Acceleration assisstance (1 Viewer)

[Dragonmaster262]

I'm planning on self accelerating for Mathematics Extension 1. Unfortunately, I use Cambridge. It has good questions but very poor explanations so I was wondering if you guys could help explain to me some things which I don't understand.

For definite integrals how are we supposed to know which shape the graph would make? What would the answer for a graph like this be:

Q) Sketch the graph and use the area formula to evaluate it:

3~6(2x+1)dx

P.S. ~ is the smooth sum.

 

[3unitz]

sketch the graph of 2x + 1 (straight line with gradient 2, cuts the y-axis at 1). area would be the area under the line, between x=3 and x=6 and the x-axis (will look like the area of a trapezium).

 

[Dragonmaster262]

sketch the graph of 2x + 1 (straight line with gradient 2, cuts the y-axis at 1). area would be the area under the line, between x=3 and x=6 and the x-axis (will look like the area of a trapezium).

Thanks, can you tell me how to solve these using the fundamental theorem?

Q) -3~1(2x^2-7x+5)dx

Q) 0~3(x+x^2+x^3)dx

 

[3unitz]

Thanks, can you tell me how to solve these using the fundamental theorem?

Q) -3~1(2x^2-7x+5)dx

integrate the function first:

(2/3)x^3 - (7/2)x^2 + 5x

then sub x=1, and subtract from when x=-3:

[(2/3) - (7/2) + 5] - [(2/3)*(-3)^3 - (7/2)*(-3)^2 + 5*(-3)]

= 200/3

Q) 0~3(x+x^2+x^3)dx

integrate first:

(1/2)x^2 + (1/3)x^3 + (1/4)x^4

then sub x=3, and subtract from when x=0:

[(1/2)*3^2 + (1/3)*3^3 + (1/4)*3^4] - 0

= 135/4

 

[Dragonmaster262]

integrate the function first:

(2/3)x^3 - (7/2)x^2 + 5x

then sub x=1, and subtract from when x=-3:

[(2/3) - (7/2) + 5] - [(2/3)*(-3)^3 - (7/2)*(-3)^2 + 5*(-3)]

= 200/3



integrate first:

(1/2)x^2 + (1/3)x^3 + (1/4)x^4

then sub x=3, and subtract from when x=0:

[(1/2)*3^2 + (1/3)*3^3 + (1/4)*3^4] - 0

= 135/4

Thanks man, how about this:

Q) Expand and evaluate:

1) 1~2 (1+x^2)/x^2

2) 1~4 (x^3-3)/x^3

 

[3unitz]

Thanks man, how about this:

Q) Expand and evaluate:

1) 1~2 (1+x^2)/x^2

= 1~2 (x^-2 + 1) dx

= (-x^-1 + x) |2,1

= (-2^-1 + 2) - (-1^-1 + 1)

= 3/2

2) 1~4 (x^3-3)/x^3

= 1~4 (1 - 3x^-3) dx

= [x + (3/2)x^-2] |4,1

= [4 + (3/2)*4^-2] - [1 + (3/2)*1^-2]

= 51/32

 

[Dragonmaster262]

These ones have got me stumped.

Q) Use the reverse chain rule:

1) a) Find (5-x^2-x)^7
b) Hence find: ~(-14x-7)(5-x^2-x)^6 and ~(2x+1)(5-x^2-x)^6
2) a) Find (2x^2+3)^1/2
b) Find ~2x/((2x^2+3)^1/2) and x/((2x^2+3)^1/2)
3) a) Find ((x)^1/2+1)^3
b) Hence find ~(3((x)^1/2+1)^2)/2(x)^1/2 and ((x)^1/2+1)^2)/(x)^1/2

Q) Write up a table of values and estimate 0~4x(x-4) using the trapezoidal rule.

 

[LordPc]

as I understand it you simply derive the function, eg (5-x^2-x)^7 becomes (-14x-7)(5-x^2-x)^6 when you derive it with respect to x ( = wrt x)

then you have an equation which reads
d[(5-x^2-x)^7]/dx = (-14x-7)(5-x^2-x)^6..............(this is in the form d(...)/dx)

integrate both sides and you get
(5-x^2-x)^7 + C = ~(-14x-7)(5-x^2-x)^6

so answer to the first is "(5-x^2-x)^7 + C"

its the same with all the rest. derive the function, re arrange it to get what they want you to find and integrate, as deriving and integrating are opposite functions.

for the next part just take out a -7 from the (-14x-7) bracket, rearrange and do the same thing

and I would suggest not using cambridge if at all possible to learn the course. I used it in year 11 and 12 for questions, but every lesson my teacher would explain the content first on the board cause cambridge are very poor at explaining concepts. for example the "reverse chain rule" is only called the reverse chain rule in cambridge, it doesnt exist anywhere else, they just made up that name. so perhaps get an excel book or one of those understand maths books from the accelerated maths learning series, they are pretty good. but yea, just cambridge for learning content is hard, best to have other support material

 

[bored of sc]

1) a) Find (5-x^2-x)^7
b) Hence find: ~(-14x-7)(5-x^2-x)^6 and ~(2x+1)(5-x^2-x)^6

(a) f(x) = (5-x²-x)⁷
f'(x) = 7(5-x²-x)⁶(-2x-1)
= (-14x-7)(5-x²-x)⁶

(b) ~f'(x)dx = f(x) +C where C is a constant
~(-14-7)(5-x²-x)⁶dx = (5-x²-x)⁷ +C

~(2x+1)(5-x²-x)⁶dx = -1/7*~(-14x-7)(5-x²-x)⁶dx
= -1/7(5-x²-x)⁷ +C

 

