# Thread: Using properties of integrals

1. ## Using properties of integrals

Use the properties of integrals to verify that

2 <= ∫ -1 to 1 (1 + x^2) ^ 1/2 dx <= 2(2) ^ 1/2

2. Originally Posted by SyNtHeSiS
Use the properties of integrals to verify that

2 <= ∫ -1 to 1 (1 + x^2) ^ 1/2 dx <= 2(2) ^ 1/2
consider the areas of the upper and lower rectangles as shown in the graph

3. where are the lines y = 2 and y = 2(2) ^ 1/2? I know that the parabola is between the graph, but I dont know how to verify it by the properties of integrals

4. Originally Posted by SyNtHeSiS
where are the lines y = 2 and y = 2(2) ^ 1/2? I know that the parabola is between the graph, but I dont know how to verify it by the properties of integrals
not the lines ... the areas.

5. I am talking about the line y = 1.4 (which should be 2(2)^1/2) and and y = 1 (which should be 2) as the parabola is squeezed between them.

6. Originally Posted by SyNtHeSiS
I am talking about the line y = 1.4 (which should be 2(2)^1/2) and and y = 1 (which should be 2) as the parabola is squeezed between them.

$f(0) = 1$ = minimum value of $y = \sqrt{x^2+1}$ on $[-1,1]$

$f(1) = f(-1) = \sqrt{2}$ = maximum value of $y = \sqrt{x^2+1}$ on $[-1,1]$

$f(0)[1 - (-1)] < \int_{-1}^{1} \sqrt{x^2+1} \, dx < f(1)[1 - (-1]
$

$2 < \int_{-1}^{1} \sqrt{x^2+1} \, dx < 2\sqrt{2}
$

7. Originally Posted by skeeter
$f(0) = 1$ = minimum value of $y = \sqrt{x^2+1}$ on $[-1,1]$

$f(1) = f(-1) = \sqrt{2}$ = maximum value of $y = \sqrt{x^2+1}$ on $[-1,1]$
Sorry but why did you sub in 0 and 1 into f ?

8. Originally Posted by SyNtHeSiS
Sorry but why did you sub in 0 and 1 into f ?
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 mf (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(ba) and M(ba), it follows that

$m(b-a) \le \int_a^b f(x) \, dx \le M(b-a)$

if you require further explanation, see your instructor.