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Problem of the Week #41 - January 7th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW!! Here's this week's problem (and the first University POTW of 2013)!

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Problem: Let $G$ be a rectangular box defined by the inequalities $a\leq x\leq b$, $c\leq y\leq d$, $k\leq z\leq l$. Show that
\[\iiint\limits_G f(x)g(y)h(z)\,dV = \left[\int_a^b f(x)\,dx\right]\left[\int_c^d g(y)\,dy\right]\left[\int_k^l h(z)\,dz\right],\]
where $\,dV=\,dx\,dy\,dz$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by Deveno and Sudharaka. You can find Deveno's solution below:

Let $H$ be an anti-derivative of $h$, $G$ be an anti-derivative of $g$ and $F$ an anti-derivative of $f$. I will call the region of integration $R$ to avoid confusion. $f,g,h$ will be assumed to be integrable over the appropriate intervals.Then:
$$\iiint\limits_R f(x)g(y)h(z)\ dV = \int_a^b \left(\int_c^d \int_k^l f(x)g(y)h(z)\ dy\ dz \right)\ dx$$
$$= \int_a^b \left( \int_c^d \left( \int_k^l f(x)g(y)h(z)\ dz \right)\ dy \right)\ dx = \int_a^b \left( \int_c^d f(x)g(y) \left( \int_k^l h(z)\ dz\right)\ dy \right)\ dx$$
$$= \int_a^b \left(\int_c^d [H(l) - H(k)]f(x)g(y)\ dy\right)\ dx = [H(l) - H(k)]\int_a^b \left(\int_c^d f(x)g(y)\ dy\right)\ dx$$
$$=[H(l) - H(k)]\int_a^b f(x) \left(\int_c^d g(y)\ dy \right)\ dx = [H(l) - H(k)]\int_a^b [G( d) - G(c)]f(x)\ dx$$
$$= [G( d) - G(c)][H(l)-H(k)]\int_a^b f(x)\ dx = [F(b)-F(a)][G( d) - G(c)][H(l) - H(k)]$$
$$= \left[\int_a^b f(x)\ dx\right]\left[\int_c^d g(y)\ dy\right]\left[\int_k^l h(z)\ dz\right]$$
 
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