Multivariable calculus

Triple Integral of y ln(1 + x) over a Box

Q: Is log(1 + x) here base 10 or base e?

Base e. In this section \ln and the machine notation \log both mean the natural logarithm, which is the convention in mathjs and most computer algebra systems. A base 10 reading would scale the answer by about 0.434.

Q: Why is the box chosen to keep x at or above 0?

Because \ln(1 + x) is undefined at x = -1 and for anything below it. Keeping x in [0, 1] leaves the integrand smooth and bounded, so no improper integral or singularity handling is needed.

The triple integral of y times the natural log of 1 + x, over the box where x runs 0 to 1, y runs 0 to 2 and z runs 0 to 3, equals 12 ln 2 minus 6, about 2.317766. The log factor needs integration by parts and gives 2 ln 2 minus 1, then the y and z edges contribute 2 and 3.

f(x,y,z) = y\ln(1 + x)

\int_0^1 \int_0^2 \int_0^3 y\ln(1 + x) \,dz\,dy\,dx = 12\ln 2 - 6

Numerically 2.317766, confirmed by quadrature on every build.

Separate, then spend the effort on one factor

Only the $x$ factor is hard, so isolate it before doing any work. The $z$ integral contributes the edge length $3$ and the $y$ integral contributes $\int_0^2 y \,dy = 2$ .

\iiint_B y\ln(1 + x) \,dV = \left(\int_0^1 \ln(1 + x) \,dx\right)\left(\int_0^2 y \,dy\right)(3) = 6\int_0^1 \ln(1 + x) \,dx

Now do the log by parts with $u = \ln(1 + x)$ and $dv = dx$ , so $du = \frac{dx}{1 + x}$ and $v = 1 + x$ . Choosing $v = 1 + x$ rather than $v = x$ makes the remaining integral collapse to a constant.

\int_0^1 \ln(1 + x) \,dx = \Big[(1 + x)\ln(1 + x) - (1 + x)\Big]_0^1 = (2\ln 2 - 2) - (0 - 1) = 2\ln 2 - 1

Multiplying by $6$ gives $12\ln 2 - 6 \approx 2.317766$ .

The mistake: guessing the antiderivative of a logarithm

The most common wrong move is writing $\int \ln(1 + x) \,dx = \frac{1}{1 + x}$ , confusing the antiderivative with the derivative. That gives a negative contribution and an answer near $-3$ , impossible for an integrand that is non negative on this box.

The second wrong move is $\int \ln(1 + x) \,dx = \frac{\left(\ln(1+x)\right)^2}{2}$ , borrowing the power rule for something that is not a power. Only $\int \frac{\ln(1+x)}{1+x} \,dx$ has that shape, because there the inner derivative is present.

Correct: $\int \ln(1 + x) \,dx = (1 + x)\ln(1 + x) - x + C$ .
Differentiate it back: $\ln(1 + x) + 1 - 1 = \ln(1 + x)$ , so the answer checks out.
The two constant forms $-(1+x)$ and $-x$ differ by $1$ , which cancels once limits are substituted.

Sanity check the size

On this box $\ln(1 + x)$ runs from $0$ to $\ln 2 \approx 0.693$ and $y$ runs from $0$ to $2$ , so the integrand never exceeds about $1.386$ . With volume $6$ the integral is at most about $8.3$ , and it should be far less because both factors vanish along two faces.

The stated value $2.317766$ is about $28$ percent of that crude bound, which is what you expect from a product of two factors that each average roughly half their maximum.

Frequently asked questions

Is log(1 + x) here base 10 or base e?

Base $e$ . In this section $\ln$ and the machine notation $\log$ both mean the natural logarithm, which is the convention in mathjs and most computer algebra systems. A base $10$ reading would scale the answer by about $0.434$ .

Why is the box chosen to keep x at or above 0?

Because $\ln(1 + x)$ is undefined at $x = -1$ and for anything below it. Keeping $x$ in $[0, 1]$ leaves the integrand smooth and bounded, so no improper integral or singularity handling is needed.