My school's calculus teacher's birthday is in a couple of days, and our class decided to give him a surprise birthday card that has a definite integral which evaluates to his initials (TAA). So far all we can think of are easy integrands, and we're stumped even when we tried ones which evaluate to $TA^{2}$ (sorry for my lack of math typography knowledge on this site). He's a really great teacher and it would mean a lot if anyone could help find such an integral. If possible, please keep the difficulty down to a moderate level like integration by parts twice or partial fractions as a relatively hard difficulty level so it won't take too much time to solve it. Any help is much appreciated. (PS: I'm relatively new to this site so my apologies if I radically broke the etiquette here.)

  • 2
    $\begingroup$ Do you mean you are looking for something like $$ \int_0^a 2tx dx = ta^2 $$ except more complex? $\endgroup$
    – gt6989b
    Feb 6, 2014 at 18:45
  • $\begingroup$ Yes, in fact that was exactly what we came up with :) $\endgroup$
    – Daccache
    Feb 7, 2014 at 3:33

1 Answer 1


The circle with radius $\sqrt{\frac{T}{\pi}}A$ has area $TA^2$. This circle can be described by the equation $x^2 + y^2 = \frac{T}{\pi}A^2$, so the upper semicircle is described by the function $y = \sqrt{\frac{T}{\pi}A^2 - x^2}$. Therefore, $$ 4\int_0^{\sqrt{\frac{T}{\pi}}A}\sqrt{\frac{T}{\pi}A^2 - x^2} dx = TA^2. $$

A more artistic version of the same concept might be to color a circle and simply indicate in your picture that the radius is $\sqrt{\frac{T}{\pi}}A$. That way the entire class can stare at him until he does the obvious computation. This also sidesteps the rather unfortunate fact that my integral has the string "$TA^2$" right there in the integrand.


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