funny title goes here

Comments

• OpenStars@startrek.website ⁨8⁩ ⁨months⁩ ago

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

  Maybe stop wasting the time of the student then!?

  You’re welcome.

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• Dave@lemmy.nz ⁨8⁩ ⁨months⁩ ago

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

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  • fossilesque@mander.xyz ⁨8⁩ ⁨months⁩ ago

    knowyourmeme.com/memes/anti-memes

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    • Rolando@lemmy.world ⁨8⁩ ⁨months⁩ ago

      Calling !peterexplainsthejoke@lemmy.world…

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    • LinkOpensChest_wav@lemmy.blahaj.zone ⁨8⁩ ⁨months⁩ ago

      Old man voice: “Aunt eye memes? In my day, we only had regular memes, and I turned out fine!”

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• driving_crooner@lemmy.eco.br ⁨8⁩ ⁨months⁩ ago

  The test:

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

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  • Usernamealreadyinuse@lemmy.world ⁨8⁩ ⁨months⁩ 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.

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• AtomfriedMegaforce@lemmy.world ⁨8⁩ ⁨months⁩ ago

  …and this student was Albert Einstein

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