Why does the indefinite integral of $x^2$ enter quadrant 3?

Question

When integrating $y=x^2$ (ending up with $y=\frac1 3 x^3$), the values while $x<0$ become negative. Why is this?

Answer 1

The integral is not the area under the curve, but the signed area under the curve. Area in quadrants I and III is defined as positive, and area in quadrants II and IV is defined as negative.

Answer 2

When looking at indefinite integrals, the value at a particular point shouldn't be given too much attention. For instance, $y = \frac{x^3}{3} - 1,000,000,000$ is also an indefinite integral of your function, and it hangs out in quadrant IV for a heck of a long time for positive $x$.

Really what we care about indefinite integrals are differences of its values. So, for $f(x) = x^2$, let's assume we have some indefinite integral $F(x) = \int f(x) dx$.

Since $f(x) \ge 0$ for all $x$, then we expect (assuming $a \lt b$) that $F(b) - F(a) = \int^b_a f(x)dx \ge 0$, which is just $F(b) \ge F(a)$, i.e. $F$ is increasing. (Okay, technically "non-decreasing", but will you let me be lazy?). And yes, we can see that all possible antiderivatives are in fact increasing: $F(x) = x^3/3 +C$ is increasing no matter what value the constant $C$ has.

Implicit in your question seems to be the assumption that "when you integrate a positive function, you should get back a positive function". And that's not always true. What is always true is "when you integrate a positive function, you get back an increasing function". It's just one of the things you'll get to learn as you develop your intuition about integrating and differentiating functions.