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g(x)=int_(2)^(x)(10 t+2)dt g^(')(3)=

g(x)=∫2x(10t+2)dt g(x)=\int_{2}^{x}(10 t+2) d t \newlineg′(3)= g^{\prime}(3)=

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Evaluate definite integrals using the power ruleright chevron icon

Full solution

Q. g(x)=∫2x(10t+2)dt g(x)=\int_{2}^{x}(10 t+2) d t \newlineg′(3)= g^{\prime}(3)=
  1. Find Derivative of Integral: We need to find the derivative of the integral from 22 to xx of (10t+2) dt(10t + 2) \, dt, which is g(x)g(x).
  2. Apply Fundamental Theorem of Calculus: By the Fundamental Theorem of Calculus, the derivative of g(x)g(x) is the integrand evaluated at xx, so g′(x)=10x+2g'(x) = 10x + 2.
  3. Evaluate at x=3x=3: Now we evaluate g′(x)g'(x) at x=3x=3, so g′(3)=10(3)+2g'(3) = 10(3) + 2.
  4. Calculate Result: Calculating g′(3)g'(3) gives us 30+230 + 2, which equals 3232.

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