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G(x)=2x^(3)-4 g(x)=G^(')(x) int_(1)^(2)g(x)dx=

G(x)=2x3−4 G(x)=2 x^{3}-4 \newlineg(x)=G′(x) g(x)=G^{\prime}(x) \newline∫12g(x)dx= \int_{1}^{2} g(x) d x=

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Evaluate definite integrals using the power ruleright chevron icon

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Q. G(x)=2x3−4 G(x)=2 x^{3}-4 \newlineg(x)=G′(x) g(x)=G^{\prime}(x) \newline∫12g(x)dx= \int_{1}^{2} g(x) d x=
  1. Set up integral of g(x)g(x): Now, set up the integral of g(x)g(x) from 11 to 22.∫12g(x) dx=∫126x2 dx\int_{1}^{2}g(x)\,dx = \int_{1}^{2}6x^2\,dx
  2. Calculate integral of 6x26x^2: Calculate the integral of 6x26x^2 from 11 to 22.∫126x2 dx=[2x3]12\int_{1}^{2}6x^2\,dx = [2x^3]_{1}^{2}
  3. Plug in upper and lower limits: Plug in the upper and lower limits of the integral. [2x3](1)(2)=2(2)3−2(1)3[2x^3]_{(1)}^{(2)} = 2(2)^3 - 2(1)^3
  4. Simplify the expression: Simplify the expression. 2(2)3−2(1)3=2(8)−2(1)=16−22(2)^3 - 2(1)^3 = 2(8) - 2(1) = 16 - 2
  5. Finish the calculation: Finish the calculation to get the final answer.\newline16−2=1416 - 2 = 14

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