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Find the interval on which the curve of y equals the integral from 0 to x of 2/(1+3t+t^2), dt is concave up.


How do I do this problem? I took notes on the lesson my course is telling me its from, but I have no idea how to go about it.

Verified Answer :

LammettHash

By the fundamental theorem of calculus, the first derivative of

\(y=\displaystyle\int_0^x\frac2{1+3t+t^2}\,\mathrm dt\)

is

\(y'=\displaystyle\frac{\mathrm d}{\mathrm dx}\int_0^x\frac2{1+3t+t^2}\,\mathrm dt=\frac2{1+3x+x^2}\)

Then the second derivative is

\(y''=\dfrac{(1+3x+x^2)(0)-2(3+2x)}{(1+3x+x^2)^2}=-\dfrac{2(2x+3)}{(1+3x+x^2)^2}\)

From here you can proceed with determining the concavity of \(y\) by finding the possible inflection points (when \(y''=0\) or undefined). Since the denominator is positive for all \(x\), you know that \(y''\) is defined for all \(x\). So you're left with solving

\(y''=-\dfrac{2(2x+3)}{(1+3x+x^2)^2}=0\)
\(\implies 2x+3=0\)
\(\implies x=-\dfrac32\)

So you have two intervals to consider, \(\left(-\infty,-\dfrac32\right)\) and \(\left(-\dfrac32,\infty\right)\). Check the sign of \(y''\) in both of these intervals; if \(y''>0\) on some interval, then \(y\) is concave upward on that interval.

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