Evaluate the triple integral of 4x + 5y - 19z with z from 7 to 1, y from 4 to 2, and x from 3 to 2.

Step-by-step solution

1. 1

   Notice that every limit is reversed. Each integral runs from a larger value to a smaller one. Since $\int_{a}^{b}=-\int_{b}^{a}$, there are three sign flips, and $(-1)^{3}=-1$: the whole answer is the negative of the same integral with all limits in ascending order. You may either flip them all up front or, as below, just evaluate carefully in place.

2. 2

   Integrate with respect to $z$ first, treating $x$ and $y$ as constants. An antiderivative in $z$ is $4xz+5yz-\tfrac{19}{2}z^{2}$, so

   $\int_{7}^{1}\left(4x+5y-19z\right)dz=\left(4x+5y-\tfrac{19}{2}\right)-\left(28x+35y-\tfrac{931}{2}\right)=-24x-30y+456$

   The order matters: the innermost differential $dz$ is integrated first, using the innermost pair of limits.

3. 3

   Integrate the result with respect to $y$. With antiderivative $(-24x+456)y-15y^{2}$:

   $\int_{4}^{2}\left(-24x-30y+456\right)dy=\left(-48x+852\right)-\left(-96x+1584\right)=48x-732$

4. 4

   Integrate with respect to $x$. With antiderivative $24x^{2}-732x$:

   $\int_{3}^{2}\left(48x-732\right)dx=\left(96-1464\right)-\left(216-2196\right)=-1368+1980=612$

5. 5

   Interpret the sign. The integrand is dominated by $-19z$ over a region where $z$ runs up to 7, so the "ascending-limits" integral is strongly negative; the three reversals turn it positive. Reversing only the $z$ limits, for instance, would give $-612$.

6. 6

   Check with a numerical Riemann sum. A $120^{3}$ midpoint sum using the signed step sizes $dz=-6/120$, $dy=-2/120$, $dx=-1/120$ returns exactly $612.0$ ✓. The agreement is exact because the integrand is linear, so the midpoint rule is exact.