where $f(x)=\min(1,x^2)$.

I thought about solving it like this:

$$\int_{-1}^1 x^3\,dx + \int_{1}^3x\,dx = \cdots = 4.$$

But the solution is $\frac{26}{3}$ and I don't understand how they got it.

  • $\begingroup$ Your answer seems correct. $\endgroup$ – enzotib Feb 1 '15 at 20:05
  • $\begingroup$ It looks ok to me. $\endgroup$ – Aaron Maroja Feb 1 '15 at 20:05
  • $\begingroup$ There is nothing wrong with your answer, see here. $\endgroup$ – Mufasa Feb 1 '15 at 20:07
  • 5
    $\begingroup$ You know, treating your solutions manual like it's The Word Of God can get you in a lot of trouble.It's put together by people-usually people who aren't really enjoying what they're doing and need the money to eat. Email them and tell them they screwed up. $\endgroup$ – Mathemagician1234 Feb 1 '15 at 20:24
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    $\begingroup$ @Mathemagician1234 I think you are being a bit unkind here. I write my manuals with answers and often I make typos and such. Luckily the good will of my classes allows me to weed these out over time. +1 for The Word of God comment though. $\endgroup$ – JP McCarthy Feb 2 '15 at 16:23

You are correct: You did just fine.

Either the intended question was misprinted, the solution is a misprint, or the exercise and its solution are incorrectly matched/numbered (e.g., misidentified: perhaps it is the solution to a different exercise?)

We can only speculate... "Why the error in the supposed solution?" But it happens.

Be reassured; you're the "winner" here, with your work and your solution.


For me it looks fine. On $[-1,1]$, the integrand is just $xf(x) = x^{3}$, which is symmetrical to the origin, which means that the integral is 0. Left over is $\int_{1}^{3} x dx = 4$. Either we are all missing something and are looking extremely stupid now or the solution is incorrect (which I think is the case).


My attempt::

Assuming $f:\Bbb R \to \Bbb R$, $$f(x) = \min\Big(1\ ,\ x^2\Big) = \begin{cases} x^2\quad,\ x \in [-1,1] \\ 1 \quad,\ x \in (-\infty,-1)\cup(1, \infty) \end{cases}$$ The graphs of $ y = \min(1, x^2)$ and $y = x\min(1,x^2)$ are respectively as follows:

Now, $$\int x\cdot f(x)\ \mathrm dx = \begin{cases} \displaystyle \int x^3 \mathrm dx = \frac{1}{4}x^4 \color{lightgray}{+ \mathcal C}\ , x \in [-1,1]\\ \displaystyle \int x \mathrm dx = \frac{1}{2}x^2 \color{lightgray}{+ \mathcal C} \ ,\ x \in \Bbb R \sim [-1,1]\end{cases}$$


$$ \int_{-1}^{3}xf(x)\mathrm dx = \int_{-1}^{1}xf(x)\mathrm dx + \int_{1}^{3} xf(x)\mathrm dx\\ = \frac{1}{4}x^4 \Big]_{-1}^{+1} + \frac{1}{2}x^{2}\Big]_{1}^{3} \\ = \frac{1}{4}\left[\Big((+1)^4 - (-1)^4\Big) + 2\Big(3^2 - 1^2\Big)\right]\\ = \frac{0 + 16}{4} = 4$$

So, yes, you're absolutely right =) The textbook was wrong. There is no room for doubt. $$\boxed{\displaystyle \int_{-1}^{3}xf(x)\mathrm dx = 4}$$


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