Integral dari $$$4 e^{- \frac{x^{2}}{2}}$$$

Temukan $$$\int 4 e^{- \frac{x^{2}}{2}}\, dx$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=4$$$ dan $$$f{\left(x \right)} = e^{- \frac{x^{2}}{2}}$$$:

$${\color{red}{\int{4 e^{- \frac{x^{2}}{2}} d x}}} = {\color{red}{\left(4 \int{e^{- \frac{x^{2}}{2}} d x}\right)}}$$

Misalkan $$$u=\frac{\sqrt{2} x}{2}$$$.

Kemudian $$$du=\left(\frac{\sqrt{2} x}{2}\right)^{\prime }dx = \frac{\sqrt{2}}{2} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \sqrt{2} du$$$.

Integralnya menjadi

$$4 {\color{red}{\int{e^{- \frac{x^{2}}{2}} d x}}} = 4 {\color{red}{\int{\sqrt{2} e^{- u^{2}} d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\sqrt{2}$$$ dan $$$f{\left(u \right)} = e^{- u^{2}}$$$:

$$4 {\color{red}{\int{\sqrt{2} e^{- u^{2}} d u}}} = 4 {\color{red}{\sqrt{2} \int{e^{- u^{2}} d u}}}$$

Integral ini (Fungsi galat) tidak memiliki bentuk tertutup:

$$4 \sqrt{2} {\color{red}{\int{e^{- u^{2}} d u}}} = 4 \sqrt{2} {\color{red}{\left(\frac{\sqrt{\pi} \operatorname{erf}{\left(u \right)}}{2}\right)}}$$

Ingat bahwa $$$u=\frac{\sqrt{2} x}{2}$$$:

$$2 \sqrt{2} \sqrt{\pi} \operatorname{erf}{\left({\color{red}{u}} \right)} = 2 \sqrt{2} \sqrt{\pi} \operatorname{erf}{\left({\color{red}{\left(\frac{\sqrt{2} x}{2}\right)}} \right)}$$

Oleh karena itu,

$$\int{4 e^{- \frac{x^{2}}{2}} d x} = 2 \sqrt{2} \sqrt{\pi} \operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)}$$

Tambahkan konstanta integrasi:

$$\int{4 e^{- \frac{x^{2}}{2}} d x} = 2 \sqrt{2} \sqrt{\pi} \operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)}+C$$

$$$\int 4 e^{- \frac{x^{2}}{2}}\, dx = 2 \sqrt{2} \sqrt{\pi} \operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)} + C$$$A