thankyouJesus:
Good morning ladies and gentlemen.
The last aspect of Integration concerning JAMB/POST JAMB.
INTEGRALS OF THE FORM [size=40]$[/size]f'(x)/f(x) dx.
Just like the last class, can I strike out? Can I manipulate to strike out?

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Example 1: Evaluate [size=40]$[/size](2x + 3)/(x² + 3x - 5) dx.
Solution:
[size=40]$[/size](2x + 3)/(x² + 3x - 5) dx
just like product, we can strike out the numerator by differentiating the denominator
let u = x² + 3x - 5
du/dx = 2x + 3
dx = du/(2x + 3)
substituting back
[size=40]$[/size]{(2x + 3)/u} {(du/(2x + 3)}
= [size=40]$[/size]1/u du
this is a standard integral right? Good, the standard is
.
.
.
.
.
.
.
Very good,
1n u + c
1n(x² + 3x - 5) + c

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Example 2: Evaluate [size=40]$[/size](4x - / (x² - 4x + 5) dx.
Solution:
[size=40]$[/size](4x - / (x² - 4x + 5) dx
what do you notice? Very good, manipulation can work.
Let u = fill in the gap
dx = fill in the gap
[size=40]$[/size](4x - /u (fill in the gap)
= 2 1n(x² - 4x + 5) + c

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Example 3: Evaluate [size=40]$[/size]cosx/sinx dx.
Solution:
[size=40]$[/size]cosx/sinx dx
allow me to call this a straight forward method i.e the differentiation of the denominator yields the numerator.
Whenever you notice that the differentiation of the denom yield the numen, solve this way
[size=40]$[/size]cosx/sinx dx
= 1nsinx + c.
Same as example 1 above.

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Example 4: Evaluate [size=40]$[/size](x - 3)/(x² - 6x + 2) dx.
Solution:
[size=40]$[/size](x - 3)/(x² - 6x + 2) dx
using direct method, but we need to manipulate the numerator by 2 (which is known from differentiation), but integration is the reverse process of differention (we will settled for the inverse of 2 which is 1/2)
1/2 1n(x² - 6x + 2) + c.
Try it with example 2.

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Any question?

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We are through with Integration concerning JAMB

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Take home:
1. No 42, 1994
[size=40]$[/size]¹_-1(2x + 1)² dx.
Ans: 14/3
2. No 42, 1997
if y = x(x⁴ + x² + 1) cal [size=40]$[/size]¹_-1y dx.
Ans: 0.
Hint: multiply out and integrate in turn.
3. No 40, 1998
[size=40]$[/size]^pi₂(sec²x - tan²x) dx.
Ans: pi - 2.
Hint: tan²x + 1 = sec²x
4. No 38, 1999
evaluate [size=40]$[/size]¹_-2(x - 1)² dx.
Ans: 9
5. No 37, 1999
evaluate [size=40]$[/size]^pi/4₀(sinx - cosx) dx.
Ans: -1 - root2
Hint: trig of pi is 180 (just as before), but pi is 22/7.

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Have a nice day.

Mathemagician1:


N = cubrt(1 - 8a/m)

Tobiee:
[email] Differentiation and integration gives me a headache [/email]

mathefaro:
@thankyouJesus, thumbs up for the wonderful with you're doing here.
Eyaah! This shouldn't be the case cos it's a very interesting topic. Trust me, you'll agree with me once you get on with it. And don't worry, JAMB will only set questions at the surface of the topic.

