show that 2 ≤ ∫ √(1+x^3)dx ≤ 6 from (0,2)?

Answer 1

Check this out... pretty cool. Not only a valid proof, but I can even narrow down the limits of the inequality. I think this is a fun approach.

One of the first things I recognized was the form it took. I dont know if you know this yet, but at least you will now.

I realized that this was a curve-length function for f(x) = (2/5)⋅x^(5/2)

The length of any continuously differentiable function, f(x), within any interval, [a,b] can be determined with this formula:
∫[a,b] √[ 1 + (∂/∂x f(x) )² ] ∂x

Well, if f(x) = (2/5)⋅x^(5/2), then the length function can be simplified into
∫ √[ 1+x³ ] ∂x
As your question is.

So, the question can be reworded as:
Show that the distance of the curve f(x) = (2/5)⋅x^(5/2) between the interval (0,2) is ≥2 but ≤ 6.

This is pretty easy to do if you consider the geometry of the function.

First, using f(x), compute the starting and ending coordinates of the function at the interval end-points.

(0,0) and (2, 8/5·√2)

The shortest distance between these two points is the direct line connecting them... using the generic distance formula, based off of the Pythagorean Theorem:
d = √[ [ f(x₁) - f(x₂) ]² + [ x₁ - x₂ ]² ]...

This distance turns out to be ≈ 3.0199...

The greatest distance is the sum of the horizontal and vertical change in the points.

This turns out to be ≈ 4.2627...

This particular curve, taking the form of cx^n, cannot be longer than or shorter than these limits

Therefore, the given problem, whatever it represents, also contains these limits

Answer 2

If 0 <= x, 1+x^3 >= 1+0 = 1
Thus ∫ √(1+x^3)dx <= ∫ √1 dx =∫ 1 dx = 2*1 - 0*1 = 2 (from 0 to 2)

If x <=2, 1 + x^3 <= 1 + 2^3 = 9
Thus ∫ √(1+x^3)dx >= ∫ √(1+2^3)dx = ∫ √9 dx = ∫ 3 dx =
= 2*3 - 2*0 = 6

Answer 3

√(1+x^3) is increasing from 0 to 2
∫ √(1+0^3)dx ≤ ∫ √(1+x^3)dx ≤ ∫ √(1+2^3)dx, x from 0 to 2
=> 2 ≤ ∫ √(1+x^3)dx ≤ 6