Use the properties of definite integrals to answer the following question. Show that: Integral b = 2, a = -1 (e^(-x)^2) dx >= 0.

Question

Not sure how to do the problem at all. Thanks.

Answer 1

The integral is the same as area under the curve.  If we evaluate the function at the bounds and examin the results, you can predict where the shaded portion of the curve lies.

 

f(-1) = e^(-(-1)²) = e^-1 = 1 / e

f(2) = e^(-(2)²) = e^-4 = 1 / e⁴

 

Now we evaluate at a point between the boundaries x=-1 and x=2.

 

f(0) = e^(-(0)²) = 1

 

The results are positive.   Therefore, the area under the curve at the boundaries is greater the zero.

Answer 2

The integrand is never negative and is defined for all Reals; therefore any definite integral in which a<b will evaluate to some positive value.

Answer 3

∫-_1² e^(-x²)dx>=0.

 

One of the simple (of naive) explanation is as following.

 Function e^-(x)^2 is positive and monotonic, it is symmetric w.r.t y axis and has max at x=0. The original function (or so called, indefinite integral F(x) of it) is monotonic and increasing. So one can state that F(2) >=F(-1). Hence the given integral cannot be negative.

 

This function under the integral is, basically the density function of the normal distribution which integral over the whole real line sums up to 1.

 

I know other ways explanations or strictly mathematical proof are available but for now I think this explanation should be satisfactory.