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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## Comments

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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 \(\dfrac{2}{3}n^{3}\).

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What if \( A = \displaystyle \int_{0}^{n^2} \lfloor \sqrt{t} \rfloor \ dt \) , and we let \( t = u^2 \implies dt = 2u \ du \ ,\) does it also mean that

\( A = \displaystyle 2 \int_{0}^{n} \lfloor u \rfloor u \ du \) ?

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

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