Belirli ve Uygunsuz İntegral Hesaplayıcı

Your input: calculate $$$\int_{\infty}^{0}\left( x e^{- x^{2}} \right)dx$$$

First, calculate the corresponding indefinite integral: $$$\int{x e^{- x^{2}} d x}=- \frac{e^{- x^{2}}}{2}$$$ (for steps, see indefinite integral calculator)

Since the upper bound is less than the lower, according to the property of integrals, we can swap them and change the sign:

$$$\int_{\infty}^{0}\left( x e^{- x^{2}} \right)dx=-\int_{0}^{\infty}\left( x e^{- x^{2}} \right)dx$$$

Since there is infinity in the upper bound, this is improper integral of type 1.

To evaluate an integral over an interval, we use the Fundamental Theorem of Calculus. However, we need to use limit if an endpoint of the interval is special (infinite).

$$$\int_{0}^{\infty}\left( x e^{- x^{2}} \right)dx=\lim_{x \to \infty}\left(- \frac{e^{- x^{2}}}{2}\right)-\left(- \frac{e^{- x^{2}}}{2}\right)|_{\left(x=0\right)}=\frac{1}{2}$$$

Do not forget about the minus sign: $$$\int_{\infty}^{0}\left( x e^{- x^{2}} \right)dx=-\int_{0}^{\infty}\left( x e^{- x^{2}} \right)dx=- \frac{1}{2}$$$

Answer: $$$\int_{\infty}^{0}\left( x e^{- x^{2}} \right)dx=- \frac{1}{2}=-0.5$$$