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

1. Oct 12, 2009

### Timiop2008

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) ∫4e2x 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=ex+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

My attempts at Solutions:

1
a) ∫4e2x 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=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

(I do not really know how to attempt the final rates of change question)
Thank You Everyone
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 12, 2009

### rock.freak667

∫eaxdx=(1/a)*eax +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

$$\int_1 ^{3} x dy$$

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

3) When integrating e2x+2ex+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

$$\frac{dV}{dt}= \frac{d?}{dt}\times{dV}{d?}$$

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

3. Oct 12, 2009

### 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

4. Oct 12, 2009

### Staff: Mentor

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:
$3~ln 3~\approx~3.30$

5. Oct 12, 2009

### Timiop2008

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

6. Oct 12, 2009

### Staff: Mentor

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

7. Oct 12, 2009

### 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

8. Oct 12, 2009

### Staff: Mentor

Error in line below. Explanation below.
$$\pi\int_0^1 e^{2x} + 2e^x + 1 dx$$
$$=~\pi((1/2)e^{2x} + 2e^x + x)\rvert_0^1$$
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.

Last edited: Oct 12, 2009
9. Oct 12, 2009

### 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?

10. Oct 12, 2009

### Staff: Mentor

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

11. Oct 12, 2009

### Staff: Mentor

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