The closest value to $\int_0^1 \sqrt{1+\frac{1}{3x}} dx$? This is a multiple choice among 1.6, 2, 1.2. So the approximation should be sufficiently accurate. The solution is 1.6 as can be verified using Taylor expansion. But Taylor expansion method takes too long and this is supposed to be answered quickly. I wonder what alternatives there are to approx integral question like this. Thanks!
 A: Let $f(x)$ be the integrand in question. Then $f^{-1}(x)=\frac1{3(x^2-1)}$ which intersects the line $y=1$ at $\left(\frac2{\sqrt3},1\right)$. The integral is thus equivalent to $$I=\int_{\frac2{\sqrt3}}^\infty\frac1{3(x^2-1)}\,dx+\frac2{\sqrt3}=\lim_{n\to\infty}\left[\frac13\cdot\frac12\ln\frac{1-x}{x+1}\right]_{2/\sqrt3}^\infty+\frac2{\sqrt3}=\frac2{\sqrt3}-\frac{\ln(7-4\sqrt3)}6$$ where the last term comes from the area of the rectangle of length $\frac2{\sqrt3}$ and height $1$. Using $1.74>\sqrt3>1.73$ we get $$\frac{200}{174}-\frac{\ln0.04}6>I>\frac{200}{173}-\frac{\ln0.08}6\implies \frac{100}{87}+\frac{\ln25}6>I>\frac{200}{173}+\frac{\ln12.5}6$$ and this rules $1.2$ and $2$ out as $e^2<3^2=9$ and similarly for $e^3$. Therefore $1.6$ is the correct answer.
A: First thought - standard numerical integration techniques. But then, we have a problem - the integrand $\sqrt{1+\frac1{3x}}$ blows up to $\infty$ at zero. It's a fairly weak singularity, but we need to neutralize it before numerical methods will have a chance. To do that, substitute $x=t^2, dx=2t\,dt$:
$$\int_0^1 \sqrt{1+\frac1{3x}}\,dx = \int_0^1 2t\sqrt{1+\frac1{3t^2}}\,dt = \int_0^1 \sqrt{4t^2+\frac43}\,dt$$
Now we can apply standard numerical integration techniques. Let $f(t)=\sqrt{4t^2+\frac43}$. We have $f(0)=\sqrt{\frac43}=\frac{2\sqrt{3}}{3}\approx 1.15$, $f(1)=\sqrt{\frac{16}{3}}=\frac{4\sqrt{3}}{3}\approx 2.31$, and $f\left(\frac12\right)=\sqrt{\frac73}=\frac{\sqrt{21}}{3}\approx 1.53$. That gets us a trapezoid rule estimate of about $1.73$ and a midpoint rule estimate of about $1.53$, so the true integral should be between them. Weighting them as in Simpson's rule, we get an estimate for the integral of $1.60$. Right on.
That wasn't quite all in my head - I've got enough digits of $\sqrt{3}$ memorized for $f(0)$ and $f(1)$, but $f\left(\frac12\right)$ called for a calculator-equivalent. I could have done it with some paper, or settled for $1.5$ from my head.
In any case, the real key is that substitution to neutralize the singularity. Once we've done that, pretty much anything will work.
Oh, and it is possible to find the integral exactly. The exact form doesn't make it easier to find a good estimate.
