Calculating the Volume of Revolution for e^x + 1 on x-axis

[Hootenanny]

If the finite region bounded by the curve $y = \text{e}^{x} +1$, the y-axis and the line $x = \ln2$ is rotatated around the x-axis by $360^{\circ}$ show that the volume of the solid formed is:
$$\frac{\pi}{2} (7 + \ln4 )$$
I did the intergral and got:
$$V = \pi \left[ (\text{e}^{4} + 2\text{e}^{2} +1) - (1 + 2 + 1) \right]$$
But I can't see how I can manipulate it to get the required answer. Any help would be much appreciated.

 

[TD]

You won't be able to rewrite it like that since they aren't the same.
Perhaps you could show how you got that?

 

[Hootenanny]

$$V = \int_{0}^{\ln2} \pi\left[ f(x) \right]^2 \;\; dx$$
$$V = \pi \int_{0}^{\ln2} \left[ e^{x} +1 \right]^2 \;\; dx = \pi \int_{0}^{\ln2} e^{2x} + 2e^{x} +1 \;\; dx$$
$$V = \pi \left[ (e^{4} + 2e^{2} +1) - (1 + 2 + 1) \right]$$

 

[TD]

$$V = \pi \int_{0}^{\ln2} e^{2x} + 2e^{x} +1 dx$$

This is still correct, the next step isn't. Show the primitive function first, then substitute the boundaries.

 

[Hootenanny]

I dear, I see that I've forgot to intergrate. That was a bit stupid wasn't it?

 

[TD]

I dear, I see that I've forgot to intergrate. That was a bit stupid wasn't it?

I think that happens at least once to everyone

 

[Hootenanny]

Have I intergrated correctly? I get:
$$V = \pi \left[ \frac{1}{2} e^{2x} + 2e^{x} + x \right]^{\ln2}_{0}$$

 

[TD]

Looks good, see if you can come to the given answer.

 

[Hootenanny]

The subsituting the values ln2 and 0 in gives:
$$V = \pi \left[ \left( \frac{1}{2}\cdot4 + 4 + \ln2 \right) - \left( \frac{1}{2} + 2 \right) \right]$$
Simplifying:
$$\pi \left[ \frac{7}{2} + \ln2 \right]$$
I can't see where I go now. I know I'm close.

 

[TD]

You are close indeed

Factor out 1/2 so you get the $\pi/2$ which is there in the given answer, then think of a property of logarithms...

 

[Hootenanny]

O yes, I forgot $2\cdot \ln2 = ln4$. My brains obviously not in gear today. Thanks very much.

 

[TD]

Well done