If \( A = \displaystyle \int_{0}^{n^2} \sqrt{t} \ dt \) , and we let \( t = u^2 \implies dt = 2u \ du \ ,\)

is it necessary to be \( A = \displaystyle 2 \int_{0}^{n} u \ du \) ?

If \( A = \displaystyle \int_{0}^{n^2} \sqrt{t} \ dt \) , and we let \( t = u^2 \implies dt = 2u \ du \ ,\)

is it necessary to be \( A = \displaystyle 2 \int_{0}^{n} u \ du \) ?

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TopNewestSince \(\sqrt{t} = u\) and \(dt = 2u du\) the last integral should be \(A = 2 \displaystyle\int_{0}^{n} u^{2} du\).

Then the integral, in either form, evaluates to \(\dfrac{2}{3}n^{3}\). – Brian Charlesworth · 5 months, 2 weeks ago

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\( A = \displaystyle 2 \int_{0}^{n} \lfloor u \rfloor u \ du \) ? – Christian Daang · 5 months, 2 weeks ago

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– Brian Charlesworth · 5 months, 2 weeks ago

Yes, that does seem to be the case. (I tried out a few values of \(n\) just to be sure.)Log in to reply

– Christian Daang · 5 months, 2 weeks ago

ok sir, thanks. :DLog in to reply