What is the surface area of the solid created by revolving $f (x) = e^{4 x} - e^{2 x}, x \in [1, 2]$ $f(x) =e^(4x)-e^(2x) , x in [1,2]$ around the x axis?

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What is the surface area of the solid created by revolving $f (x) = e^{4 x} - e^{2 x}, x \in [1, 2]$ $f(x) =e^(4x)-e^(2x) , x in [1,2]$ around the x axis?

1 Answer

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Answer 1 · 2016-01-29T07:14:46

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Explanation:

see above!

Just soze you know....I did a word=>pdf=>snapshot=>jpg with this answer....that's why you get those weird html thingy-like tags....
Also I meant to say "square inside that ""radicand"", not integrand.."

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What is the surface area of the solid created by revolving $f (x) = e^{4 x} - e^{2 x}, x \in [1, 2]$ $f(x) =e^(4x)-e^(2x) , x in [1,2]$ around the x axis?

1 Answer

Explanation:

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What is the surface area of the solid created by revolving f(x)=e4x−e2x,x∈[1,2]f(x) =e^(4x)-e^(2x) , x in [1,2] around the x axis?

1 Answer

Explanation:

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Impact of this question

What is the surface area of the solid created by revolving $f (x) = e^{4 x} - e^{2 x}, x \in [1, 2]$ $f(x) =e^(4x)-e^(2x) , x in [1,2]$ around the x axis?