Answer
In the $MLT system$, the requested dimensions are:
a)$\frac{∂u}{∂t}=LT^{-2}$
b)$\frac{∂^{2}u}{∂x∂t}=T^{-2}$
c)$∫(\frac{∂u}{∂t})dx=L^{2}T^{-2}$
Work Step by Step
Integration and derivative signs are mathematical operators and not physical quantities, so have no dimensions. If $u$ is velocity, $x$ is length, and $t$ is time, then:
a)$\frac{∂u}{∂t}=\frac{m/s}{s}=\frac{m}{s^{2}}=LT^{-2}$
b)$\frac{∂^{2}u}{∂x∂t}=\frac{m/s}{m\times s}=\frac{1}{s^{2}}=T^{-2}$
c)$∫(\frac{∂u}{∂t})dx=(\frac{m/s}{s})\times m=\frac{m^{2}}{s^{2}}=L^{2}T^{-2}$
