How to find this?

Note by Rajath Krishna R
7 years, 9 months ago

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  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

13[x]cos(π2(x[x])dx=12[x]cos(π2(x[x]))dx+23[x]cos(π2(x[x]))dx=121cos(π2(x1)dx+232cos(π2(x2)dx=1π2+21π2=6π \int_1^{3}[x]cos(\frac{\pi}{2}(x - [x]) dx \\ = \int_1^{2}[x]cos(\frac{\pi}{2}(x - [x]))dx + \int_2^{3}[x]cos(\frac{\pi}{2}(x - [x]))dx \\ = \int_1^{2} 1cos(\frac{\pi}{2}(x - 1) dx + \int_2^{3} 2cos(\frac{\pi}{2}(x - 2) dx \\ = \frac{1}{\frac{\pi}{2}} + 2 \frac{1}{\frac{\pi}{2}} \\ = \frac{6}{\pi}

jatin yadav - 7 years, 9 months ago
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