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F(x)=int_(1)^(2x)sqrt(1+t^(3))dt F^(')(x)=

F(x)=∫12x1+t3dt F(x)=\int_{1}^{2 x} \sqrt{1+t^{3}} 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)=∫12x1+t3dt F(x)=\int_{1}^{2 x} \sqrt{1+t^{3}} d t \newlineF′(x)= F^{\prime}(x)=
  1. Apply Fundamental Theorem of Calculus: Use the Fundamental Theorem of Calculus Part 11, which states that if F(x)F(x) is defined as an integral with an upper limit of xx, then F′(x)F'(x) is the integrand evaluated at xx. However, since the upper limit here is 2x2x, we need to use the chain rule as well.
  2. Differentiate upper limit: Differentiate the upper limit 2x2x with respect to xx to get 22. This is the outside function's derivative when applying the chain rule.
  3. Use chain rule: Now, multiply the derivative of the upper limit by the integrand evaluated at the upper limit. So, F′(x)=2×1+(2x)3F'(x) = 2 \times \sqrt{1 + (2x)^3}.
  4. Simplify integrand: Simplify the expression inside the square root: (2x)3=8x3(2x)^3 = 8x^3. So, F′(x)=2⋅1+8x3F'(x) = 2 \cdot \sqrt{1 + 8x^3}.
  5. Final derivative expression: The final simplified expression for the derivative is F′(x)=2⋅1+8x3F'(x) = 2 \cdot \sqrt{1 + 8x^3}. This is the exact, simplified answer.

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