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G(x)=2x^(3)-4

g(x)=G^(')(x)

int_(1)^(2)g(x)dx=

$G(x)=2 x^{3}-4$ $\newline$ $g(x)=G^{\prime}(x)$ $\newline$ $\int_{1}^{2} g(x) d x=$

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Evaluate definite integrals using the power rule

Full solution

Q.

G(x)=2 x^{3}-4

\newline

g(x)=G^{\prime}(x)

\newline

\int_{1}^{2} g(x) d x=

Set up integral of $g(x)$ : Now, set up the integral of $g(x)$ from $1$ to $2$ . $\int_{1}^{2}g(x)\,dx = \int_{1}^{2}6x^2\,dx$
Calculate integral of $6x^2$ : Calculate the integral of $6x^2$ from $1$ to $2$ . $\int_{1}^{2}6x^2\,dx = [2x^3]_{1}^{2}$
Plug in upper and lower limits: Plug in the upper and lower limits of the integral. $[2x^3]_{(1)}^{(2)} = 2(2)^3 - 2(1)^3$
Simplify the expression: Simplify the expression. $2(2)^3 - 2(1)^3 = 2(8) - 2(1) = 16 - 2$
Finish the calculation: Finish the calculation to get the final answer. $\newline$ $16 - 2 = 14$

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