Question

How does one do 1296?

Answer 1

Frankly I thought 'a' was also applicable as a lower limit and just take the integral, so I got somewhere close to an exact number unless that's not what they want you to do

And yeah I didn't understand the second portion of the ? either

Answer 2

\(\int\limits_0^{2 a} \frac{2 }{a^2}x - \frac{1}{a^3} x^2 ~ dx = \frac{4}{3}\)

Complete Sq:

\(\frac{2 }{a^2}x - \frac{1}{a^3} x^2 = \frac{1}{a} - \frac{(x - a)^2}{a^3}\)

Vertex aways at \((a, \frac{1}{a})\)

ie on curve \(y = \frac{1}{x}\)

Answer 3

Lul thanks yall, I actually finished this yesterday before class, but forgot I had this question up. Thanks for the effort.