## Calculus: Early Transcendentals 8th Edition

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.