Why you can't integrate this as it appears in the solution? And why this value is 0?

Question

I know what they do, but I want to understand why cannot use the silly way that is integrating e. Also, I don't understand why f(3) is 0, I mean, it would be close to 0, but not cero right?

Answer 1

You can't obtain the antiderivative of $e^{-x^2}$, as it requires a special function.

Now you are free to define

$$F'(x)=e^{-x^2}\leftrightarrow F(t):=\int e^{-x^2}dx+C$$ even if you don't have an analytical expression. From this, the definite integral

$$\int_t^3 e^{-x^2}dx=F(3)-F(t)$$

and the derivative

$$\frac d{dt}\int_t^3 e^{-x^2}dx=\frac d{dt}F(3)-\frac d{dt}F(t).$$

The rest is yours.

Answer 2

Yves Daoust is correct. This is just an alternate description.

Without knowing how to integrate f(x), another function, say, F(t), can be defined as being the result of that unknown integration: $F(t)\text{ is }\int f(x) \, dx+C_1$, where $C_1$ is some unknown constant.

The inverse operation of integration is differentiation.

Now, let us do the problem using that special "undefined" function.

$\frac{\partial \left(\left(C_1+F(3)\right)-\left(C_1+F(t)\right)\right)}{\partial t}$

$\frac{\partial (F(3)-F(t))}{\partial t}$

$\frac{\partial F(3)}{\partial t}-\frac{\partial F(t)}{\partial t}$

The first term is $0$ because the derivative of a constant is $0$.

$-\frac{\partial F(t)}{\partial t}$

Since integration and differentiation are inverse operations except that integration generates unknown constants (kind of like one may not divide by zero). This becomes:

$-e^{-x^2}$ evaluated for $x$ at $t$. Which gives the answer $-e^{-t^2}$.

In this case, the special function, with a normalization so that the area under the curve to the x-axis is $1$ is known as the error function and commonly written as: $\int e^{-x^2} \, dx=\frac{1}{2} \sqrt{\pi } \text{erf}(x)$. The curve is known as the Gaussian and the function and curve are commonly encountered in statistics.

I know $\LaTeX$ and MathJax did some rearrangement of my input. I regret that result.

Answer 3

I think the solution, though correct, is contributing to your confusion. You can do all this without even mentioning the $3$. One part of the fundamental theorem of calculus says that when you differentiate an integral with respect to it's upper limit you recover the integrand: $$ \frac{d}{dt} \int_c^t f(x)dx = f(t) . $$ This result does not depend on or mention the constant lower limit $c$.

You solve your problem by replacing $\int_t^3$ by $-\int_3^t$.