Can someone explain this to me? Not the maths, the joke.
Science Memes
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
- Don't throw mud. Behave like an intellectual and remember the human.
- Keep it rooted (on topic).
- No spam.
- Infographics welcome, get schooled.
This is a science community. We use the Dawkins definition of meme.
Research Committee
Other Mander Communities
Science and Research
Biology and Life Sciences
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- !reptiles and [email protected]
Physical Sciences
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
Humanities and Social Sciences
Practical and Applied Sciences
- !exercise-and [email protected]
- [email protected]
- !self [email protected]
- [email protected]
- [email protected]
- [email protected]
Memes
Miscellaneous
Hrm, it sounds like the teacher already knew the answer.
Maybe stop wasting the time of the student then!?
You're welcome.
The test:
Solve e^{int_0 1 x2+x+1 dx}^
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.
...and this student was Albert Einstein