@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, …

Nov 29, 2013 · Wolfram Mathworld says that an indefinite integral is "also called an antiderivative". This MIT page says, "The more common name for the antiderivative is the …

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.

Sep 27, 2013 · My HW asks me to integrate $\sin (x)$, $\cos (x)$, $\tan (x)$, but when I get to $\sec (x)$, I'm stuck.

If by integral you mean the cumulative distribution function $\Phi (x)$ mentioned in the comments by the OP, then your assertion is incorrect.

One possible interpretation: a "normal" integral is simply a line integral where the path is straight and oriented along a particular axis. Thus, as soon as you perform a transformation to the …

Feb 4, 2018 · 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 …

Nov 11, 2015 · The integral from $-\infty$ to $\infty$ is just twice this. So boom. If you want, you can rewrite $e^ {ix^2}=\cos (x^2)+i\sin (x^2)$ and equate the real and imaginary parts in the …

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:

Aug 9, 2017 · So, I can intuitively grasp that the derivative of the integral of a given function brings you back to that function. Is this also the case with the integral of the derivative?