$$\int_{-99}^{99} (ax^3+bx^2 + cx) \,dx= 2\int_{0}^{99} bx^2 \,dx$$
Is this assertion true or false? I would say this is only true if $a$ and $c$ are zero, so that $bx^2$ is even. But this is false if $a$ and/or $c$ are not zero since the function is not even then. Am I correct?
You can split the integral $\int_{-99}^{99}(ax^3+bx^2+cx)\ dx=\int_{-99}^{99}(ax^3+cx)\ dx+\int_{-99}^{99}bx^2\ dx=\int_{-99}^{99}bx^2\ dx=2\int_0^{99}bx^2\ dx$, since the first summand is odd and the second is even