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(int_(1)^(3)3x^(2)-2x-2dx)/(2)

$\frac{\int_{1}^{3} 3 x^{2}-2 x-2 d x}{2}$

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Find derivatives of using multiple formulae

Full solution

Q.

\frac{\int_{1}^{3} 3 x^{2}-2 x-2 d x}{2}

Integrate function: We need to find the integral of the function $3x^2 - 2x - 2$ from $1$ to $3$ and then divide the result by $2$ . The first step is to integrate the function. $\newline$ The integral of $3x^2$ with respect to $x$ is $x^3$ , the integral of $-2x$ is $-x^2$ , and the integral of $-2$ is $-2x$ . $\newline$ So, $1$ $1$ , where $1$ $2$ is the constant of integration.
Evaluate definite integral: Now we need to evaluate the definite integral from $1$ to $3$ . $\int_{1}^{3}(3x^2 - 2x - 2)dx = [x^3 - x^2 - 2x]$ from $1$ to $3$ . We plug in the upper limit of $3$ into the antiderivative and then subtract the result of plugging in the lower limit of $1$ . $= (3^3 - 3^2 - 2\cdot3) - (1^3 - 1^2 - 2\cdot1)$ $= (27 - 9 - 6) - (1 - 1 - 2)$ $= (27 - 9 - 6) - (-1)$ $3$ $0$ $3$ $1$
Divide by $2$ : The last step is to divide the result of the definite integral by $2$ . $\newline$ $(\int_{1}^{3}(3x^2 - 2x - 2)\,dx) / 2 = \frac{13}{2}$ $\newline$ $= 6.5$

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