$$$\frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}}$$$'nin integrali

Bulun: $$$\int \left(\frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}}\right)\, dy$$$.

Her terimin integralini alın:

$${\color{red}{\int{\left(\frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}}\right)d y}}} = {\color{red}{\left(\int{\frac{1}{2 \sqrt{y}} d y} + \int{\frac{\sqrt{y}}{2} d y}\right)}}$$

Sabit katsayı kuralı $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(y \right)} = \sqrt{y}$$$ ile uygula:

$$\int{\frac{1}{2 \sqrt{y}} d y} + {\color{red}{\int{\frac{\sqrt{y}}{2} d y}}} = \int{\frac{1}{2 \sqrt{y}} d y} + {\color{red}{\left(\frac{\int{\sqrt{y} d y}}{2}\right)}}$$

Kuvvet kuralını $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=\frac{1}{2}$$$ ile uygulayın:

$$\int{\frac{1}{2 \sqrt{y}} d y} + \frac{{\color{red}{\int{\sqrt{y} d y}}}}{2}=\int{\frac{1}{2 \sqrt{y}} d y} + \frac{{\color{red}{\int{y^{\frac{1}{2}} d y}}}}{2}=\int{\frac{1}{2 \sqrt{y}} d y} + \frac{{\color{red}{\frac{y^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{2}=\int{\frac{1}{2 \sqrt{y}} d y} + \frac{{\color{red}{\left(\frac{2 y^{\frac{3}{2}}}{3}\right)}}}{2}$$

Sabit katsayı kuralı $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(y \right)} = \frac{1}{\sqrt{y}}$$$ ile uygula:

$$\frac{y^{\frac{3}{2}}}{3} + {\color{red}{\int{\frac{1}{2 \sqrt{y}} d y}}} = \frac{y^{\frac{3}{2}}}{3} + {\color{red}{\left(\frac{\int{\frac{1}{\sqrt{y}} d y}}{2}\right)}}$$

Kuvvet kuralını $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=- \frac{1}{2}$$$ ile uygulayın:

$$\frac{y^{\frac{3}{2}}}{3} + \frac{{\color{red}{\int{\frac{1}{\sqrt{y}} d y}}}}{2}=\frac{y^{\frac{3}{2}}}{3} + \frac{{\color{red}{\int{y^{- \frac{1}{2}} d y}}}}{2}=\frac{y^{\frac{3}{2}}}{3} + \frac{{\color{red}{\frac{y^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{2}=\frac{y^{\frac{3}{2}}}{3} + \frac{{\color{red}{\left(2 y^{\frac{1}{2}}\right)}}}{2}=\frac{y^{\frac{3}{2}}}{3} + \frac{{\color{red}{\left(2 \sqrt{y}\right)}}}{2}$$

Dolayısıyla,

$$\int{\left(\frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}}\right)d y} = \frac{y^{\frac{3}{2}}}{3} + \sqrt{y}$$

Sadeleştirin:

$$\int{\left(\frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}}\right)d y} = \frac{\sqrt{y} \left(y + 3\right)}{3}$$

İntegrasyon sabitini ekleyin:

$$\int{\left(\frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}}\right)d y} = \frac{\sqrt{y} \left(y + 3\right)}{3}+C$$

$$$\int \left(\frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}}\right)\, dy = \frac{\sqrt{y} \left(y + 3\right)}{3} + C$$$A