help please, a number of questions (1 Viewer)

[Pace_T]

These have built up over the week and im wondering u have can answer them for me please


1) S(b=5, a=1) xdx/(sqrt(3x+1) using u^2=3x + 1


2) S(b=1, a=0) dx/[x^1/4(1+x^3/4] using u=1 + x^3/4


3) S(b=2, a=0) dx/[(5-2x)^2] using u=5 - 2x

4) Using u = 3x01, find the area bounded by the graph of f(x) = ^3sqrt(3x - 1), the x axis and the ordinates x=2/3, x=3. find the volume when this area is rotated about the x axis to form a solid of revolution.

5) S (b= pi, a=0) (2sin(x)Cos^2(x) dx

6) S(b=4, a =0) x*sqrt(16-x^2) using x=4sinY

thanx

 

[Trev]

Question 1.
u = (3x+1)^1/2; x=(u²-1)/3
du/dx=3/[2(3x+1)]
2du/3 = dx/sqrt(3x+1)
Therefore
S(b=5, a=1) xdx/(sqrt(3x+1) equals [when x=5, u=4; when x=1, u=2]
S(b=4, a=2) 2/9*(u²-1)du

 

[KFunk]

∫(b=5, a=1) xdx/(sqrt(3x+1) using u^2=3x + 1

x=(1/3)u² - 1/3
dx= (2/3)u.du


and u = √(3x +1) so when x=1 --> u=2 and when x=5 --> u=4 making the integral:

∫(b=4, a=2) [(1/3)u² - 1/3](2/3)u.du/u

= ∫(b=4, a=2) (2/9)u² - 2/9 du

= [ (2/27)u³ - (2/9)u] between 2 & 4

= 8 8/27

 

[KFunk]

3) ∫(b=2, a=0) dx/[(5-2x)^2] using u=5 - 2x

du = -2dx and when x=0 --> u=5 and when x=2 --> u=1

makign it:

-1/2∫(b=1, a=5) u^-2 du

= -1/2[-u^-1] (between 5 and 1) = -2/5 or somethign

 

[KFunk]

Sorry for the funny order, I don't like linearity

2) ∫(b=1, a=0) dx/[x^1/4(1+x^3/4] using u=1 + x^3/4

du = (3/4)x^-1/4dx and when x=0 --> u=1 when x=1 --> u=2

4/3 ∫(b=1, a=0) 3dx/[4x^1/4(1+x^3/4]

= 4/3 ∫(b=2, a=1) u^-1 du

= 4/3 ln(2)

 

[Trev]

Question 5)
5) S (b= pi, a=0) (2sin(x)Cos^2(x) dx
=∫2sinxcos²x dx
=∫ [-2cos^3(x)]/3 where a =0, b=pi
you can substitute that yourself....

 

[KFunk]

4) Using u = 3x01, find the area bounded by the graph of f(x) = ^3sqrt(3x - 1), the x axis and the ordinates x=2/3, x=3. find the volume when this area is rotated about the x axis to form a solid of revolution.

--->f(x) = ^3sqrt(3x - 1)

is that cube root? or should it be x^3?

 

[Pace_T]

the cube root
thanks for ur help

 

[KFunk]

Ok, cool. Well what you're doing is summing 2-d circular segments around the x-axis instead of 1-d strips. The radius of each circular segment is of course your y value which gives us π∫y²dy

and the integral is (between 2/3 and 3)

π∫(3x-1)^2/3dx ..... you don't really need a substitution here (the rule you use for ∫(ax +b)ⁿ automates the process) but if you're going to do it manually then:

u = 3x-1 so du = 3dx

π/3.∫u^2/3du

= π/3[3u^5/3/5]
= π/3[3(3x-1)^5/3/5] between 2/3 and 3
= 93π/15 = 31π/5 ? [sorry, I'm not in a good calculating mood today]

but as you can see it's the same deal as ∫(ax +b)ⁿ = (ax +b)ⁿ⁺¹/a(n+1)

 

[KFunk]

6) S(b=4, a =0) x*sqrt(16-x^2) using x=4sinY

between (0-->4) ∫ x√(16-x²) dx where x=4sinθ

dx = 4cosθdθ when x=0 then θ=0 and when x=4 then θ= π/2
so the integral is between (0-->π/2)
= ∫ 4sinθ√(16-16sin²θ )4cosθdθ
= ∫ 16sinθcosθ4√(1-sin²θ )dθ
= ∫ 64sinθcos³θdθ

then you do a quick substitution of u=cosθ du=-sinθdθ
yielding:

-64[(1/4)cos⁴θ] between 0 and π/2
= 16

Maybe there's a quicker way to do this?