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g(x)=int_(-10)^(x)(10-3t)dt g^(')(-4)=

g(x)=∫−10x(10−3t)dt g(x)=\int_{-10}^{x}(10-3 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)=∫−10x(10−3t)dt g(x)=\int_{-10}^{x}(10-3 t) d t \newlineg′(−4)= g^{\prime}(-4)=
  1. Apply Fundamental Theorem of Calculus: We have g(x)=∫−10x(10−3t) dtg(x) = \int_{-10}^{x}(10-3t)\,dt. To find g′(x)g'(x), we use the Fundamental Theorem of Calculus which tells us that the derivative of an integral with respect to its upper limit is the integrand evaluated at that upper limit.
  2. Calculate g′(x)g'(x): So, g′(x)=10−3xg'(x) = 10 - 3x. Now we just need to plug in x=−4x = -4 into this derivative.
  3. Evaluate g′(−4)g'(-4): g′(−4)=10−3(−4)=10+12=22g'(-4) = 10 - 3(-4) = 10 + 12 = 22.

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