Calculus of several variables

Question

I have problem in solving the definite integral of the following function:

$$f(x, y, z) = \int_0^z \int_0^y \int_0^x e^{-x-y-z} dx dy dz, $$

All I know is the calculus for single variable. But with the addition of another variables, things get complicated.

Answer 1

As mentioned, it is sort of "bad manners" to use a variable both as a dummy variable and as a bound. So your integral... $$f(x, y, z) = \int_0^z \int_0^y \int_0^x e^{-x-y-z} \,dx \,dy \,dz$$ might be better written as... $$f(x, y, z) = \int_0^z \int_0^y \int_0^x e^{-u-v-w} \,du \,dv \,dw$$

Now that said, this is an iterated integral meaning you tackle each integral, working from the inside outward, one at a time. First, with respect to $u$, then $v$, then $w$.

So first $\displaystyle \int_0^x e^{-u-v-w} \,du$. Treating $v$ and $w$ as constants, this is essentially like integrating something like $\int e^{-u+5}\,du$. So the antiderivative is $-e^{-u+5}$ or in your case $-e^{-u-v-w}$. Thus $$ \int_0^x e^{-u-v-w} du = -e^{-u-v-w} \Bigg|_0^x = -e^{-x-v-w}-(-e^{0-v-w})$$ $$= e^{-v-w}-e^{-x-v-w} = (1-e^{-x})e^{-v-w}$$

This means that $$f(x, y, z) = \int_0^z \int_0^y \int_0^x e^{-u-v-w} \,du \,dv \,dw = \int_0^z \int_0^y (1-e^{-x})e^{-v-w} \,dv \,dw$$

We can then integrate with respect to $v$ and then $w$. These steps go essentially the same way so in the end we get $$f(x, y, z) = \int_0^z \int_0^y \int_0^x e^{-u-v-w} \,du \,dv \,dw = (1-e^{-x})(1-e^{-y})(1-e^{-z})$$

Alternatively you can notice that the bounds of integration are independent from each other. So each can be treated as a constant. Because of this, this particular iterated integral can be "factored"... $$f(x, y, z) = \int_0^z \int_0^y \int_0^x e^{-u-v-w} \,du \,dv \,dw = \int_0^z \int_0^y \int_0^x e^{-u}e^{-v}e^{-w} \,du \,dv \,dw$$ $$ = \int_0^z e^{-w} \int_0^y e^{-v} \int_0^x e^{-u} \,du \,dv \,dw =\left(\int_0^x e^{-u} \,du\right) \left(\int_0^y e^{-v} \,dv\right) \left(\int_0^z e^{-w} \,dw\right) $$ $$= (1-e^{-x})(1-e^{-y})(1-e^{-z})$$

Answer 2

$$f(x, y, z) = \int_0^z \left(\int_0^y\left( \int_0^x e^{-x'-y'-z'} dx'\right) dy'\right) dz' \\= \int_0^z \left(\int_0^y [-e^{-x'-y'-z'}]_{x'=0}^x dy'\right) dz' \\=\int_0^z \left(\int_0^y (1-e^{-x})e^{-y'-z'}dy'\right) dz' \\=\int_0^z \left((1-e^{-x})[-e^{-z'-y'}]_{y'=0}^y \right) dz' \\= \int_0^z (1-e^{-x})(1-e^{-y})e^{-z'} dz'= (1-e^{-x})(1-e^{-y})(1-e^{-z}) $$