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Oxana [17]
4 years ago
9

Evaluate the triple integral. t 5x2 dv, where t is the solid tetrahedron with verticies (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0,

0, 1)
Mathematics
1 answer:
pshichka [43]4 years ago
5 0
The face of the tetrahedron in the first octant, not coinciding with any of the planes generated by the coordinate axes, is given by the plane x+y+z=0. So the integral can be set up as

\displaystyle\iiint_{\mathcal T}5x^2\,\mathrm dV=\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}\int_{z=0}^{z=1-x-y}5x^2\,\mathrm dz\,\mathrm dy\,\mathrm dx=\frac1{12}
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So, there is something known as a removable discontinuity, and it's essentially where you can define f(x) using the most simplified fraction, where you could normally not define f(x).

So we have the following equation:

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

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Original Equation

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

Keep, change, flip

f(x) = (\frac{x+5}{x+1}*\frac{(x-4)(x+1)}{(x+3)(x-2)})-\frac{1}{x-2}

Multiply the two fractions

f(x) = (\frac{(x+5)(x-4)(x+1)}{(x+1)(x+3)(x-2)})-\frac{1}{x-2}

Notice how the x+1 is in the numerator and fraction? That means we can cancel it out!

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In this simplified version of the fraction, we can technically define f(-1), but in the original version, since it's not defined there is a removable discontinuity at x=-1, meaning there is no vertical asymptote, but the function is still not defined at f(-1), and there will be a hole at that point.

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