Answer
$$\textbf{True}$$
Work Step by Step
$$\int^{10}_{-10} (ax^3+bx^2+cx+d)dx=\int^{10}_{-10} (ax^3+cx)dx+\int^{10}_{-10} (bx^2+d)dx$$
The first term is integral of an odd function from from -10 to 10 and is hence zero. The second term is the integral of an even function from -10 to 10 and can be written as twice the integral from 0 to 10.
So,
$$\int^{10}_{-10} (ax^3+bx^2+cx+d)dx=\int^{10}_{-10} (ax^3+cx)dx+\int^{10}_{-10} (bx^2+d)dx$$$$= 0+2\int^{10}_{0} (bx^2+d)dx=2\int^{10}_{0} (bx^2+d)dx$$
