Question

Why is the integral of x^2 dx from x=-2 to 2 positive? I plugged it into wolfram alpha and it says it is positive 16/3. Shouldnt it be x^3/3 evaluated from -2 and 2? that gives me -8/3 - 8/3 = -16/3

Answer 1

\[\Large \int\limits_{-2}^{2} x^2 dx = \left[\frac{ x^3 }{ 3 } \right]_{-2}^{2} = \] \[\Large \frac{ 2^3 }{ 3 } -\frac{ (-2)^3 }{ 3 }\]

Answer 2

\[\Large = \frac{ 8 }{ 3 } +\frac{ 8 }{ 3 }\]

Answer 3

Also the area has to be positive, since it's entirely above the x-axis. Area will only be negative when below the x-axis.

Answer 4

The explanation on area makes good sense to me, after looking at the graph of it. I think I just got the order backward, no idea what I was thinking. Thank you.