Integration Question. (1 Viewer)

[m0ofin]

Hello,
I've not a clue as to what the questions are actually asking of me, the attachment is my teacher's hard copy and so nothing's been added or taken out.

"I" = Integration.

Does anyone have an idea of what to do? Or how to interpret the questions themselves for that matter? "Calculate I_m,₀", would that mean find the integration from M to 0 or something of the like?

Thanks.

 

[bos1234]

that makes 2 of us :S

 

[Yip]

These are double-variate reduction formula questions, i think its off-syllabus.
(b) Let t=1-x
dt=-dx
I[m,n]=∫₁⁰(1-t)^mtⁿ(-dt)
=∫₀¹tⁿ(1-t)^mdt=I[n,m]
(c)I[m,n]=1/(m+1)[x^m+1(1-x)ⁿ]₀¹-[n/(m+1)]∫₀¹x^m+1(1-x)^n-1(-dx)
=n/(m+1)I[m+1,n-1]
(d) I[m,n]=n/(m+1)I[m+1,n-1]
=[n/(m+1)][(n-1)/(m+2)]I[m+2,n-2]
=...
=[n/(m+1)][(n-1)/(m+2)][(n-2)/(m+3)]...[2/(m+n-1)][1/(m+n)]I[(m+n),0]
=n!/(m+n+1)!/m!
=m!n!/(m+n+1)!

 

[m0ofin]

Thanks for your help Yip, though it still means nothing to me.

Iruka, so it's safe to say I shouldn't even bother understanding it if I'm only doing 3u?

 

[pLuvia]

It's good to be exposed to a variety of questions, these questions seem familiar to me somewhere in the 4u course, but nonetheless they may put this in a 3u exam or something less difficult but using the same principles

But other than that, don't worry about it

 

[m0ofin]

Thanks for the great advice Pluvia and Iruka. I'll give it a go if it were to come up again someway down the track.

 

[m0ofin]

Just another question.
1. The velocity of a particle is given by v= 2x + 1 cm^-1. If the particle is initially at the origin, find:
a) What the displacement will be when the acceleration is 5 cm^-2
b) The exact displacement after 5 secs

The answer is probably painfully obvious so :S

Thanks for any help.

 

[ianc]

Just another question.
1. The velocity of a particle is given by v= 2x + 1 cm^-1. If the particle is initially at the origin, find:
a) What the displacement will be when the acceleration is 5 cm^-2

b) The exact displacement after 5 secs

Hope this helps!

 

[Nodice]

Haven't done much maths in a while so here goes:
v = 2x + 1 at t=0, x=0
a) d(0.5v^2)/dx = acceleration (a)
0.5v^2 = 0.5(2x+1)^2
= 0.5 (4x^2 + 4x + 1)
= 2x^2 + 2x + 0.5
d(0.5v^2)/dx = 4x + 2
So a = 4x + 2... at a=5...
5 = 4x + 2
x = 3/4 cm

b) dx/dt = 2x + 1
dx/2x+1 = dt
0.5ln(2x+1) + C = t
But when t=0, x=0, hence C=0 (ln1=0)
Rearranging...
0.5(e^2t -1) = x
So displacement after 5 seconds is 0.5(e^10 -1)

Edit: Looks like im a little slow!

 

[m0ofin]

A little late in reply but thanks to both of you for your kind help.

 

[m0ofin]

Anyone mind helping on a whim?

A region is bounded by the curve y= sin^-1x, the y-axis and the tangent to the curve at point (√3/2 , pi/3)

a) Show that the area of the region is 1/4 u²

b) Show that the volume of the solid formed when the region is rotated about the y-axis is:

pi/34 times (9√3 - 4pi) u³

Thanks

PS: √3/2 is root 3 on 2

 

[Z66R2V]

Here's a start:

You know

y(x) = arcsin(x)

so

y'(x) = (1) / ( (1-x^2)^0.5)

y'(sqrt(3)/2) = 2.00

and

int(y(x)) = x*asin(x) + (1-x^2)^0.5

 

[pinkly]

MOofin. I'm very very rusty, mind you, haven't done this since the exams last year lol.

y= sin^-1x
y'= 1/√(1-x²)
Sub pt x= √3/2 into y' for gradient
y'= 2 = m

y - y₁ = m(x - x₁)
y - pi/3 = 2(x - √3/2)
y - pi/3 = 2x - √3
Therefore, y = 2x - √3 + pi/3 is the equation of the tangent at pt (√3/2 , pi/3)

Draw the graph.

I is "Integration"

A = Integration from √3/2 to 0 [y= sin[sup]-1[/sup]x - (2x - √3 + pi/3)] dx

= [xsin^-1x + √(1 - x²) - (x² - √3x + (pi/3)x)] from √3/2 to 0.

And so on.

b) Rearrange the values so that:
y= sin^-1x becomes x= sin y and
y= 2x - √3 + pi/3 becomes x= (y + √3 = pi/3) / 2

V= pi times I from pi/3 to 0 [sin y - (y + √3 - pi/3) / 2]² + V(Cone)

You'll see where the Cone is from if you split the shaded area with the x - axis. I haven't actually finished the second part so it might not go down fine.

EDIT: I think part A was done wrongly but it still came up with 1/4 u² though.

 

[m0ofin]

Thanks for your help Pinkly.