Understanding integral de exponencialelevado a x requires examining multiple perspectives and considerations. calculus - Is there really no way to integrate $e^ {-x^2 .... @user599310, I am going to attempt some pseudo math to show it: $$ I^2 = \int e^-x^2 dx \times \int e^-x^2 dx = Area \times Area = Area^2$$ We can replace one x, with a dummy variable, move the dummy copy into the first integral to get a double integral. $$ I^2 = \int \int e^ {-x^2-y^2} dA $$ In context, the integrand a function that returns ... solving the integral of $e^ {x^2}$ - Mathematics Stack Exchange. The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions. It's important to note that, for example, you can express $\int x^2 \mathrm {d}x$ in elementary functions such as $\frac {x^3} {3} +C$.
In this context, however, the indefinite integral from $ (-\infty,\infty)$ does exist and it is $\sqrt {\pi}$ so explicitly: $$\int^ {\infty}_ {-\infty} e^ {-x^2} = \sqrt {\pi}$$ Note ... What is the integral of 1/x? - Mathematics Stack Exchange. Answers to the question of the integral of $\frac {1} {x}$ are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers. What is the difference between an indefinite integral and an ....
Wolfram Mathworld says that an indefinite integral is "also called an antiderivative". Furthermore, this MIT page says, "The more common name for the antiderivative is the indefinite integral." One is free to define terms as you like, but it looks like at least some (and possibly most) credible sources define them to be exactly the same thing. What does the dx mean in an integral?
I know dy/dx for example means "derivative of y with respect to x," but there's another context that confuses me. You will generally just see a dx term sitting at the end of an integral equation an... Building on this, how to calculate the integral in normal distribution?. If by integral you mean the cumulative distribution function $\Phi (x)$ mentioned in the comments by the OP, then your assertion is incorrect. This perspective suggests that, how do I integrate $\\sec(x)$?
My HW asks me to integrate $\sin (x)$, $\cos (x)$, $\tan (x)$, but when I get to $\sec (x)$, I'm stuck. In relation to this, the integral of 0 is C, because the derivative of C is zero. Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because any function f (x)=C will have a slope of zero at point on the function. What does it mean for an "integral" to be convergent?.
The improper integral $\int_a^\infty f (x) \, dx$ is called convergent if the corresponding limit exists and divergent if the limit does not exist. While I can understand this intuitively, I have an issue with saying that the mathematical object we defined as improper integrals is "convergent" or "divergent". Moreover, differentiating Definite Integral - Mathematics Stack Exchange.
For an integral of the form $$\tag {1}\int_a^ {g (x)} f (t)\,dt,$$ you would find the derivative using the chain rule. As stated above, the basic differentiation rule for integrals is:
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