Funktion $$$- 5^{x} + 13 x^{2} - 38$$$ integraali

Määritä $$$\int \left(- 5^{x} + 13 x^{2} - 38\right)\, dx$$$.

Integroi termi kerrallaan:

$${\color{red}{\int{\left(- 5^{x} + 13 x^{2} - 38\right)d x}}} = {\color{red}{\left(- \int{38 d x} - \int{5^{x} d x} + \int{13 x^{2} d x}\right)}}$$

Sovella vakiosääntöä $$$\int c\, dx = c x$$$ käyttäen $$$c=38$$$:

$$- \int{5^{x} d x} + \int{13 x^{2} d x} - {\color{red}{\int{38 d x}}} = - \int{5^{x} d x} + \int{13 x^{2} d x} - {\color{red}{\left(38 x\right)}}$$

Apply the exponential rule $$$\int{a^{x} d x} = \frac{a^{x}}{\ln{\left(a \right)}}$$$ with $$$a=5$$$:

$$- 38 x + \int{13 x^{2} d x} - {\color{red}{\int{5^{x} d x}}} = - 38 x + \int{13 x^{2} d x} - {\color{red}{\frac{5^{x}}{\ln{\left(5 \right)}}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=13$$$ ja $$$f{\left(x \right)} = x^{2}$$$:

$$- \frac{5^{x}}{\ln{\left(5 \right)}} - 38 x + {\color{red}{\int{13 x^{2} d x}}} = - \frac{5^{x}}{\ln{\left(5 \right)}} - 38 x + {\color{red}{\left(13 \int{x^{2} d x}\right)}}$$

Sovella potenssisääntöä $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=2$$$:

$$- \frac{5^{x}}{\ln{\left(5 \right)}} - 38 x + 13 {\color{red}{\int{x^{2} d x}}}=- \frac{5^{x}}{\ln{\left(5 \right)}} - 38 x + 13 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- \frac{5^{x}}{\ln{\left(5 \right)}} - 38 x + 13 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Näin ollen,

$$\int{\left(- 5^{x} + 13 x^{2} - 38\right)d x} = - \frac{5^{x}}{\ln{\left(5 \right)}} + \frac{13 x^{3}}{3} - 38 x$$

Lisää integrointivakio:

$$\int{\left(- 5^{x} + 13 x^{2} - 38\right)d x} = - \frac{5^{x}}{\ln{\left(5 \right)}} + \frac{13 x^{3}}{3} - 38 x+C$$

$$$\int \left(- 5^{x} + 13 x^{2} - 38\right)\, dx = \left(- \frac{5^{x}}{\ln\left(5\right)} + \frac{13 x^{3}}{3} - 38 x\right) + C$$$A