# Is this necessary to be true?

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 $$A = \displaystyle 2 \int_{0}^{n} u \ du$$ ?

Note by Christian Daang
1 year, 4 months ago

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Since $$\sqrt{t} = u$$ 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 $$\dfrac{2}{3}n^{3}$$.

- 1 year, 4 months ago

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

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

- 1 year, 4 months ago

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

- 1 year, 4 months ago

ok sir, thanks. :D

- 1 year, 4 months ago