[bored of sc]

2) a) Find (2x^2+3)^1/2
b) Find ~2x/((2x^2+3)^1/2) and x/((2x^2+3)^1/2)

(a) f(x) = (2x²+3)^1/2
f'(x) = (1/2)(2x²+3)^-1/2*4x
= 2x/(2x²+3)^1/2

(b) ~2x/(2x²+3)^1/2dx = (2x²+3)^1/2 +C

~x/(2x²+3)^1/2 = 1/2*~2x/(2x²+3)^1/2dx = 1/2(2x²+3)^1/2 +C

 

[bored of sc]

3) a) Find ((x)^1/2+1)^3
b) Hence find ~(3((x)^1/2+1)^2)/2(x)^1/2 and ((x)^1/2+1)^2)/(x)^1/2

(a) f(x) = (x^1/2+1)³
f'(x) = 3(x^1/2+1)²*(1/2)x^-1/2
= (3(x^1/2+1)²)/2x^1/2

(b) ~[3(x^1/2+1)²/2x^1/2]dx = (x^1/2+1)³ +C

~[(x^1/2+1)²/x^1/2] dx = (2/3)~[(3(x^1/2+1)²)/2x^1/2]dx = 2/3(x^1/2+1)³ +C

 

[bored of sc]

Q) Write up a table of values and estimate 0~4x(x-4) using the trapezoidal rule.

Using 5 function values [sub integer values of x into y = x(x-4) from 0-4]:
x 0 1 2 3 4
y 0 -3 -4 -3 0

Area:
₀~⁴x(x-4)dx approximately = h/2[y₀+y₄+2(y₁+y₂+y₃)] where h = b-a/n = (4-0)/4 = 1 {n = number of trapezia} and y values are from the table

Area approximately = 1/2[0+0+2(-3-4-3)]
= |-10| units² since it is area
= 10 units²

Is this right?

 

[Dragonmaster262]

(a) f(x) = (5-x²-x)⁷
f'(x) = 7(5-x²-x)⁶(-2x-1)
= (-14x-7)(5-x²-x)⁶

(b) ~f'(x)dx = f(x) +C where C is a constant
~(-14-7)(5-x²-x)⁶dx = (5-x²-x)⁷ +C

~(2x+1)(5-x²-x)⁶dx = -1/7*~(-14x-7)(5-x²-x)⁶dx
= -1/7(5-x²-x)⁷ +C

So just how did you get -1/7? That's the only thing I don't get.

 

[Dragonmaster262]

Using 5 function values [sub integer values of x into y = x(x-4) from 0-4]:
x 0 1 2 3 4
y 0 -3 -4 -3 0

Area:
₀~⁴x(x-4)dx approximately = h/2[y₀+y₄+2(y₁+y₂+y₃)] where h = b-a/n = (4-0)/4 = 1 {n = number of trapezia} and y values are from the table

Area approximately = 1/2[0+0+2(-3-4-3)]
= |-10| units² since it is area
= 10 units²

Is this right?

Yeah, that's correct.

 

[bored of sc]

So just how did you get -1/7? That's the only thing I don't get.

(2x+1)(5-x²-x)⁶ = (-1/7)*[-7(2x+1)(5-x²-x)⁶] = (-1/7)(-14x-7)(5-x²-x) = (-1/7)(derivative of initial equation)

Note that (-1/7)*-7 = 1

You can put the -1/7 out the front of the integration.

That making sense?

 

[Dragonmaster262]

Using 5 function values [sub integer values of x into y = x(x-4) from 0-4]:
x 0 1 2 3 4
y 0 -3 -4 -3 0

Area:
₀~⁴x(x-4)dx approximately = h/2[y₀+y₄+2(y₁+y₂+y₃)] where h = b-a/n = (4-0)/4 = 1 {n = number of trapezia} and y values are from the table

Area approximately = 1/2[0+0+2(-3-4-3)]
= |-10| units² since it is area
= 10 units²

Is this right?

Why did you divide by 4? Cambridge's formula is:

(b-a)/2*(f(a)+f(b))

 

[bored of sc]

Why did you divide by 4? Cambridge's formula is:

(b-a)/2*(f(a)+f(b))

I did the general formula for the trapezoidal rule. You will learn this soon enough. Dividing by 4 in this case is the number of trapezia I used.

Wait, isn't that formula (b-a)/2*{f(a)+4f[(a+b)/2]+f(b)} or is that Simpson's?

 

[Trebla]

If we let h = (x_n - x₁)/n and for each strip between x_i and x_i+1 we have: (think of a = x₁ and b = x_n)
x₁ < x₂ < x₃ < ..... < x_n-1 < x_n then

- Trapezoidal rule is

You can easily derive this by taking strips of trapeziums and finding the approximate area of each individual trapezium before summing them up.

- Simpsons rule is

 

[Dragonmaster262]

Complete the square:

y=2x^2+5x-3

How do you use that LaTex Equator to put equations into messages?

 

[Timothy.Siu]

Complete the square:

y = 2x^2+5x-3

y=2(x^2+5/2-3/2)=2((x+5/4)^2-25/16-3/2)=2((x+5/4)^2-49/16)

=2((x+5/4+7/4)(x+5/4-7/4)=2(x+3)(x-1/2)