Domination property of integrals

[mathkillsalot]

Homework Statement

prove that

2√2 <= ∫(from 0 to 1) (√x+8) dx <= 3

Homework Equations

The Attempt at a Solution

well...my only idea on how to solve this would be to evaluate the middle term, but my prof says it's not allowed. Do I just assign functions to the left and right numbers??

if yes, do I just grab them from out of nowhere?

please help...don't know where to start

 

[Mark44]

I can't tell what your integral is. This is what you wrote:
$$\int_0^1 (\sqrt{x} + 8)~dx$$

Did you mean this?
$$\int_0^1 \sqrt{x + 8}~dx$$

 

[Mark44]

Whichever integral it is, the integrand is an increasing function, so its smallest function value will be at the left endpoint of the interval, and the largest function value will be at the right endpoint.

I would start by approximating the integral by using two rectangles, one whose area is smaller than that under the integrand function, and one whose area is larger than that under the integrand functions.

 

[LCKurtz]

Think about this property of integrals:

If m ≤ f(x) ≤ M on [a,b] then

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

 

[mathkillsalot]

I can't tell what your integral is. This is what you wrote:
$$\int_0^1 (\sqrt{x} + 8)~dx$$

Did you mean this?
$$\int_0^1 \sqrt{x + 8}~dx$$

I meant the second one x+8 under the sqrt sign

okaaay. Wait, I'll try to use the solution you suggested. Not quite sure if I can do it, thanks for the help

 

[mathkillsalot]

Think about this property of integrals:

If m ≤ f(x) ≤ M on [a,b] then

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

hmm...I'm sorry, I don't really get how that could help, could you please elaborate?? I'm not really good at this

 

[dextercioby]

That's a theorem which places the value of the definite integral (which has the significance of an area under the graph of a continuous function on the interval [a,b]) between the areas of 2 rectangles.

To use it, you must find m and M. Who are they equal to ?