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$$F(x') = \int_{1}^{x'}(1+x)e^{1-x^2}\ \text{d}x$$ Hello everyone, I'm trying to study an integral fuction, so I can plot the most of it, without integrate it. I know that,

  • For $x->+\infty$ the function F(x') converges to a positive value (f(x) is positive and the range is growing)
  • For $x->-\infty$ the function F(x') converges again to a positive value (f(x) is not positive, but the range is decreasing so the value is >0)

I don't care about other points (like x=1), but they ask me to know the minima and maxima, both local. I know that the first derivate of F(x') is f(x), the integrand function, and it is $f(x)=0$ just when $x=-1$, that is clearly a minima. But, I know for a fact that in the graph I have to draw the F(-1) is under the x-axis. I really don't understand why. I know that the interval is decreasing (from 1 to -1), but how can I know that the function too have to be positive? I know that decreasing areas are negative, but only if these areas are positive! I don't undertand this step, made without integrating the function.

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    Your description is not really clear. Is the variable of $F$ the upperbound of the integral ? In the case you should write the integrand with another variable and explicitly specify the $dx$ or $dt$ term which, for once, makes things clearer ! – 2017-01-24

1 Answers 1

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Let

$$F(t):= \int_{1}^{t}(1+x)e^{1-x^2}\ \text{d}x$$

Then

$$\frac{dF(t)}{dt}=(1+t)e^{1-t^2},$$ which only cancels for $t=-1$.

The function value at this minimum is

$$F(-1)=\int_{1}^{-1}(1+x)e^{1-x^2}\ \text{d}x$$ which can be expressed analytically using the error function. This is clearly a negative value, as the integrand is positive in the given range.