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g(x)=int_(0)^(x)(t^(2)-t)/(sqrtt)dt g^(')(4)=

g(x)=∫0xt2−ttdt g(x)=\int_{0}^{x} \frac{t^{2}-t}{\sqrt{t}} d t \newlineg′(4)= g^{\prime}(4)=

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Evaluate definite integrals using the power ruleright chevron icon

Full solution

Q. g(x)=∫0xt2−ttdt g(x)=\int_{0}^{x} \frac{t^{2}-t}{\sqrt{t}} d t \newlineg′(4)= g^{\prime}(4)=
  1. Apply Fundamental Theorem: First, let's use the Fundamental Theorem of Calculus to find g′(x)g'(x). g′(x)=t2−ttg'(x) = \frac{t^2 - t}{\sqrt{t}} evaluated at t=xt=x.
  2. Find g′(x)g'(x): Now plug in x=4x=4 into the derivative.g′(4)=42−44g'(4) = \frac{4^2 - 4}{\sqrt{4}}.
  3. Evaluate at x=4x=4: Simplify the expression.g′(4)=16−42.g'(4) = \frac{16 - 4}{2}.
  4. Simplify the expression: Finish the calculation. g′(4)=122g'(4) = \frac{12}{2}.
  5. Finish the calculation: Get the final answer.\newlineg′(4)=6g'(4) = 6.

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