- Integration, Rate of Change, and Volume of Revolution Questions

[Timiop2008]

Please Help! - Integration, Rate of Change, and Volume of Revolution Questions

Hi
I have completes these following questions but am not sure if I have done them correctly as it is a long time since i studied these topics. I would really appreciate any help. :-)

1 Simplify the following as far as Possible
(a) ∫4e^2x dx (2 marks)
(b) ∫(4/x) dx between 1 and 3 (3 marks)
(c) ∫(3x+4)⁵ dx (2 marks)
(d) ∫((x/2)+1)² dx (2 marks)

2 Find the area bound by the graph y=3/x, the lines y=1, y=3, and the y-axis (3 marks)

3 Find the Volume of the solid formed by the line y=e^x+1 between 0≤x≤1 when it is rotated through 360 degrees about the x axis. Leave your answer in terms of e. ( 5 marks)

4
(a) The radius of a sphere is increasing at a rate of 0.8mm/s. Find the rate at which the volume of the sphere is increasing when r=10mm. Leave your answer in terms of pi (3 marks)

(b) The Volume of the same sphere then starts to decrease at a rate of 27m³/s. Find the rate at which the radius is decreasing when the volume is 3m³
Leave your answer in terms of Pi. (5 marks)

My attempts at Solutions:

1
a) ∫4e^2x dx
=(4/2)e² X x²
= 2e²x² +K

b) (4/x) = 4x^-1
4x^-1 dx
[4lnx] between 3 and 1
4ln(3) - 4ln(1)
4ln(3)
=4.39

c) ∫(3x+4)⁵ dx

= ∫243x⁵+1620x⁴+4320x³+5760x²+3072x

= (81/2)x⁶+324x⁵+1080x⁴+1920x³+1920x²+1024x +K

d) ∫((x/2)+1)² dx
= ∫(x²/4)+x+1 dx
= (x³+6x²+12x)/12
= (x/12)³+(1/2x²)+x +K

2) ∫(between 1 and 3) of (3/x) dy
=∫(between 1 and 3) of 3x^-1 dy
=[3x^-1y] for 1 and 3
= (3x^-1X3) - (3x^-1)
= 9/x - 3/x
= 6/x +K

3) y=e^x+1 0≤x≤1
=∫(between 0 and 1) of pi y² dx
=∫(between 0 and 1) of pi (e^x+1)² dx
=∫(between 0 and 1) of (e^2x+2e^x+1) dx
= Pi[1/2e^2x+e^2x+x] between 0 and 1
= Pi[1/2e²+e²+1] - [1/2e⁰+e⁰+0]
= Pi[1/2e²+e²+1] - [1+1+0]
= Pi[1/2e²+e²+1] - 2
= 1/2e²Pi+e²Pi+Pi-2
= 3/2e²Pi+Pi-2


(I do not really know how to attempt the final rates of change question)
Thank You Everyone

 

[rock.freak667]

∫e^axdx=(1/a)*e^ax +C

b) is correct though you shouldn't put it as 4x^-1. As soon as you know you have to integrate 1/x with respect to x, just write ln(x). Putting it as 4x^-1 sort of implies that you are going to integrate it in the traditional way.

c) looks correct, but there is an easier way to do it rather than having to expand it out. It is known as a substitution method.

∫(3x+4)⁵ dx

let t=3x+4 so dt/dx = 3 and so dx=dt/3

∴∫(3x+4)⁵ dx ≡ (1/3)∫t⁵ dt

and when you finish integrating you just replace t by 3x+4.

d) looks correct.

2) try this again

Your integral is supposed to be

[tex]\int_1 ^{3} x dy[/tex]

but we know that y= 3/x so what is x equal to?

3) When integrating e^2x+2e^x+1, why did you leave out the 2 in the next line?

4) we are told the radius is increasing at a rate of 0.8mm/s so dr/dt=+0.8mm/s. We want to get dV/dt

Using the chain rule we can get

[tex]\frac{dV}{dt}= \frac{d?}{dt}\times{dV}{d?}[/tex]

and for a sphere how does one find the volume V?
what should the '?' be replaced given our information?

