Use the properties of integrals to verify that
2 <= ∫ -1 to 1 (1 + x^2) ^ 1/2 dx <= 2(2) ^ 1/2
$\displaystyle f(0) = 1$ = minimum value of $\displaystyle y = \sqrt{x^2+1}$ on $\displaystyle [-1,1]$
$\displaystyle f(1) = f(-1) = \sqrt{2}$ = maximum value of $\displaystyle y = \sqrt{x^2+1}$ on $\displaystyle [-1,1]$
$\displaystyle f(0)[1 - (-1)] < \int_{-1}^{1} \sqrt{x^2+1} \, dx < f(1)[1 - (-1]
$
$\displaystyle 2 < \int_{-1}^{1} \sqrt{x^2+1} \, dx < 2\sqrt{2}
$
Upper and lower bounds. An integrable function f on [a, b], is necessarily bounded on that interval. Thus there are real numbers m and M so that m ≤ f (x) ≤ M for all x in [a, b]. Since the lower and upper sums of f over [a, b] are therefore bounded by, respectively, m(b − a) and M(b − a), it follows that
$\displaystyle m(b-a) \le \int_a^b f(x) \, dx \le M(b-a)$
if you require further explanation, see your instructor.