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𝘶-substitution warmup

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
Before diving into our practice exercise, gain some risk-free experience performing 𝘶-substitution.
Find each indefinite integral.

Problem 1

integral, cosine, left parenthesis, x, squared, right parenthesis, 2, x, d, x, equals
plus, space, C

Problem 2

integral, start fraction, 3, x, squared, divided by, left parenthesis, x, cubed, plus, 3, right parenthesis, squared, end fraction, d, x, equals
plus, space, C

Problem 3

integral, e, start superscript, 4, x, end superscript, d, x, equals
plus, space, C

Problem 4

integral, x, dot, square root of, start fraction, 1, divided by, 6, end fraction, x, squared, plus, 1, end square root, d, x, equals
plus, space, C

Want to join the conversation?

  • aqualine ultimate style avatar for user Roberto Dias
    Well, in the problem #2 what happens to the "square power" of "u" when you substitue back the equation x^3 + 3 please? I did not catch what happened to get the given answer. Thank you. :)
    (8 votes)
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    • female robot grace style avatar for user Brielle of Earth
      1/(u^2) == u^(-2). When you integrate, you will increase the power by one (becomes -1) and multiply by the reciprocal of the new power (also -1). Your integral is -1*(u^-1) ==(-1/u).
      This problem is tricky because of the properties of exponents, just try rewriting the factors to understand where the exponent went to.
      (11 votes)
  • hopper cool style avatar for user Vítor Watanabe
    =∫1/u^2 ​du shoudn't be = ln(|u^2|)?
    (5 votes)
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  • blobby green style avatar for user Avalon Z
    In problem 2, why the negative?

    1/u^2 * du
    -1/u
    (1 vote)
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  • male robot hal style avatar for user scott.d.corwin
    I was given the problem:
     ∫ sin³(x)cos(x)dx = ? + C

    I entered (sin(x)^4)/4 the first time & was marked wrong. Then I tried entering the exact solution given which is impossible to do as far as I know on my mobile phone: (1/4)sin(x)^4. There is no process or command available to enter it as (1/4)sin^4(x). I'm assuming that is the reason I'm being marked wrong. Is it possible to enter the exponent before a trig function's parentheses? Please advise.
    (1 vote)
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  • blobby green style avatar for user Najee Hillman
    in problem 4 why is xdx= 3?
    (1 vote)
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  • blobby green style avatar for user Laurie Eyre
    how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)
    (1 vote)
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  • purple pi teal style avatar for user Nischal Chandur
    In the first question, is it right to take cos(x^2) as u?
    (0 votes)
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    • piceratops ultimate style avatar for user Samuli Niemi
      If you choose cos(x^2) as your u, your du ends up being -sin(x^2)*2x*dx. You could rearrange the equation as du/-sin(x^2) = 2x*dx and replace the 2x*dx in the original equation accordingly, but you're still left with the x^2 inside the sine-function. For the u-substitution to work, you need to replace all variables with u and du, so you're not getting far with choosing u = cos(x^2). If you choose, as you should, u = x^2 and your du = 2*x*dx, you'll get int(cos(u)*du) and that's pretty straight-forward to integrate.
      (4 votes)
  • blobby green style avatar for user RussellSH519
    Actually the problem 3 can already be solved by using the integration formula of e.
    (1 vote)
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  • spunky sam blue style avatar for user Avi Perl
    In order for most of these to work, the constant multiple rule must apply to integrals in exactly the same way that it applies to derivatives. Is that assumption correct?
    (1 vote)
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  • hopper cool style avatar for user Vítor Watanabe
    =∫1/u^2 ​du = −1/u ​+C

    Anyone could explain why appeared a negative signal on −1/u ​+C?
    (1 vote)
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    • stelly green style avatar for user The #1 Pokemon Proponent
      Think about it this way. Let's say we have the function y=1/x. Now let's think about the graph in the 1st quadrant. The slope is always negative. Therefore, the derivative of the curve at any point on it in the first quadrant should be a negative number. d/dx (1/x) < 0 for x > 0 . This should hopefully provide some intuition for the negative sign.

      For the rigorous proof, try finding d/dx (-1/x).
      (1 vote)