How is $\int_{-2}^2 (x^3\cos \frac{x}{2})\sqrt{4-x^2} \, dx$ equal to $\pi$?

Question

You know those jokes where someone has need for a number like for a WiFi password then they're given an integral so they'd just give up? People like sending me those cause I love math. But today I got a unique one.

$$\int_{-2}^2 \left(x^3\cos \frac{x}{2}\right)\sqrt{4-x^2} \, dx$$

Why is it different? Cause I evaluated it on Desmos and it's equal to $\pi$!

How is that possible? I'm trying to work around it and it's proving tricky. The much I know is if I can separate the square root term as an integral then it'll be the area of a semicircle with radius 2.

You forgot a term from there. And yes it's a wi-fi joke:https://math.stackexchange.com/q/3427744/515527 — Zacky, Jan 01 '20 at 11:35
Yeah. Definitely. Thanks. Also got a hint of how even and odd functions intertwine so I've learnt something more — Nεo Pλατo, Jan 01 '20 at 15:56

score 0 · Answer 1 · answered Jan 01 '20 at 11:37

0

As Lord said, it's an odd function which means the integral over the given region is equal to 0. Many functions don't have an elementary antiderivative.

https://www.desmos.com/calculator/80bfudcv7h

answered Jan 01 '20 at 11:37

Alessio K

10,599

How is $\int_{-2}^2 (x^3\cos \frac{x}{2})\sqrt{4-x^2} \, dx$ equal to $\pi$?

1 Answers1