disk method to find the volume

[kidmo87]

Hello, I have been doing this problem and understand everything except for why its pi/2? i circled it in red. Can someone please explain this to me. Thanks.

 

[Deleted member 4993]

Hello, I have been doing this problem and understand everything except for why its pi/2? i circled it in red. Can someone please explain this to me. Thanks.

\(\displaystyle \int e^{2x} dx \ = \ \frac{1}{2}e^{2x} + C \)

 

[kidmo87]

thanks for your help, but Im still struggling with understanding.

 

[DrPhil]

Hello, I have been doing this problem and understand everything except for why its pi/2? i circled it in red. Can someone please explain this to me. Thanks.

Your next-to-last line is rather peculiar. You can't replace x with numbers until after the integral is found, then there won't be an integration sign any more. AH - just a typo? the exponent should be 2(x)-2, not 2(2)-2

So back the the integral itself. Subhotosh Khan showed you the anti-derivative.

\(\displaystyle \int{e^{2x} dx} = \frac{1}{2}e^{2x} +C \)

The (1/2) comes from the coefficient of x in the exponent. In general, the derivative of e^(ax) is ae^(ax), and the integral of e^(ax) is (1/a)e^(ax).

 

[lookagain]

The second to last step should be:

\(\displaystyle \pi\displaystyle\int_1^2{e^{2x - 2}}dx\)


Let u = 2x - 2

du = 2dx


So, to complete the du, multiply the integrand by 2. But simultaneously divide the constant pi by 2
to compensate:


\(\displaystyle \dfrac{\pi}{2}\displaystyle\int_1^2({e^{2x - 2}})(2dx)\)


And your last step is missing a pair of grouping symbols to encompass the entire difference:



\(\displaystyle \dfrac{\pi}{2}\bigg\{\bigg[e^{2(2) - 2}\bigg] \ -\ \bigg[e^{2(1) - 2}\bigg]\bigg\}\)