P3 Integration

2) Find, in terms of e and pi, the volume generatre when the region bounded by the lines x=1, x=3 and y=0 and the curve with equation y=e^[3x/2] is rotated completely about the x-axis.

Thank you all so much - i figured you wouldn't mind too much as you'll enjoy the P3 practise ready for next week!

For 2), I get V = (1/3)(pi)e³[ e³ - 1] ???

Hello again! Could someone suggest what method I should use to solve this question:

1) integral (0.1 -> 0) x^2.e^x^3 dx

well integrating e^x^3 is ok since u can let u = x^3 and du/dx = 3x^2 therefore reaaraging gives 1/3du = x^2 dx.

so this means integrating 1/3e^u which is 1/3e^u (u = x^3) so this becomes 1/3e^x^3. sticking the limits in gives 1/3e^1/1000 - 1/3. put 1/3 outside into a bracket and becomes 1/3(e^0.001 - 1).

?? plizzz say dats ryt!

Hello again! Could someone suggest what method I should use to solve this question:

1) integral (0.1 -> 0) x^2.e^x^3 dx

I tried using by parts twice but I got stuck the second time round as it didnt seem to have made the integral any easier. (i am right in thinking that integral of e^x^3 = [1/3x^2]e^x^3?)

Secondly, could someone work through the below problem as I end up with something completely different to the answers! arghh:

2) Find, in terms of e and pi, the volume generatre when the region bounded by the lines x=1, x=3 and y=0 and the curve with equation y=e^[3x/2] is rotated completely about the x-axis.

Thank you all so much - i figured you wouldn't mind too much as you'll enjoy the P3 practise ready for next week!

the first integrates perfectly to 1/3 e^x^3. differentiate it and see for yourself.

the second one is basically integrate between 1 and 3, pi y^2 dx.

so it's the integral of pi e^3x = 1/3 pi e^3x, between limits 1 and 3. so your answer should be 1/3 pi (e^9 - e^3).

i hope that's right

I think bezza and mockel have accidently put e^3 in the brackets, but e^9 = e^3 x e^6, and not e^3 x e^3!

I can indeed.

The volume of a rotated shape is the (pi) x ∫y² dx

This gives (pi) x ∫(e^(3x/2))^2 dx (Between limits 1 and 3)

= (pi) x ∫e^(3x) dx

= (pi)[(1/3)(e^3x)] between 1 and 3

Take out the factors 1/3 and e^3 gives::

(1/3)(pi)e³[ e^6 - 1]

I think bezza and mockel have accidently put e^3 in the brackets, but e^9 = e^3 x e^6, and not e^3 x e^3!

Hope this helps

Thanks for all your help. However, i am slightly confused. surely 3^[3x/2] all squared would give e^[9x^2/4]? Oh god, how am I going to get my A in P3 if i can't even do simple indices!!!

Bezza is right. e^(3x/2) all squared is the same as:

e^(3x/2) x e^(3x/2)

which is e^((3x/2) x 2)

which is e^(3x).

You can think of the indice theory in terms of 2s.

2^4 = 16

so (2^4) all squared is 16 x 16 which equals 256.

and this is equal to 2^8, which is of course the same as 2^(4x2).

But don't worry you'll do fine in the exam if you just practice a bit of algebra over the next week.

This is related to integration (i promise!!! !!!)

Ok in the above question, you have to integrate e^(3x), so i was wondering what is the integral of something like e^(3(x^2))

Thanks

u cannot express the integral of e^x^2 in terms of elementary functions,but u can find the exact value of the area under the graph of y=e^-x^2.