# E^x^3 > 0

1. Sep 9, 2006

### 413

http://i5.tinypic.com/4hionb9.jpg

a) I know that e^x^3 > 0 and for all x ∈ [-1, 1], so we have ∫[-1 to 1] e^x^3 dx > 0. But how would you explain in words?

b) For all x ∈ [0, 1] , 0 ≤ x^2 ≤ x , so e^0 ≤ e^x^2 ≤ e^x
Therefore
∫[0 to 1] e^0 dx ≤ ∫[0 to 1] e^x^2 dx ≤ ∫[0 to 1] e^x dx
where do i go from here?

thanks.

2. Sep 10, 2006

### HallsofIvy

One definition of the definite integral of such a function (positive on the interval) is that it is the "area under the graph". And area is always positive.

A more rigorous proof is this: f(0)= 1 and the function is continuous so there exist some neighborhod of x= 0, say $-\delta< x< \delta$ such that f(x)> 1/2. We must have $\int_{-\delta}^{\delta}f(x)dx> (2\delta)(1/2)= \delta> 0$. Since f(x) is never negative, other parts of the integral cannot cancel that: $\int_{-1}^1 e^{x^3}dx> \delta> 0. Well, you know that e0= 1 so [itex]\int_0^1 e^0dx= \int_0^1 dx= 1$.
You also know that $\int e^x dx= e^x$ so $\int_0^1 e^x dx= e^1- e^0= e- 1$
What does that tell you about $\int_0^1 e^{x^2}dx$?

3. Sep 10, 2006

### 413

then like this?...

[x] [0 to 1] ≤ ∫[0 to 1] e^x^2 dx ≤ [e^x] [0 to 1]
1 ≤ ∫[0 to 1] e^x^2 dx ≤ e - 1
Since 0 < 1 and e - 1 < 3 so we must have 0 < ∫[0 to 1] e^x^2 dx < 3.

4. Sep 10, 2006

### 413

and also where do all those numbers come from?...8712,08804,8804?