Exercise · ZW-TRIPLE-INTEGRAL-009 · ★★★ Advanced

The average value \(\tfrac1{V(E)}\iiint_E f\,dV\) of \(f=x+y+z\) over the unit cube \(E=[0,1]^3\) is:

Choose one option.

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  • C. \(\tfrac32\)

Explanation

The volume is \(1\) and the integral is \(\tfrac12+\tfrac12+\tfrac12=\tfrac32\), so the average is \(\tfrac32\). The value \(3\) would be the maximum of \(f\), not its average.

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Review the concept: Triple integral

The triple integral of a function \(f\) over a solid \(E\) of space is the limit of the sums of the values of \(f\) times the volume \(\Delta V\) of ever finer cells covering \(E\).

\[\iiint_E f\,dV=\lim_{\max\Delta V_i\to0}\sum_i f(x_i^*,y_i^*,z_i^*)\,\Delta V_i\]

More practice problems: triple integral

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