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What is the Integral of 2^(x)? - Physics Forums The integral of 2^x with respect to x is (1 ln (2)) * 2^x + C, where C is the constant of integration This result is derived using the property that the derivative of a^x is ln (a) * a^x, allowing for the straightforward application of integration techniques
Solving Integrals for e^-ax^2: (i), (ii) (iii) - Physics Forums The discussion focuses on solving integrals of the form \ (\int_ {0}^ {\infty} e^ {-ax^2} x^n dx\) for \ (n = 2, 3, 4\) using differentiation and integration by parts The integral \ (\int_ {0}^ {\infty} e^ {-ax^2} dx\) is established as \ (\frac {\sqrt {\pi}} {2\sqrt {a}}\) Participants suggest using differentiation with respect to the parameter \ (a\) for integrals (i) and (iii), while
Integrate exp(-x^2), dx - Physics Forums This integral can be done the same way that the integral of exp (-x 2) can be done First, write the integral of x 2 exp (-x 2) from zero to infinity Then write the integral of y 2 exp (-y 2) from zero to infinity (they're both exactly the same as your integral) Now multiply the integrands together double integrate over x and y
How can I solve integrals of the form x^n e^ (-x^2) by hand? This discussion focuses on solving integrals of the form \int x^n e^ {-x^2} dx, particularly in the context of quantum mechanics The integral for n=2 simplifies to \int_ {-\infty}^ {\infty} x^2 e^ {-x^2} dx = \frac {\sqrt {\pi}} {2}, utilizing integration by parts and symmetry arguments The discussion highlights that odd powers of x^n yield zero due to the odd nature of the integrand, while
Integrating (x^2-1)^n: How to Get to the Answer? - Physics Forums The integral of the expression \ ( (-1)^n (2n)! (2^ {2n} (n!)^2) \int_ {-1}^ {1} (x^2-1)^n dx\) simplifies to \ (\frac {2} {2n+1}\) through a series of transformations and substitutions The discussion highlights the use of integration by parts and the binomial expansion to derive the solution Key techniques include substituting \ (x\) with \ (\cos (x)\) and applying trigonometric reduction