How do I calculate $\int_1^5\int_0^3\int_0^2 xy^2e^{xyz}\,dx\,dy\,dz$?

Question

$$\int_1^5\int_0^3\int_0^2 xy^2e^{xyz}\,dx\,dy\,dz$$

Can I count so?

$$\int_0^1 xdx\int_0^3 y^2dy\int_1^5 e^{xyz}dz$$

Or, how can I pull $(x)$ and $(y)$ out of degree $(e^{xyz})$?

Answer 1

Your approach is valid, but the $z$ integral is a function of $x$ and $y$ and must be included as a factor in the integrand of the $y$ integral, leaving a function of $x$ to include as a factor in the $x$ integral. In other words, $$\int_0^1dx\left[x\int_0^3dyy^2\left[\int_1^5dze^{xyz}\right]\right]=\int_0^1dx\left[\int_0^3dyy(e^{5xy}-e^{xy})\right]\\=\int_0^1\frac{(25-75x)e^{3x}+(15x-1)e^{15x}-24}{25x^2}dx,$$which looks demanding. So let's try a different tactic. Define $f(x):=\sum_{n\ge3}\frac{x^n}{n!}=e^x-1-x-\frac12x^2$ so the triple integral is$$\sum_{n\ge0}\frac{1}{n!}\int_0^1x^{n+1}dx\int_0^3y^{n+2}dy\int_1^5z^ndz=\sum_{n\ge0}\frac{3^{n+3}(5^{n+1}-1)}{(n+3)!}\\=\frac{f(15)}{25}-f(3)=\frac{e^{15}-128.5}{25}-(e^3+8.5)=0.04e^{15}-e^3-13.64.$$You'll want to double-check my arithmetic, though.

Answer 2

Order the triple integral below to integrate $z$ first, then $x$, and lastly $y$, $$\int_1^5\int_0^3\int_0^2 xy^2e^{xyz}\,dx\,dy\,dz=\int_0^3 dy\int_0^2 dx\int_1^5 xy^2e^{xyz}dz$$ $$=\int_0^3 dy\int_0^2 dx \int_1^5 yd(e^{xyz})$$ $$=\int_0^3dy\int_0^2 y(e^{5xy}-e^{xy})dx$$ $$=\int_0^3dy\int_0^2 \left[\frac15d(e^{5xy})-d(e^{xy})\right]$$ $$=\int_0^3 \frac15(e^{10y}-1)dy-\int_0^3 (e^{2y}-1)dy$$