Asked by Anonymous
A solid generated by rotating about the x-axis the region under the curve of f(x) from x=0 to x=b, is b^2 FOR ALL b>0. Find the function f.
Answers
Answered by
Steve
Let's assume that f(0) = 0. Then
∫[0,b] f'(x) dx = b^2
f(b) = b^2
f(x) = x^2
f'(x) = 2x
∫[0,b] 2x dx = x^2 [0,b] = b^2
There are lots of other possible functions. Such as
∫[0,b] e^x dx = 1/k e^(kb) - 1
1/k e^(kb) - 1 = b^2
e^(kb) = k(1+b^2)
∫[0,b] f'(x) dx = b^2
f(b) = b^2
f(x) = x^2
f'(x) = 2x
∫[0,b] 2x dx = x^2 [0,b] = b^2
There are lots of other possible functions. Such as
∫[0,b] e^x dx = 1/k e^(kb) - 1
1/k e^(kb) - 1 = b^2
e^(kb) = k(1+b^2)
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