I'm trying to solve this:
Which of the following is the 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.
I've found a lower bound by manually calculating $\int_{0}^{1} \sqrt{1+\frac{1}{3}} \ dx \approx 1.1547$. This eliminates option (A). I also see no reason why the integral shouldn't converge. However, to pick an option out of (B), (C) and (D) I need to find an upper bound too. Ideas? Please note that I'm not supposed to use a calculator to solve this.
From GRE problem sets by UChicago
Best Answer
$$\int_0^1\sqrt{1+\dfrac1{3x}}dx=2\int_0^1\sqrt{t^2+\dfrac13}dt$$ proves convergence.
Then
$$\frac1{\sqrt 3}\le\sqrt{t^2+\frac13}\le t+\frac1{\sqrt3}$$ implies
$$\frac2{\sqrt 3}\approx 1.155\le I\le1+\frac2{\sqrt 3}\approx2.155$$
A tighter upper bound is obtained by noting that the function is convex and
$$\sqrt{t^2+\frac13}\le \frac1{\sqrt3}+t\left(\sqrt{\frac 43}-\frac1{\sqrt3}\right),$$ giving $$I\le\sqrt3\approx1.732$$ A tighter lower bound could be found by considering the tangents at both endpoints up to their intersection, but we can already conclude C.
The exact value is $$1.5936865\cdots$$ The bounds can be computed by hand, by squaring to avoid square roots.