By the definition of definite integral, the integral of a function in a interval is the sum of the areas above the x-axis subtracted by the areas below the x-axis inside of the interval.
Given the function
[tex]f(x)=4x^2-16[/tex]
Since our function is an upward parabola and our interval is [-2, 2], and those are the roots of our function, the whole region is under the x-axis, therefore, the area we want is minus the integral of our function on this interval.
[tex]A=-\int ^2_{-2}(4x^2-16)dx[/tex]
Since we're dealing with an integral of a pair function in a symmetric interval, we can rewrite our integral as
[tex]-\int ^2_{-2}(4x^2-16)dx=-2\int ^2_0(4x^2-16)dx[/tex]
Multiplying an integral by (-1) is the same as inverting the order of the limits.
[tex]-2\int ^2_0(4x^2-16)dx=2\int ^0_2(4x^2-16)dx[/tex]
To find the area, we can just solve the integral
[tex]\begin{gathered} A=2\int ^0_2(4x^2-16)dx=2\lbrack\frac{4}{3}x^3-16x\rbrack^0_2_{}^{} \\ A=2(32-\frac{32}{3}) \\ A=64-\frac{64}{3} \\ A=\frac{128}{3} \end{gathered}[/tex]
And this is our answer.
