Waste less time on Facebook — follow Brilliant.
×

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
8 months, 2 weeks ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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

Brian Charlesworth - 8 months, 2 weeks ago

Log in to reply

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

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

Christian Daang - 8 months, 2 weeks ago

Log in to reply

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

Brian Charlesworth - 8 months, 2 weeks ago

Log in to reply

@Brian Charlesworth ok sir, thanks. :D

Christian Daang - 8 months, 2 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...