Question

a) In what way Fubini and Clairault’s theorem are similar?

b) If \(f(x,y)\) is continuous on \(R = (a,b) \times (c,d)\) and \(g(x,y) = \int\limits_0^1 {\int\limits_0^1 {f(s,t)dtds} } \) for \(a < x < b\), \(c < y < d\), Show that \({g_{xy}} = {g_{yx}} = f(x,y)\).

Short answer

a) The similar thing between Fubini and Clairault’s theorem is that the function\(f(x,y)\) must be continuous on respective domains.

b) Let \(f(x,y)\)is continuous on rectangle \((a,b) \times (c,d)\).

Consider \(g(x,y) = \int\limits_0^1 {\int\limits_0^1 {f(s,t)dtds} } \) with respect to x.

Step-by-step solution

Differentiating \({g_y}(x,y)\) with respect to x.

\(\begin{array}{l}{g_x}(x,y) = \frac{d}{{dx}}\left( {\int\limits_a^x {\left( {\int\limits_c^y {f(s,t)dt} } \right)} ds} \right)\\ = \int\limits_c^y {f(x,t)dt} \end{array}\)

Differentiating with respect to y.

\({g_{xy}}(x,y) = \frac{d}{{dy}}\left( {\int\limits_c^y {f(x,t)dt} } \right)\)

Therefore,\({g_{xy}} = f(x,y)\).

Similarly, we get\({g_{yx}} = f(x,y)\).

Hence, \({g_{xy}} = {g_{yx}} = f(x,y)\).