@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 ...
The integral is also known (less commonly) as the anti-derivative, because integration is the inverse of differentiation (loosely speaking). Integrals are indefinite when there are no bounds imposed, and the result is a family of functions (dependent on the variable of integration) and separated only by an arbitrary additive constant.
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:
This integral is one I can't solve. I have been trying to do it for the last two days, but can't get success. I can't do it by parts because the new integral thus formed will be even more difficult to solve. I can't find out any substitution that I can make in this integral to make it simpler. Please help me solve it.
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.
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.
I rolled back the previous edit of the title because the use of "primitive" to mean "indefinite integral" is not universally understood in the mathematical literature.
What is the integral of $$\int_ {-\infty}^ {\infty}e^ {-x^2/2}dx\,?$$ My working is here: = $-e^ (-1/2x^2)/x$ from negative infinity to infinity. What is the value of this?
