= Solution
{c}
The <volumetric flow rate> is the <integral> of axial <velocity> over the cross-section:
$$
Q=\int_{-a}^{a}\int_{-b}^{b}w(x,y)\,dy\,dx.
$$
The <plane Poiseuille flow> contribution is
$$
\int_{-a}^{a}\int_{-b}^{b}\frac{G}{2\mu}(b^2-y^2)\,dy\,dx
=\frac{4ab^3G}{3\mu}.
$$
For each correction <eigenfunction>,
$$
\int_{-a}^{a}\frac{\cosh(k_nx)}{\cosh(k_na)}\,dx
=\frac{2\tanh(k_na)}{k_n},\qquad
\int_{-b}^{b}\cos(k_ny)\,dy=\frac{2(-1)^n}{k_n}.
$$
<Uniform convergence> permits <termwise integration>. Multiplying these two <integrals> by $A_n$ gives
$$
A_n\frac{4(-1)^n}{k_n^2}\tanh(k_na)
=-\frac{256Gb^4}{\mu\pi^5}
\frac{\tanh(k_na)}{(2n+1)^5}.
$$
Consequently
$$
\boxed{Q=\frac{4ab^3G}{3\mu}
-\frac{256b^4G}{\pi^5\mu}
\sum_{n=0}^{\infty}\frac{1}{(2n+1)^5}
\tanh\!\left(\frac{(2n+1)\pi a}{2b}\right).}
$$
Every summand in the subtracted quantity is positive. Thus \b[the side walls reduce the flux] below the infinite-width <plane Poiseuille flow> prediction. Its dimensions are <volume> per unit <time>, as required for a <volumetric flow rate>.
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