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F(x)=int_(0)^(sqrtx)2tdt where x > 0. F^(')(x)=

F(x)=∫0x2tdt F(x)=\int_{0}^{\sqrt{x}} 2 t d t \newlinewhere x>0 x>0 .\newlineF′(x)= F^{\prime}(x)=

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Evaluate definite integrals using the power ruleright chevron icon

Full solution

Q. F(x)=∫0x2tdt F(x)=\int_{0}^{\sqrt{x}} 2 t d t \newlinewhere x>0 x>0 .\newlineF′(x)= F^{\prime}(x)=
  1. Apply Fundamental Theorem: Use the Fundamental Theorem of Calculus Part 11 to find F′(x)F'(x). This theorem states that if F(x)F(x) is defined as an integral from aa to g(x)g(x) of f(t) dtf(t) \, dt, then F′(x)F'(x) is f(g(x))×g′(x)f(g(x)) \times g'(x).
  2. Find Derivative of x\sqrt{x}: First, find the derivative of the upper limit of integration, which is x\sqrt{x}. The derivative of x\sqrt{x} with respect to xx is (1/2)x(−1/2)(1/2)x^{(-1/2)}.
  3. Evaluate Integrand at x\sqrt{x}: Now, apply the Fundamental Theorem of Calculus. The integrand is 2t2t, and we need to evaluate it at the upper limit, which is x\sqrt{x}. So we replace tt with x\sqrt{x} to get 2x2\sqrt{x}.
  4. Multiply Result by Derivative: Multiply the result from the previous step by the derivative of the upper limit. So we have 2x×(12x−12)2\sqrt{x} \times \left(\frac{1}{2}x^{-\frac{1}{2}}\right).
  5. Simplify the Expression: Simplify the expression. The (12)x(−12)(\frac{1}{2})x^{(-\frac{1}{2})} cancels out the x\sqrt{x} in 2x2\sqrt{x}, leaving us with just 22.

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