By Math Vault | Calculus

Hehe…we are always amused by calculus students who manage to mis-integrate a function to the point of yielding nonsensical answers. To illustrate what we mean by this, here is Steve, who — at the point of this writing at least — still haven’t got a clue where his computations went awry:

## How do you evaluate $\displaystyle \int_{-5}^5 x – \sqrt{25-x^2} \, dx$ algebraically? When I take the areas geometrically, I get $\displaystyle -\frac{25\pi}{2}$, as $\displaystyle \int_{-5}^5 x \, dx=0$ and $\displaystyle \int_{-5}^5 \sqrt{25-x^2} \, dx$ is the area of a half-circle with radius $5$. But when I try to evaluate it algebraically, I don’t seem to get the same answer: $\displaystyle \int^5_{-5}\left(x-\sqrt{25-x^2}\right)\,dx=\left[\frac{x^2}{2}-\frac{2}{3}(25-x^2)^{3/2}+C\right]^{5}_{-5} \dots$

OK…What’s going on here? Well, keep reading!

Well. $\displaystyle\int_{-5}^5(x-\sqrt{25-x^2} \, dx=\int_{-5}^5 x \, dx-\int_{-5}^5\sqrt{25-x^2}\, dx$. For the 2^{nd} integral, we can’t integrate it as if it were $\sqrt{x}$. In fact, if we try to “integrate” $\displaystyle\sqrt{25-x^2}$ this way, we get:

\[ \frac{2}{3}\left({25-x^2}\right)^{\frac{3}{2}} \]

However,

\[ \frac{d}{dx}\left[ \frac{2}{3}\left({25-x^2}\right)^{\frac{3}{2}} \right] = \sqrt{25-x^2}(-2x) \ne\sqrt{25-x^2} \]

Oops. Looks like the obnoxious *Chain Rule* just destroyed our whole scheme.

We have just seen how misusing *Power Rule* can create abnormal antiderivatives. For the 2^{nd} integral, what we need to do is to use a *back substitution* with $x=5\sin(u)$. In which case, $dx=5\cos(u)du$, and the limits of the integral now run from $\displaystyle -\frac{\pi}{2}$ to $\displaystyle \frac{\pi}{2}$. Putting everything together, we get:

\begin{align*} \int_{-5}^5 \sqrt{25-x^2}\, dx =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{25-\left(5\sin(u)\right)^2} \ 5\cos(u) \, du \end{align*}

Simplifying further, we get:

\begin{align*} \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}5\cos(u) \, 5\cos(u) \, du=25\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2(u) \, du \end{align*}

By using the *half-angle formula* for cosine $\displaystyle \left(\text{i.e., } \cos^2(u) =\frac{1 + \cos(2u)}{2} \right)$, we can continue to work on the antiderivative:

\begin{align*} 25\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2(u) \, du & = 25\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos(2u)}{2} \, du \\ & = 25\left [\frac{1}{2}x+ \frac{\sin(2u)}{4}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\ & = 25\left(\frac{\pi}{2}\right) \end{align*}

which means that our original integral, $\large \displaystyle \int_{-5}^5 x – \sqrt{25-x^2} \, dx$, would be $\displaystyle -\frac{25\pi}{2}$, as Steve has previously inferred using a geometric argument.

And if there is a moral to this story, it would be that there would be no paradox at all — as long as we don’t go for an anarchic free-for-all! 😉

**Math Vault and its Redditbots** has the singular goal of advocating for education in higher mathematics through *digital publishing* and the *uncanny* use of technologies. Head to the **Vault** for more math cookies. :)

Desmos: A Definitive Guide (How to Perform Cool Computations and Creating Great Graphs Using Online Graphing Calculator)

Integration Technique Series — How to Make Use of the Overshooting Method and Integrate with Ease

Infinite Limits and the Behaviors of Polynomials at the Infinities — A Theoretical Musing

Chain Rule for Derivative — Venturing Into The Dark Side Beneath Applied Calculus…

Add Your Reply