anyone know how to solve for the highlighted question? im stuck

[gaunt spade]

i also attached my working out but i don't know where to go from there.. i tried substitution (letting u = ln(x)) but it gets super messy and doesn't reduce down

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[odd scroll]

Derivative of ln(4x)/x is (1 - ln(4x))/x^2, not what you wrote. Other than that, not quite sure how to solve the integral.

[sullen brook]

Since you are asked for the area bounded by the curve when rotated about the x-axis, you should use the formula for the surface area of a solid of revolution:
[ A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ]

[limpid ermine]

Arith

[gaunt spade]

i did use that equation

[sullen brook]

You have given ( f(x) = \log_e(4x) ). The derivative is:
[ \frac{dy}{dx} = \frac{d}{dx} \log_e(4x) = \frac{1}{4x} \cdot 4 = \frac{1}{x} ]

The integral for the surface area is:
[ A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ]
Plugging in the values:
[ A = 2\pi \int_{\frac{e}{4}}^{\frac{e^2}{4}} \log_e(4x) \sqrt{1 + \left(\frac{1}{x}\right)^2} , dx ]

[ \sqrt{1 + \left(\frac{1}{x}\right)^2} = \sqrt{1 + \frac{1}{x^2}} ]

I think your work correctly shows this step:
[ A = 2\pi \int_{\frac{e}{4}}^{\frac{e^2}{4}} \log_e(4x) \sqrt{1 + \frac{1}{x^2}} , dx ]

[limpid ermine]

Arith

[sullen brook]

These are the possible mistakes..

[sullen brook]

Just noting on that, I think you made you're mistake in simplifying the integrand.

[odd scroll]

But isn't f(x) = ln(4x)/x, rather?

Not that it would really matter, as I doubt that any of these two cases can be integrated using elementary functions.

[sullen brook]

Looking at the work, we can see that the integral involves the function ( \log_e(4x) ) and the term ( \sqrt{1 + \frac{1}{x^2}} ), right. As I said before, your work shows a manipulation error when simplifying the integrand. In the final integral setup, you should have:
[ A = 2\pi \int_{\frac{e}{4}}^{\frac{e^2}{4}} \log_e(4x) \sqrt{1 + \frac{1}{x^2}} , dx ]

To simplify the term under the square root:
[ \sqrt{1 + \frac{1}{x^2}} ]

Your work shows that you might have overcomplicated the simplification.... I'd let Darpinger verify this

[limpid ermine]

Arith

[odd scroll]

As I said above, neither (ln(4x)/x)√(1 + (1 - ln(4x))^2/x^4), which we have for this exercise, nor ln(4x)√(1 + 1/x^2), which we would get if we had f(x) = ln(4x), seem to have an elementary antiderivative.

[gaunt spade]

ohhh okok thanks guys

[gaunt spade]

thank you

[simple ore]

@gaunt spade has given 1 rep to @odd scroll

@gaunt spade has given 1 rep to @sullen brook

[gaunt spade]

i did it without altering the initial formula and got the solution

+close