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Problem Set 4, Number 3. Dan Heflin. The Problem at Hand. Well, to begin, let us try and find an a that fits the description we need. a = ?. So, these curves intersect at every value for which , and . But, we will look at , since it is the smaller value.
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Problem Set 4, Number 3 Dan Heflin
The Problem at Hand Well, to begin, let us try and find an a that fits the description we need.
a = ? • So, these curves intersect at every value for which , and . But, we will look at , since it is the smaller value. • So, , since we need the smallest positive value where the curves next intersect after 0.
Now We Must Integrate! • Well, now that we know the top and bottom curve, all we must simply do is integrate.
Final Answer….. • After plugging in our values, we get • Mathematica comes up with the same answer!! =