, m except for
n, m = −1.
Say ’the integral of xn, dx, is x to the n plus 1, over n plus
1, plus C.’
8. Examples
xn
dx =
xn+1
n + 1
+ C
What is the integral of x4 + x3 + x2 with respect to x?
Answer:
x4
+ x3
+ x2
dx =
x5
5
+
x4
4
+
x3
3
+ C
What is the integral of x2 + x + 1 with respect to x?
Answer:
x2
+ x + 1 dx =
x3
3
+
x2
2
+ x + C.
9. Examples
axn
dx =
axn+1
n + 1
+ C
What is the integral of 5x4 + 4x3 + 3x2 with respect to x?
Answer:
5x4
+4x3
+3x2
dx =
5x5
5
+
4x4
4
+
3x3
3
+C = x5
+x4
+x3
+C
What is the integral of 3x2 + 2x + 1 with respect to x?
Answer:
3x2
+ 2x + 1 dx =
3x3
3
+
2x2
2
+ x + C = x3
+ x2
+ x + C
10. Examples
axn
dx =
axn+1
n + 1
+ C
What is the integral of 2x−3 + 5 with respect to x?
Answer:
2x−3
+ 5 dx =
2x−2
−2
+ 5x + C = −x−2
+ 5x + C
What is the integral of 6x
1
2 with respect to x?
Answer:
6x
1
2 dx =
6x1/2+1
1/2 + 1
+C =
6x3/2
3/2
+C = 6x
3/2
×
2
3
+C =
4x3/2
3
+C
11. The same question asked in three ways
Question 1: If dy
dx = x6 + 10x + 3, what is y?
Question 2: Integrate x6 + 10x + 3 with respect to x.
Question 3: Find x6 + 10x + 3 dx.
Answer 1: y = x7
7 + 5x2 + 3x + C
Answer 2: x7
7 + 5x2 + 3x + C
Answer 3: x7
7 + 5x2 + 3x + C
12. Finding C, the constant of integration
Question:
A curve with gradient function
dy
dx = 3x2 − 12x + 9 passes through
the point (1, 5). Find the equation of
the curve.
1 2 3 4
2
4
6
•
(1, 5)
Answer: Integrate dy
dx to find y
y = 3 × x3
3 − 12 × x2
2 + 9x + C
y = x3 − 6x2 + 9x + C.
We know that the point (1, 5) is on the curve, so substitute
x = 1 and y = 5 to find C.
5 = 13 − 6(1)2 + 9(1) + C
5 = 4 + C
C = 1
So the equation of the curve is y = x3 − 6x2 + 9x + 1.
13. Finding C, the constant of integration
Question:
A line with gradient dy
dx = 6 passes
through the point (1, 5). Find the
equation of the line.
−2 2 4 6 8−2
2
4
6
8
•(1, 5)
Answer: Integrate dy
dx to find y
y = 6x + C
We know that the point (1, 5) is on the curve, so substitute
x = 1 and y = 5 to find C.
5 = 6 × 1 + C
C = −1
So the equation of the line is y = 6x − 1.
14. Use integration to finding the area under a curve
b
a
f(x) dx
is area under the curve y = f(x) between x = a and x = b.
a b
x
y
a and b are the limits of integration
a is the lower limit and b is the upper limit
is the integral sign (introduced by Leibniz in 1675)
15. The Fundamental Theorem of Calculus Part II
If f(x) dx = F(x), then
b
a
f(x) dx = [F(x)]b
a = [F(b)] − [F(a)]
16. Example
Question:
What is the area under the
curve y = 3x3 + 2 between
x = 0 and x = 2?
−1 1 2
10
20
30
Answer:
2
0 3x3 + 2 dx = [3x4
4 + 2x]2
0
= [3×(2)4
4 + 2 × 2] − [3×04
4 + 2 × 0]
= [12 + 4] − [0 + 0]
= [16] − [0] = 16
17. Example 2
Question:
What is the area under the
curve y = 6x2 + 2x−3 between
x = −2 and x = −1?
−2 −1
10
20
Answer:
−1
−2 6x2 + 2x−3 dx = [2x3 − x−2]−1
−2
= [2(−1)3 − (−1)−2] − [2(−2)3 − (−2)−2]
= [−2 − 1] − [−16 − 1
4 ]
= [−3] − [−64−1
4 ]
= 53
4 = 131
4 = 13.25
18. Differentiating and integrating - summary
Differentiate to find the derivative
If y = xn
, then
dy
dx
= nxn−1
Integrate to find the integral
xn
dx =
xn+1
n + 1
+ C If
dy
dx
= xn
, then y =
xn+1
n + 1
+ C
Differentiate
y = ax2 + bx + c
dy
dx = 2ax + b
d2y
dx2 = 2a
Integrate
Differentiate
y = ax3 + bx2 + cx + d
dy
dx = 3ax2 + 2bx + c
d2y
dx2 = 6ax + 2b
Integrate
