Với mỗi số $k$, đặt ${I_k} = \int\limits_{ - \sqrt k }^{\sqrt k } {\sqrt {k - {x^2}} dx}$. Khi đó ${I_1} + {I_2} + {I_3} + ... + {I_{12}}$ bằng:

Đặt $x = \sqrt k \sin t$ $\Rightarrow dx = \sqrt k \cos tdt$.

Đổi cận: $\left\{ \begin{array}{l}x =  - \sqrt k  \Leftrightarrow \sin t =  - 1 \Leftrightarrow t =  - \dfrac{\pi }{2}\\x = \sqrt k  \Leftrightarrow \sin t = 1 \Leftrightarrow t = \dfrac{\pi }{2}\end{array} \right.$.

Khi đó ta có

$\begin{array}{l}{I_k} = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sqrt {k - k{{\sin }^2}t} .\sqrt k \cos tdt} \\{I_k} = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {k{{\cos }^2}tdt} \\{I_k} = k\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{1 + \cos 2t}}{2}dt} \\{I_k} = \dfrac{k}{2}\left. {\left( {t + \dfrac{1}{2}\sin 2t} \right)} \right|_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}}\\{I_k} = \dfrac{k}{2}\left( {\dfrac{\pi }{2} + \dfrac{1}{2}\sin \pi  + \dfrac{\pi }{2} - \dfrac{1}{2}\sin \left( { - \pi } \right)} \right)\\{I_k} = \dfrac{k}{2}.\pi  = \dfrac{{k\pi }}{2}\end{array}$

$\begin{array}{l} \Rightarrow {I_1} + {I_2} + {I_3} + ... + {I_{12}}\\ = \dfrac{\pi }{2}\left( {1 + 2 + 3 + ... + 12} \right)\\ = \dfrac{\pi }{2}.\dfrac{{12.13}}{2} = 39\pi \end{array}$