Asked by saud
Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=x^8/9 about the x-axis over the interval [1,2].
Answers
Answered by
saud
Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=(2)/(x+1) about the x-axis over the interval [0,4].
Answered by
saud
Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=e^x about the x-axis over the interval [0,2].
Answered by
Reiny
is that f(x) = x^(8/9) or f(x) = (x^8)/9
is will assume the latter.
Volume = π[integral) (x^16)/81 dx from 1 to 2
= π[x^17)/(17(81)) ] from 1 to 2
= π(2^17/(17(81)) - 1/(17(81))
= ....
is will assume the latter.
Volume = π[integral) (x^16)/81 dx from 1 to 2
= π[x^17)/(17(81)) ] from 1 to 2
= π(2^17/(17(81)) - 1/(17(81))
= ....
Answered by
Reiny
second one:
volume = π[integral] 4/(x+1)^2 from 0 to 2
= π((-4/(x+1)) from 0 to 2
= π(-4/3 - (-4/1)
= ....
third one:
you try it, the integral of (e^x)^2 is (1/2)e^(2x)
volume = π[integral] 4/(x+1)^2 from 0 to 2
= π((-4/(x+1)) from 0 to 2
= π(-4/3 - (-4/1)
= ....
third one:
you try it, the integral of (e^x)^2 is (1/2)e^(2x)
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