Problem with integration by substitution.

Question

Using the substitution $u=1-x$, compute the integral of $\int{x(1-x)^2}dx.$

My Work: Let $u=1-x.$ Then $\mathrm dx=-\mathrm du$ and $x=1-u,$ so

$$\int{x(1-x)^2}\,\mathrm dx\\ =-\int{(1-u)u^2}\,\mathrm du\\ =\frac{u^4}{4}-\frac{u^3}{3}+C\\ =\frac{(1-x)^4}{4}-\frac{(1-x)^3}{3}+C\\ =\frac{3x^4-8x^3+6x^2-1}{12}+C.$$

However, the answer gotten by expanding the brackets first then integrating does not seem to match that of the answer above arrived by substitution:

$$\int{x(1-x)^2}\,\mathrm dx\\ =\int{x-2x^2+x^3}\,\mathrm dx\\ =\frac{x^2}{2}-\frac{2x^3}{3}+\frac{x^4}{4}+C.$$

I cannot account for the $-\frac{1}{12}$ from the method using substitution.

Can you post the two answers you get? the one by substitution and the one by expanding the brackets? — Alessandro, May 19 '21 at 13:57
Hi Joe. Alessandro is right, you should show us exactly what you have done. Click "Edit" above, and show us your mathematics. — 311411, May 19 '21 at 13:58
Hope it is a little clearer now, was struggling with some of the formatting, accustom to using Lyx. — Joe Watkins, May 19 '21 at 15:00
An indefinite integral is determined only up to a constant summand. That is what the "$+C$" is for. In your two computations, the two values of $C$ differ by $-\frac{1}{12}$. This is not a problem. — GEdgar, May 19 '21 at 15:47

score 2 · Answer 1 · answered May 19 '21 at 14:02

2

Well

\begin{align} \int{x(1-x)^2}dx &= \int x^3-2x^2+x \mathrm{d}x \\ &=\frac{1}{4}x^4-\frac{2}{3}x^3+\frac{x^2}{2} +C_1 \end{align}

Whereas with $u=1-x \, \implies \mathrm{d}u=-\mathrm{d}x$ and \begin{align} \int (1+u)^2u (- \mathrm{d}u) &= -\int u^3+2u^2+u \mathrm{d}u \\ &=-\frac{u^4}{4}-\frac{2}{3}u^3-\frac{u^2}{2}+C_2 \end{align} Now substitute in $x=1-u$ to see equivalence.

answered May 19 '21 at 14:02

Thanks, I see what you did. I edited my post and hope my question is a little clearer now. – Joe Watkins May 19 '21 at 15:03

ryang · Accepted Answer · 2023-01-24T01:11:06.267

1

Let the domain of $f$ be an interval.

Then every pair of the infinitely many antiderivatives of $f$ differ by a constant. Since the indefinite integral $$\int f(x)\,\mathrm dx$$ gives the general specification of these antiderivatives, it has the form $F(x)+C,$ where the constant of integration $C$ is arbitrary.

Both your answers are correct, and their specified constants of integration differ by $C_2-C_1=\frac1{12}.$ To simplify the first answer, substitute $$C:=D+\frac1{12}$$ so that $$\int{x(1-x)^2}\,\mathrm dx=\frac{3x^4-8x^3+6x^2}{12}+D.$$

edited Jan 24 '23 at 01:11

answered May 19 '21 at 15:31

ryang

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Thanks for the edit and the answer. I was thinking along those lines but wanted to be 100% sure before I brought it to my students. They asked if everything then wouldn't be shifted by $-\frac{1}{12}$ and the plot of the solutions online seems to have shown that. Usually when doing some of these textbook questions things work out exactly or no one bothers to check it both ways. Kudos to them. – Joe Watkins May 19 '21 at 16:54

Problem with integration by substitution.

2 Answers2