 

[Timiop2008]

So for Q2, is it correct to say:

³₁∫x dy
y= 3/x so x=3/y
=³₁∫(3/y)dy
=³₁∫3lny
=3ln3-3ln1
=3ln3
=3.30

 

[Mark44]

So for Q2, is it correct to say:

³₁∫x dy
y= 3/x so x=3/y
=³₁∫(3/y)dy
=³₁∫3lny
=3ln3-3ln1
=3ln3
=3.30

3 ln3 = ln 27 is the exact answer, but 3.30 is only a rough approximation. You should indicate that when you go from the exact answer to an approximation, like this:
[itex]3~ln 3~\approx~3.30[/itex]

 

[Timiop2008]

Yes but is it ln(27) or ln(27)+C

 

[Mark44]

For a definite integral, you don't need the constant of integration.

 

[Timiop2008]

Do you agree that this is the correct solution for Question 3?

y=ex+1 0≤x≤1
=∫(between 0 and 1) of pi y2 dx
=∫(between 0 and 1) of pi (ex+1)2 dx
=∫(between 0 and 1) of (e2x+2ex+1) dx
= Pi[1/2e2x+e2x+x] between 0 and 1
= Pi[1/2e2+e2+1] - [1/2e0+e0+0]
= Pi[1/2e2+e2+1] - [1+1+0]
= Pi[1/2e2+e2+1] - 2
= 1/2e2Pi+e2Pi+Pi-2
= 3/2e2Pi+Pi-2

 

[Mark44]

Do you agree that this is the correct solution for Question 3?

y=ex+1 0≤x≤1
=∫(between 0 and 1) of pi y2 dx
=∫(between 0 and 1) of pi (ex+1)2 dx
=∫(between 0 and 1) of (e2x+2ex+1) dx

Error in line below. Explanation below.

= Pi[1/2e2x+e2x+x] between 0 and 1
= Pi[1/2e2+e2+1] - [1/2e0+e0+0]
= Pi[1/2e2+e2+1] - [1+1+0]
= Pi[1/2e2+e2+1] - 2
= 1/2e2Pi+e2Pi+Pi-2
= 3/2e2Pi+Pi-2

[tex]\pi\int_0^1 e^{2x} + 2e^x + 1 dx[/tex]
[tex]=~\pi((1/2)e^{2x} + 2e^x + x)\rvert_0^1[/tex]
Can you take it from here? If you click on what I wrote, you can see the LaTeX script I wrote to format this stuff. You can copy it and paste it into what you're doing.

 

[Timiop2008]

so is it:

y=ex+1 0≤x≤1
=∫(between 0 and 1) of pi y2 dx
=∫(between 0 and 1) of pi ((e^x)+1)2 dx
=∫(between 0 and 1) of (e^2x+2e^x+1) dx
= Pi[(1/2)e^2x+(2e)^x+x] between 0 and 1
= [1/2e^(2x)pi+(2e)^xpi+xpi] between 0 and 1
=((1/2e)^2pi+2epi+pi) - (1/2)pi+2pi)
= (1/2)e+2epi-(3/2)pi

Also does anybody know how to do that rate of change question?

 

[Mark44]

so is it:

y=ex+1 0≤x≤1
=∫(between 0 and 1) of pi y2 dx
=∫(between 0 and 1) of pi ((e^x)+1)2 dx
=∫(between 0 and 1) of (e^2x+2e^x+1) dx
= Pi[(1/2)e^2x+(2e)^x+x] between 0 and 1
= [1/2e^(2x)pi+(2e)^xpi+xpi] between 0 and 1
=((1/2e)^2pi+2epi+pi) - (1/2)pi+2pi)
= (1/2)e+2epi-(3/2)pi

This is very difficult to read with everything jammed together. Please try to format it so that it's easier to read.

Also does anybody know how to do that rate of change question?

 

[Mark44]

Also does anybody know how to do that rate of change question?

I'm sure someone does. What have you tried to do on it?