Before we start let's remember two important properties of integrals
[tex]\begin{gathered} \int (f(x)\pm g(x))_{}dx=\int f(x)dx\pm\int g(x)dx \\ \\ \int c\cdot f(x)dx=c\cdot\int f(x)dx \end{gathered}[/tex]Using that, we can rewrite
[tex]\int ^2_{-2}(7x^2-28)dx[/tex]as
[tex]7\int ^2_{-2}x^2dx-28\cdot\int ^2_{-2}dx[/tex]The integral of a monomial is
[tex]\int x^ndx=\frac{x^{n+1}}{n+1},n\ne-1[/tex]Using it let's integrate the two monomials
[tex]F(x)=\int (7x^2-28)dx=\frac{7x^3}{3}-28x+C[/tex]Using that and the Fundamental Theorem of Calculus:
[tex]\int ^b_af(x)dx=F(b)-F(a)[/tex]We just gotta evaluate F(x) at 2 and -2.
[tex]\begin{gathered} \int ^2_{-2}(7x^2-28)dx=F(2)-F(-2) \\ \\ \int ^2_{-2}(7x^2-28)dx=(\frac{7\cdot2^3}{3}-28\cdot2)-(\frac{7\cdot(-2)^3}{3}-28\cdot(-2)_{}) \\ \\ \int ^2_{-2}(7x^2-28)dx=-\frac{56}{3}-56 \\ \\ \int ^2_{-2}(7x^2-28)dx=-56(1+\frac{1}{3}) \end{gathered}[/tex]Therefore, the final result is
[tex]\int ^2_{-2}(7x^2-28)dx=-56(1+\frac{1}{3})[/tex]That's the signed area.