integration Q (1 Viewer)

[pLuvia]

∫(2x^3/2 - 1)^1/3 √x dx

Let u = 2x^3/2 - 1
du/dx = 3√x
du/3 = √x.dx
∴ ∫(2x^3/2 - 1)^1/3 √x dx = ∫(u)^1/3(du/3) etc etc.
Do you understand it now?

no no no sorry bout the first post its this Question now

∫x(3x-2)^3 dx

 

[FinalFantasy]

∫x(3x-2)^3 dx
let u=3x-2
du\dx=3
dx=(1\3) du
u+2=3x
x=(1\3)(u+2)
so ∫x(3x-2)^3 dx=∫(1\3)(u+2)(u)^3 (1\3) du
=(1\9)∫(u^4+2u^3) du
=(1\9) [(1\5)u^5+(2\4)u^4]+C
=(1\45)u^5+(1\18)u^4+C
=(1\45)(3x-2)^5+(1\18)(3x-2)^4+C

 

[lucifel]

no no no sorry bout the first post its this Question now

∫x(3x-2)^3 dx

on 2nd thought, if we let u = 3x-2:
du = 3dx

and also: x = (u+2)/3

so int = ∫(u+2)u³/9du
= ∫u⁴/9 + 2u³/9 du
=u⁵/45 +u⁴/18 +c
=(3x-5)⁵/45 + (3x-5)⁴/18 + c

though the constants may be wrong, the method holds.

 

[haboozin]

hey if you only started integration last saturday...

maybe start with some easier ones?

also, ask your teacher.. nothing better than face to face interation when you are learning something for the first time.

 

[pLuvia]

hey if you only started integration last saturday...

maybe start with some easier ones?

also, ask your teacher.. nothing better than face to face interation when you are learning something for the first time.

nar integration is easy its just that the answer i got was not the same as the real answer, your answers were exaclty the same as mine.. so that means the answers might be wrong

 

[lucifel]

whoops Final Fantasy beat me to it again. blargh.

 

[pLuvia]

another question

At any point on the curve, y" = 1-x², the equation of the tangent to the curve at the point (1,4) is y = 5-x. Find the equation of the curve.

how do i do this question.?

 

[KFunk]

At any point on the curve, y" = 1-x², the equation of the tangent to the curve at the point (1,4) is y = 5-x. Find the equation of the curve.

how do i do this question.?

You just use integration to find the curve. What they've done is give you enough information to work out the value of the constant in each integration step.

For the first integration you know that the tangent at (1,4) is y=5-x hence the gradient at that point is -1 so:

f'(1) = -1 , or y'=-1 when x=1 so you can work out what f'(x) is.

Then you know that f(1) = 4 , or y=4 when x=1 so you can work out what f(x) is.

 

[pLuvia]

You just use integration to find the curve. What they've done is give you enough information to work out the value of the constant in each integration step.

For the first integration you know that the tangent at (1,4) is y=5-x hence the gradient at that point is -1 so:

f'(1) = -1 , or y'=-1 when x=1 so you can work out what f'(x) is.

Then you know that f(1) = 4 , or y=4 when x=1 so you can work out what f(x) is.

can you show me the working out please?

 

[KFunk]

can you show me the working out please?

Alright, if that's what you're after.

f"(x) = 1-x²
---> f'(x) = ∫ 1-x² dx
f'(x) = x - 1/3.x³ + c

The tangent at (1,4) is y=5-x , i.e. the gradient at that point is -1 --> f'(1) = -1

-1 = 1 - 1/3 + c ----> c = -5/3

f'(x) = x - 1/3.x³ - 5/3
f(x) = ∫ x - 1/3.x³ - 5/3 dx
f(x) = 1/2.x² - 1/12.x⁴ - 5/3.x + c

(1,4) is a point on the curve so f(1) = 4

4 = 1/2 - 1/12 - 5/3 +c ---> c = 21/4

∴ f(x) = 1/2.x² - 1/12.x⁴ - 5/3.x + 21/4

 

[pLuvia]

Alright, if that's what you're after.

f"(x) = 1-x²
---> f'(x) = ∫ 1-x² dx
f'(x) = x - 1/3.x³ + c

The tangent at (1,4) is y=5-x , i.e. the gradient at that point is -1 --> f'(1) = -1

-1 = 1 - 1/3 + c ----> c = -5/3

f'(x) = x - 1/3.x³ - 5/3
f(x) = ∫ x - 1/3.x³ - 5/3 dx
f(x) = 1/2.x² - 1/12.x⁴ - 5/3.x + c

(1,4) is a point on the curve so f(1) = 4

4 = 1/2 - 1/12 - 5/3 +c ---> c = 21/4

∴ f(x) = 1/2.x² - 1/12.x⁴ - 5/3.x + 21/4

ok thx mate