I was finding the area below the x-axis and above $y = x^2 – 4x$. My outcome was a negative number (fair enough, it's under the x-axis), but wolframalpha for instance gives me a positive number (I get the same number but negative) which one is correct?
[Math] Negative area below x-axis and above $f(x)$
areacalculusintegration
Related Solutions
Your thinking is correct, but the book is wrong in general.
Assuming that $t\in [a,b]$ with $a<b$, then considering $\displaystyle \int_a^b y \frac{dx}{dt}dt$, we can split into two cases:
(i) The integrand $y \dfrac{dx}{dt} > 0$, which can happen (a) if $y$ and $\dfrac{dx}{dt}$ are both positive, or (b) both are negative.
(ii) The integrand $y \dfrac{dx}{dt} < 0$, which can happen if either $y > 0$ and $\dfrac{dx}{dt} < 0$, vice-versa.
Both cases above correlate with your interpretation, which is correct.
Of course, the integrand can change sign within $[a,b]$, in which case the negative and positive areas will cancel each other out. Let's forget about that for now.
It is not true to say that $y \dfrac{dx}{dt} > 0$ corresponds to clockwise motion about the origin. This is a really important point.
Consider the parametric curve $x=t, y=t^2$ for $t\in [0,1]$. This is just the parabola $y=x^2$. Intuitively, one can see that in moving from $(0,0)$ ($t=0$) to $(1,1)$ ($t=1$) the curve moves anti-clockwise relative to the origin. Yet, computing the integral $\displaystyle \int_a^b y \frac{dx}{dt}dt = \dfrac{1}{3}$, which is positive.
More rigorously, we can convert the problem above to polar coordinates using the transformations $x=r\cos\theta, y=r\sin\theta$. We find that $r\geq0$ and $0\leq \theta < \dfrac{\pi}{2}$. It's quite easy to show that $\dfrac{d\theta}{dr} > 0$, thus the 'motion' is counter-clockwise throughout.
Compare this to $x=\cos\theta, y=\sin\theta$ for $\theta\in\left[0,\dfrac{\pi}{2}\right]$. The 'motion' is also counter-clockwise. However, if you compute the integral, you'll get the negative result $-\dfrac{\pi}{4}$.
The conclusion? Forget clockwise vs anticlockwise motion. What's important is the sign of $y \dfrac{dx}{dt}$, nothing more.
In general $$ \int_{a}^bf(x)\, dx $$ gives us the signed area of the region in the plane bounded by the graph of $f$, the $x$-axis and the lines $x=a$, $x=b$. See the picture at the bottom for an illustration.
In particular $$ \int_{-1}^{1}\sqrt{1-x^2}\, dx\tag{0} $$ gives us the signed area bounded by the upper semicircle, the $x$-axis, and the lines $x=1$ and $x=-1$ and hence gives us the area of the upper half disk.
In a similar way, $$ \int_{-1}^{1}-\sqrt{1-x^2}\, dx\tag{1} $$ gives us the signed area bounded by the lower semicircle, the $x$-axis, and the lines $x=1$ and $x=-1$ and hence gives us the negative of the area of the lower half disk.
If we add the two previous integrals together, we get zero.
Best Answer
An area is always positive, occasionally an integral is negative. It depends on whether you want the area or the value of the integral.