I got this:

$0 = \int_{-1}^{1} x^3(x^2+1) dx = \frac{1}{2} \int_{-1}^{1} x^2(x^2+1) dx^2 = \frac{1}{2} \int_{-1}^{1} x(x+1) dx \neq 0$

I think the third equation is wrong, but what is the correct answer?

  • $\begingroup$ Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. $\endgroup$ Dec 6, 2021 at 9:59

1 Answer 1


Your mistake is the last "$=$" sign: $$ \frac{1}{2} \int_{-1}^{1} x^2(x^2+1) dx^2 = \frac{1}{2} \int_{-1}^{1} x(x+1) dx \neq 0$$ In fact, we did a substitution $t=x^2$ here, explicitly.-- But there is a mistake, as the inverse function of $t=x^2$ is not a one-value function on the interval $x\in[-1,1]$.

The correct is as below: $$ \begin{align} \int_{-1}^1{x^2}(x^2+1)\mathrm{d}x^2=&\int_0^1{x^2}(x^2+1)\mathrm{d}x^2+\int_{-1}^0{x^2}(x^2+1)\mathrm{d}x^2 \\ (\color{green}{\text{use}\;t=x^2})=&\int_0^1{t}(t+1)\mathrm{d}t+\int_1^0{t}(t+1)\mathrm{d}t \\ =&\int_0^1{t}(t+1)\mathrm{d}t-\int_0^1{t}(t+1)\mathrm{d}t \\ =&0 \end{align} $$ which avoids that $ x=\begin{cases} -\sqrt{t}\,\,/; x<0\\ \sqrt{t}\,\,/; x\ge 0\\ \end{cases} $.

  • $\begingroup$ Thank you very much! $\endgroup$ Dec 7, 2021 at 6:34
  • $\begingroup$ @Jingjian Lu , you are welcome. $\endgroup$
    – SHBooKP
    Dec 7, 2021 at 7:46

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