I am an undergraduate studying for the GRE.
I found this problem on a mock practice exam. It states the following.
"Which of the following is closest to the value of this integral:
$\int_0^1 \sqrt{1+\frac{1}{3x}} dx$
(A) 1
(B) 1.2
(C) 1.6
(D) 2
(E) The integral doesn't converge."
The correct answer turns out to be (C), and that the fastest way is to estimate from above and below. It seems to me like the best way to do a lot of GRE integration problems is by estimating in order to save time. The only problem is the integrand is unbounded on $(0,1)$, so I need a different approach and plugging in a upper bound. Any tips for this problem and any more tricks to estimate integrals?
Best Answer
You want the area sandwiched between $x=0, y=0, x=1$ and $ y=\sqrt{1+\frac{1}{3x}}$. So instead of integrating $0<x<1$, integrate $\sqrt{\frac{4}{3}}<y<\infty$ and add the rectangle below. So your answer is
$$1\times\sqrt{\frac{4}{3}} + \int_{\sqrt{\frac{4}{3}}}^{\infty}\frac{1}{3y^2-3}dy$$
The integral can be done with partial fractions and equals
$$\frac{-1}{6} \ln(7 - 4 \sqrt3)$$
To estimate these by hand, note $\frac{4}{3}\approx 1.3$, and since $(1+x)^2\approx 1+2x$ for small $x$, we have $\sqrt\frac{4}{3}\approx 1.15$.
Further, recall $\sqrt{3}\approx 1.73$. Hence $7-4\sqrt{3}\approx .08\approx\frac{1}{12}$. If we recall $e^2\approx7$ and $e^3\approx 20$, then $e^{2.5}\approx\sqrt{7*20}=\sqrt{140}\approx12$. So $\ln(7 - 4 \sqrt3)\approx-2.5$. So $\frac{-1}{6} \ln(7 - 4 \sqrt3)\approx.416$
Hence the total is $\approx1.15+.416\approx1.6$.