I'll be available for any assistance you guys might need. And for more effective and faster response, contact me on whatsapp:

vickvan:
Where are the math gurus?
You guys should treat Binary Operations, Matrices, Determinants, Permutations and Combinations. Pls explain further on Integration.
@ Umartins1
Orezy5
Geofavor
Thankyoujesus. Thanks

Goldmaxx:
i scored 248 in jamb and 327 in post jamb. Unn precisely

vickvan:
The binary operation * is defined by x * y= xy - y - x for all real values of x and y. If x * 3= 2 * x, find the value of x.
@ Umartins1 Orezy5 Geofavor Thankyoujesus. Pls i need you guys back on this thread.

vickvan:
Where are the math gurus?
You guys should treat Binary Operations, Matrices, Determinants, Permutations and Combinations. Pls explain further on Integration.

parkparky:
Solve d eqn. m^2 + n^2 =29; m + n=7

Umartins1:


M² + n² = 29 ...... (1)
M+n=7....... (2)


From eqn (2), m+n= 7

M= 7-n


Sub for m in eqn (1)


(7-n)² + n² = 29


(7-n)(7-n) + n² = 29


49-7n-7n+n²+n² = 29

49-14n+2n² = 29

2n²-14n+49-29= 0


2n²-14n+20= 0


n²-7n+10= 0


n²-5n-2n+10= 0


n(n-5)-2(n-5) = 0


(n-2)(n-5) = 0


n=2 or n=5


When n=2,


Sub for n in eqn (2)


M+n=7


M+2=7


M= 7-2


M= 5



When n=5


Sub for n in eqn (2)


M+n= 7


M+5= 7


M= 7-5


M= 2



(2,5) (5,2)

Umartins1:


M² + n² = 29 ...... (1)
M+n=7....... (2)


From eqn (2), m+n= 7

M= 7-n


Sub for m in eqn (1)


(7-n)² + n² = 29


(7-n)(7-n) + n² = 29


49-7n-7n+n²+n² = 29

49-14n+2n² = 29

2n²-14n+49-29= 0


2n²-14n+20= 0


n²-7n+10= 0


n²-5n-2n+10= 0


n(n-5)-2(n-5) = 0


(n-2)(n-5) = 0


n=2 or n=5


When n=2,


Sub for n in eqn (2)


M+n=7


M+2=7


M= 7-2


M= 5



When n=5


Sub for n in eqn (2)


M+n= 7


M+5= 7


M= 7-5


M= 2



(2,5) (5,2)

vickvan:
Please math gurus, you guys should tutor us on binary operations, integration, matrices and determinants. Exam is just 4 weeks away. @ Umartins1 Orezy5 Thankyoujesus. God bless you all.

vickvan:
Please math gurus, you guys should tutor us on binary operations, integration, matrices and determinants. Exam is just 4 weeks away. @ Umartins1 Orezy5 Thankyoujesus. God bless you all.

Umartins1:


M² + n² = 29 ...... (1)
M+n=7....... (2)


From eqn (2), m+n= 7

M= 7-n


Sub for m in eqn (1)


(7-n)² + n² = 29


(7-n)(7-n) + n² = 29


49-7n-7n+n²+n² = 29

49-14n+2n² = 29

2n²-14n+49-29= 0


2n²-14n+20= 0


n²-7n+10= 0


n²-5n-2n+10= 0


n(n-5)-2(n-5) = 0


(n-2)(n-5) = 0


n=2 or n=5


When n=2,


Sub for n in eqn (2)


M+n=7


M+2=7


M= 7-2


M= 5



When n=5


Sub for n in eqn (2)


M+n= 7


M+5= 7


M= 7-5


M= 2



(2,5) (5,2)

parkparky:
dis one na surd.
2/1-₹3

Ikennablue:
May God bless you all, geofavour, ThankyouJesus and all our teachers, we are very grateful, I've been following this thread religiously, and I tell u, coupled with my textbook and past questions I've been doing well..





please dnt stop

vickvan:
The binary operation * is defined by x * y= xy - y - x for all real values of x and y. If x * 3= 2 * x, find the value of x.
@ Umartins1 Orezy5 Geofavor Thankyoujesus. Pls i need you guys back on this thread.

Geofavor:
This thread is too dull for my liking. I will soon be bombarding all of you with questions. No dulling!

vickvan:
If y= 3sin (-4x), dy/dx is