Can someone explain this to me? Not the maths, the joke.
funny title goes here
Submitted 1 month ago by fossilesque@mander.xyz to science_memes@mander.xyz
https://mander.xyz/pictrs/image/d4912ee7-9484-4f11-99f9-95897d719db3.jpeg
Comments
Dave@lemmy.nz 1 month ago
fossilesque@mander.xyz 1 month ago
Rolando@lemmy.world 1 month ago
Calling !peterexplainsthejoke@lemmy.world…
LinkOpensChest_wav@lemmy.blahaj.zone 1 month ago
Old man voice: “Aunt eye memes? In my day, we only had regular memes, and I turned out fine!”
CubbyTustard@reddthat.com 1 month ago
at the core of humor is the twist, where you expected one thing but get another. In an anti joke or meme the twist is usually that there is no joke, but the format set you up to expect one.
driving_crooner@lemmy.eco.br 1 month ago
The test:
Solve e^{int_0^1 x2+x+1 dx}
Usernamealreadyinuse@lemmy.world 1 month ago
No, copiloot 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@lemmy.world 1 month ago
…and this student was Albert Einstein
OpenStars@startrek.website 1 month ago
Hrm, it sounds like the teacher already knew the answer.
Maybe stop wasting the time of the student then!?
You’re welcome.