Integral of $$$3 x^{2} - 4 x$$$
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Find $$$\int \left(3 x^{2} - 4 x\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(3 x^{2} - 4 x\right)d x}}} = {\color{red}{\left(- \int{4 x d x} + \int{3 x^{2} d x}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = x$$$:
$$\int{3 x^{2} d x} - {\color{red}{\int{4 x d x}}} = \int{3 x^{2} d x} - {\color{red}{\left(4 \int{x d x}\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:
$$\int{3 x^{2} d x} - 4 {\color{red}{\int{x d x}}}=\int{3 x^{2} d x} - 4 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\int{3 x^{2} d x} - 4 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=3$$$ and $$$f{\left(x \right)} = x^{2}$$$:
$$- 2 x^{2} + {\color{red}{\int{3 x^{2} d x}}} = - 2 x^{2} + {\color{red}{\left(3 \int{x^{2} d x}\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$- 2 x^{2} + 3 {\color{red}{\int{x^{2} d x}}}=- 2 x^{2} + 3 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- 2 x^{2} + 3 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$
Therefore,
$$\int{\left(3 x^{2} - 4 x\right)d x} = x^{3} - 2 x^{2}$$
Simplify:
$$\int{\left(3 x^{2} - 4 x\right)d x} = x^{2} \left(x - 2\right)$$
Add the constant of integration:
$$\int{\left(3 x^{2} - 4 x\right)d x} = x^{2} \left(x - 2\right)+C$$
Answer
$$$\int \left(3 x^{2} - 4 x\right)\, dx = x^{2} \left(x - 2\right) + C$$$A