Find $\int_{-1}^3xf(x)\,dx$ where $f(x)=\min(1,x^2)$

Find:

$$\int_{-1}^3xf(x)\,dx,$$

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.

Solution 1:

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.

Solution 2:

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).