this post was submitted on 21 Mar 2024
65 points (66.8% liked)

Science Memes

10559 readers
2743 users here now

Welcome to c/science_memes @ Mander.xyz!

A place for majestic STEMLORD peacocking, as well as memes about the realities of working in a lab.



Rules

  1. Don't throw mud. Behave like an intellectual and remember the human.
  2. Keep it rooted (on topic).
  3. No spam.
  4. Infographics welcome, get schooled.


Research Committee

Other Mander Communities

Science and Research

Biology and Life Sciences

Physical Sciences

Humanities and Social Sciences

Practical and Applied Sciences

Memes

Miscellaneous

founded 2 years ago
MODERATORS
 
all 8 comments
sorted by: hot top controversial new old
[–] [email protected] 52 points 6 months ago

Hrm, it sounds like the teacher already knew the answer.

Maybe stop wasting the time of the student then!?

You're welcome.

[–] [email protected] 52 points 6 months ago (2 children)

Can someone explain this to me? Not the maths, the joke.

[–] [email protected] 17 points 6 months ago* (last edited 6 months ago) (1 children)

The test:

Solve e^{int_0 1 x2+x+1 dx}^

[–] Usernamealreadyinuse 2 points 6 months ago* (last edited 6 months ago)

No, copilot solve this test

Answer:

To solve the expression ( e^{\int_0^1 x^2+x+1 , dx} ), we first need to calculate the integral of ( x^2+x+1 ) from 0 to 1.

The integral of ( x^2+x+1 ) is: [ \int x^2+x+1 , dx = \frac{x^3}{3} + \frac{x^2}{2} + x + C ]

Evaluating this from 0 to 1 gives us: [ \left[ \frac{x^3}{3} + \frac{x^2}{2} + x \right]_0^1 = \left( \frac{1}{3} + \frac{1}{2} + 1 \right) - \left( 0 + 0 + 0 \right) = \frac{11}{6} ]

Now, we raise ( e ) to the power of this result: [ e^{\frac{11}{6}} ]

This is the solution to the given expression. If you need a numerical approximation, ( e^{\frac{11}{6}} ) is approximately 4.953.

[–] AtomfriedMegaforce 4 points 6 months ago

...and this student was Albert Einstein