Asked by Mike
Find the area of the following shaded region, an unbounded region and demarcated by the equation of the curve
y = 10/(x^2 -10x +29) and the curve x=3
In the figure, the shaded area doesn't have bounds and when I plot the curves I can see that the first equation is clearly divergent towards +inf. Does it mean I can only give an approximate area?
y = 10/(x^2 -10x +29) and the curve x=3
In the figure, the shaded area doesn't have bounds and when I plot the curves I can see that the first equation is clearly divergent towards +inf. Does it mean I can only give an approximate area?
Answers
Answered by
oobleck
what is the shaded region, to the left or right of x=3? I assume the region is bounded by the x-axis and one of the asymptotes. Let's take the region to the left of x=3 and to the right of x = 5 - 3√6 = -2.3485
That makes the area
A = ∫[-2.3485,3] -10/(x^2-10x-29) dx
You can do that using partial fractions, and you get
∫ 10/(x^2-10x-29) dx = 5/3∫6 log (x-(5 + 3√6))/(x-(5 - 3√6))
That makes the area
A = ∫[-2.3485,3] -10/(x^2-10x-29) dx
You can do that using partial fractions, and you get
∫ 10/(x^2-10x-29) dx = 5/3∫6 log (x-(5 + 3√6))/(x-(5 - 3√6))
Answered by
oobleck
Hmmm. I just took a look at that integral, and it is infinite, because at either of the asymptotes you get log(0) or log(∞). It's just like trying to find the area under the curve y = 1/x^2 where the y-axis is one boundary.
I suspect you have left something out.
I suspect you have left something out.
Answered by
Mike
Yes sorry, the shaded region is to the right of x=3
Answered by
oobleck
yeah, well, the area is still unbounded.
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