calculus

  1. N

    I built The Daily Derivative, a world and dailyintegral.com inspired daily calculus site.

    Link: "dailyderivative.dev" I first wanted to say that my inspiration came mostly from @jamie1234567890 who created the amazing site, "dailyintegral.com" -It give 3 new derivatives daily, (easy, medium, hard) to fit everyone from beginner to advanced, I will say however that this site is...
  2. J

    Daily Integral - A study tool for HSC maths students

    “https://dailyintegral.com” (This is the URL for the website) Hey guys! I built The Daily Integral, a site that updates every day with three integrals—Easy, Medium, Hard—so it’s genuinely useful for all experience levels. It’s inspired by daily games like Wordle and Nerdle: quick...
  3. waltssillyhat

    displacement-time and velocity-time graphs

    For part (iv) of this question, I understand the shape of the displacement-time graph in the solution, but how do you use the graph given in the question to find that the displacement at 4 and 6 seconds is 6m, as well as the displacement being 8m at five seconds?
  4. A

    I dont understand the working out in the first line

    Like I understand both curves intersect at x as they are the inverse and its way too hard to find the inverse but using x instead of the inverse wont give u the correct shaded area
  5. SB257426

    Challenge integral

    does anyone know how to do this ? I tried doing it and i got it wrong
  6. SB257426

    Integration Question

    I came across an integral that my teacher gave me to do. I have an answer but am unsure if my answer is correct. Other softwares (wolfram alpha and symbolab) are giving me different answers, but I suspect they are algebraically equivalent since they probably have different constants of...
  7. SB257426

    Integration Question

    I am pretty confused with my solution to this integration proof question. I feel like it isn't the right way to prove it. d/dx (uv) = u(dv/dx)+v(du/dx) take the integral with respect to x of both sides: uv = ∫ u(dv/dx ) dx + ∫ v(du/dx) dx therefore, ∫ u(dv/dx ) dx = uv - ∫ v(du/dx)...
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