Answer
TRUE
Work Step by Step
Since,$\int _{-a}^{a}f(x) dx=2\int _{0}^{a}f(x) dx $ if $f(x)$ is an even function and continuous on $ [-a,a]$ and $\int _{-a}^{a}f(x) dx=0$ if $f(x)$ is an odd function and continuous on $ [-a,a]$.
$\int _{-5}^{5}(ax^{2}+bx+c) dx=\int _{-5}^{5}(ax^{2}+c) dx+\int _{-5}^{5}(bx) dx$
Because $(ax^{2}+c)$ is an even function and $bx$ is an odd function.
Therefore,
$\int _{-5}^{5}(ax^{2}+bx+c) dx=2\int _{0}^{5}(ax^{2}+c) dx$
Hence , the statement is true.