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F(x)=int_(-2)^(2x)(3t^(2)+2t)dt F^(')(x)=

F(x)=∫−22x(3t2+2t)dt F(x)=\int_{-2}^{2 x}\left(3 t^{2}+2 t\right) d t \newlineF′(x)= F^{\prime}(x)=

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Evaluate definite integrals using the power ruleright chevron icon

Full solution

Q. F(x)=∫−22x(3t2+2t)dt F(x)=\int_{-2}^{2 x}\left(3 t^{2}+2 t\right) d t \newlineF′(x)= F^{\prime}(x)=
  1. Use Fundamental Theorem: Use the Fundamental Theorem of Calculus Part 11, which states that if F(x)F(x) is defined as the integral from aa to xx of f(t)f(t)dt, then F′(x)F'(x) is f(x)f(x).
  2. Differentiate Upper Limit: Differentiate the upper limit of the integral, which is 2x2x, with respect to xx. The derivative of 2x2x is 22.
  3. Plug Upper Limit: Plug the upper limit into the integrand and multiply by the derivative of the upper limit. So, F′(x)=2×(3(2x)2+2(2x))F'(x) = 2 \times (3(2x)^2 + 2(2x)).
  4. Simplify Expression: Simplify the expression. F′(x)=2×(12x2+4x)=24x2+8xF'(x) = 2 \times (12x^2 + 4x) = 24x^2 + 8x.

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