Multivariable calculus

Triple Integral of e^(x+y+z) over the Unit Cube

Q: Is (e - 1)^3 the same as e^3 - 1?

No. Expanding gives (e-1)^3 = e^3 - 3e^2 + 3e - 1 \approx 5.073214, while e^3 - 1 \approx 19.086. The cube of a difference keeps the cross terms, and those cross terms are most of the gap.

Q: What is the average value of e^(x+y+z) on the cube?

The cube has volume 1, so the average value equals the integral itself, about 5.073214. That sits well above the midpoint value e^{1.5} \approx 4.4817, which is what convexity of the exponential predicts.

The triple integral of e raised to the power x plus y plus z, over the unit cube with each variable from 0 to 1, equals the cube of e minus 1, which is about 5.073214. That exponential factors into one exponential per variable, so the answer is the cube of a single integral worth e minus 1.

f(x,y,z) = e^{x+y+z}

\int_0^1 \int_0^1 \int_0^1 e^{x+y+z} \,dz\,dy\,dx = (e - 1)^3

Numerically 5.073214, confirmed by quadrature on every build.

A sum in the exponent is a product of exponentials

The integrand looks tangled because all three variables sit in one exponent, but the exponent law untangles it before any calculus happens.

e^{x+y+z} = e^x \, e^y \, e^z

Now it is a product of one function per variable over a box, so the triple integral is a product of three single integrals, and all three are the same.

\left(\int_0^1 e^x \,dx\right)\left(\int_0^1 e^y \,dy\right)\left(\int_0^1 e^z \,dz\right) = (e - 1)^3

Each factor is $\left[e^t\right]_0^1 = e - 1 \approx 1.718282$ , and cubing that gives about $5.073214$ .

Doing it the long way, as a check

The iterated version reaches the same place. Integrating in $z$ first, $e^{x+y}$ comes out as a constant.

\int_0^1 e^{x+y+z} \,dz = e^{x+y}\left[e^{z}\right]_0^1 = (e - 1)\,e^{x+y}

The constant $(e - 1)$ now rides along through the $y$ integral and then the $x$ integral, each of which contributes another $(e - 1)$ .

(e-1)\int_0^1 \int_0^1 e^{x+y} \,dy\,dx = (e-1)\cdot(e-1)\cdot(e-1)

The mistake: writing e^3 - 1 instead

Students who spot that the exponent runs from $0$ to $3$ across the cube sometimes write $e^3 - 1 \approx 19.086$ . That is the answer to a single integral of $e^u$ from $0$ to $3$ , not a triple integral, and it is nearly four times too big.

The difference is that $u = x + y + z$ is not a change of variable you can make for free. Substituting it would require the correct Jacobian and new limits describing how much of the cube sits at each value of $u$ , which is a genuinely different and harder computation.

A quick bound settles it. On the unit cube $e^{x+y+z}$ ranges from $e^0 = 1$ to $e^3 \approx 20.09$ , and the volume is $1$ , so the integral must lie between $1$ and $20.09$ . Both candidates pass that test, so use the factoring argument rather than the size check to decide.

Frequently asked questions

Is (e - 1)^3 the same as e^3 - 1?

No. Expanding gives $(e-1)^3 = e^3 - 3e^2 + 3e - 1 \approx 5.073214$ , while $e^3 - 1 \approx 19.086$ . The cube of a difference keeps the cross terms, and those cross terms are most of the gap.

What is the average value of e^(x+y+z) on the cube?

The cube has volume $1$ , so the average value equals the integral itself, about $5.073214$ . That sits well above the midpoint value $e^{1.5} \approx 4.4817$ , which is what convexity of the exponential predicts.