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g(x)=int_(0)^(x)(5t^(2)-t)dt g^(')(2)=

g(x)=∫0x(5t2−t)dt g(x)=\int_{0}^{x}\left(5 t^{2}-t\right) d t \newlineg′(2)= g^{\prime}(2)=

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Evaluate definite integrals using the power ruleright chevron icon

Full solution

Q. g(x)=∫0x(5t2−t)dt g(x)=\int_{0}^{x}\left(5 t^{2}-t\right) d t \newlineg′(2)= g^{\prime}(2)=
  1. Find Derivative of g(x)g(x): First, let's find the derivative of g(x)g(x) using the Fundamental Theorem of Calculus.\newlineg′(x)=ddx∫0x(5t2−t)dtg'(x) = \frac{d}{dx} \int_{0}^{x} (5t^2 - t) dt\newlineg′(x)=5x2−xg'(x) = 5x^2 - x
  2. Evaluate g′(x)g'(x) at x=2x=2: Now, let's evaluate g′(x)g'(x) at x=2x=2.
    g′(2)=5(2)2−2g'(2) = 5(2)^2 - 2
    g′(2)=5(4)−2g'(2) = 5(4) - 2
    g′(2)=20−2g'(2) = 20 - 2
    g′(2)=18g'(2) = 18

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