Substitution help please (1 Viewer)

[MolestLawzy]

1. Find the area bounded by the curve y = x^3(x^4-2), the x axis and the lines x = 0 and x = 1

2. Find the volume of the solid of revolution formed when the curve
y = x(x^3-3)^2 is rotated about the x-axis between x = 0 and x = 1

3. How do I write in subscript!?

Cheers.

 

[Mountain.Dew]

1. Find the area bounded by the curve y = x^3(x^4-2), the x axis and the lines x = 0 and x = 1

2. Find the volume of the solid of revolution formed when the curve
y = x(x^3-3)^2 is rotated about the x-axis between x = 0 and x = 1

3. How do I write in subscript!?

Cheers.

for 1) the area is =
1
∫-x³(x⁴-2) dx =
0

1
∫-x⁷-2x³ dx = -[x⁸/8 - x⁴/2]¹₀ =
0

= -(1/8 - 1/2) = 0.375 units²

now, i placed a minus sign in front of the function because i know the area bounded is BELOW the x-axis, so i would of gotten a 'negative' answer.

3) how to write in superscript/ subscript is as follows

from x^3 ==> x³...use this code: (sup)<insert number for power>(/sup), replace the ( ) with [ ]. the same code applies for subscript, except replace 'sup' with 'sub'

 

[pLuvia]

2. Find the volume of the solid of revolution formed when the curve
y = x(x^3-3)^2 is rotated about the x-axis between x = 0 and x = 1

Using the same method as the first one, expanding it

you should end up with

y=x⁷-6x⁴+9x
find the y² then sub it into the equation

pi ∫ y² dx and you should get your answer

 

[MolestLawzy]

Thanks both of you for your answers but what I really need is fully worked proofs. Yes I'm annoying. These q's are from ex 3:11 of MIF, I've been trying all weekend to get them and have used a fair few resources to try and find the method to solve but I can never get it

I can get 4), as a negative answer, but the answer in the back of the book is positive, perhaps this means after absolute value as area > 0?

I'm not dumb, I got 100% in my last 2U exam and 88% in my last 3U. WHY CAN'T I FKING DO THIS !@!

Anyways, thanks again.

EDIT: Why don't you use substitution for Q2 pLuvia?

I'll show my working out for Q1:

1. Find the area bounded by the curve y = x^3(x^4-2), the x axis and the lines x = 0 and x = 1

x = (u + 2)^.25
u = x⁴-2
du/dx = 4x³
dx = du/4x³
x = 0, u = -2
x = 1, u = -1

I = 1/4 S u du
= 1/4 [u²/2] with limits as -1 to -2
= -3/8

END

Shit i just realised I can do this one, it's the next one I'm screwed with but crits of this 1 will help with the next I'm sure. Thanks!

 

[Riviet]

Why don't you use substitution for Q2 pLuvia?

It's an alternate and probably faster way of integrating it, although the substitution is a neat method as well. You can do it either way, whichever you prefer.

 

[pLuvia]

EDIT: Why don't you use substitution for Q2 pLuvia?

Not necessary for this question, and it'll probably take longer than the conventional way, whatever works for you