The given curve is
[tex]f(x)=x^2[/tex]
To find the area under the curve we will use the integration
[tex]A=\int_0^2x^2dx[/tex][tex]\begin{gathered} A=[\frac{x^{2+1}}{2+1}]_0^2 \\ \end{gathered}[/tex]
Simplify it
[tex]A=[\frac{x^3}{3}]_0^2[/tex]
Substitute x by 2 and 0
[tex]\begin{gathered} A=[\frac{2^3}{3}]-[\frac{0^3}{3}] \\ \\ A=\frac{8}{3}-\frac{0}{3} \\ \\ A=\frac{8}{3} \end{gathered}[/tex]
The area under the curve is 8/3 square units
The answer is A
