APPLICATION OF INTEGRATION
THIRD TERM E-LEARNING NOTE
SUBJECT: FURTHER MATHEMATICS CLASS: SS 2
SCHEME OF WORK
WEEK NINE
TOPIC : APPLICATION OF INTEGRATION II : SOLID REVOLUTION AND TRAPEZOIDAL RULE
A solid whichb has a central axis of symmetry is a solid of revolution.Forexample, a cone, a cylinder , a vase etc.
y
Consider the area under a portion AB of the curve y = f(x) revolved about the x axis through four right angle or 3600, each point of the curve describes a circle centered on the x axis. A solid revolution can be thought of as created in this way with the circular plane ends cutting the x – axis at x = a and x = b.
Let v be the volume of the solid for x = a up to an arbitrary value of x between a and b. Given abincreament dx in x , and y takes an increamentdy and v increases by dv.
The figure shows a section through the x axis, from this it is seen that the slice dv of bthickness dx is enclosed between two cylinders of outer radius y +dy and inner radius y .
Then ,πy2dx< dv < π (y + dy)2dx;
With appropriate modification, if the curve is falling at this point.
π y2 < dv/dx < π (y + dy)2
if dx 0, dy 0 as dv/dx dv/dx
: . dv/dx = πy2 or V =
Where y = f(x) and v = volume of solid revolution of the curve where y = f(x) is rotation completely and x – axis between limits x = a and x =b.
Examples;The portion ofthe curve y = x2 between x = 0 and x = 2 is rotated complrtely around the x axis, find the volume of the solid generated?
V =
=
V =
V = =
Put x = 2 and x = 0 then substitute into the expression above
V =
THE TRAPEZOIDAL RULE
There are many definite integrals which can’t be evaluated and thus required advance techniques e.g
etc
We can find an approximate value for such integralsbyb finding the area approximately. There are many methods methods of doing this and such methods include the Trapezium rule.
Y
Y = f(x)
y1 y2y3 yn-1yn
x1 x2 x3 xn-1xn
= ½ (y1 +y2)h + ½ (y2 +y3)h + ½ (yn-1 +yn)h
½ h(y1+2y2+2y3+……….+2yn-1 + yn)
½ (width of each trap. ) × (first ordinate + last ordinate )+ 2( sum of all other ord.)
F(x)dx = ½ h{y1 + yn} +2{y2 +y3 + …Yn}.
Example
Find the approximate value of at interval 0.5
| X | 1 | 1.5 | 2 | 2.5 | 3.0 |
|---|---|---|---|---|---|
| Y = 1/x | 1 | 0.67 | 0.5 | 0.4 | 0.33 |
Applying the rule;
{ ½ . ½ { (1 +0.33) + 2 ( 0.67 + 0.5 + 0.4)}
¼ {(1.33) +2(1.57)}
¼ (4.47) = 1.12.
1 0.67 0.5 0.4 0.33
Ex (2). Make a table of value of y for which y = for which x =2 t0 x = 3 at interval of 0.2.
| X | 2 | 2.2 | 2.4 | 2.6 | 2.8 | 3 |
|---|---|---|---|---|---|---|
| X2-1 | 3 | 3.84 | 4.76 | 5.76 | 6.84 | 8 |
| 1.732 | 1.956 | 2.182 | 2.4 | 2.615 | 2.828 | |
| 0.5754 | 0.5703 | 0.4583 | 0.4167 | 0.3824 | 0.3536 |
Using the rule.
= ½ *0.2 (0.5774 +0.3536 + 3.5354)
0.44664 *2
0.89 correct to 2 dp
APPLICATION OF INTEGRATION TO KINEMATICS
If the ve;locityis given as a function of time, the displacement is the integral of the velocity function with respect to the time .
ds/dt = f(t)
then S =
= f(t) + C
Similarly, if the acceleration is a function of time, the velocity is the integral of the acceleration function.
Ex. A particle is projected in a straight line from O until a speed of 6m/s is attained. At time t secs.Later,its acceleration is (1 + 2t) m/s2 for the value of t = 4. Calculate for the particle (i) its velocity (ii) its distance from O
dv/dt = 1+ 2t
v = = t + t2 + C
when t = 0
v = 6m/sand c = 6.
V = (t2 + t + 6) m/s
When t = 4, v = 16 + 4 + 6 = 26m/s.
(ii) distance (s) = ds/dt = t2 + t + 6
S =
S = {t3/3 + 16/2 6t}4
= 160/3
m.
Evaluation
1. Find the area enclosed byb y = x2 – x -2 and the x axis
2. find the area under the curve y = x2/3 between x = 2 and x = k is 8 times the area under the same curve between x = 1 and x =2, hence find the value of k.
GENERAL EVALUATION
A particle moves in astraight line from O until the initial velocity was 2m/s. its acceleration is given by (2t -3)m/s2. Calc. (i) its velocity after 3 secs. (ii) the distance from O when it is momentarily at rest.
Find the volume of solid revolution when a is the region bounded by the cuerve y = 2x. and the ordinate at x = 2,and x = 4 and the x axis is revolved by 2π.
Reading Assignment :F/Matrhs Project, pg 47 – 63
WEEKEND ASSIGNMENT
1. Integrate 2√x A. B.4x3/2 + C C. + C D.
2. Integrate A. + C B. x3/2 + C C. x2 + C D. + C
3. The gradient of a curve is 6x + 2 and it passes through the point (1,3), find its equation A. 3x2 – 2x + 2 B. 3x2 -2x -2 + 2x + 2 C. 3x2 – 2x + 2 D. 3x2 – 2x -2
4. Evaluate A. – + x2 – 2x + C B. x3/3 – x3/2 + x2 + 2x +C C. x3/3 + x2/2 +x3-2x + C D. x4/4+x3/3-x2+2x+C
5. Eval. A. 9 +C B. 8/3 + C C. 24 D. 18 + C
THEORY
Evaluate
Using trapezoidal rule, with ordinate x = -3,-2,-1,0,1,2,3 and 4. Calc correct to 3 dp an approximate value of
