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FundamentalTheoremofCalculus(Part2) Created by Barbara Forrest and Brian Forrest COPYRIGHTED FundamentalTheoremofCalculus(Part2) Example: Evaluate Z 2 3 t dt. Observation: If 0 Z x 3 G(x) = t dt, 0 then Z 2 3 G(2) = t dt. WeknowfromtheFTC1that 0 G′(x) = x3 3 so Gis an antiderivative of x . Hence there exists a constant C such that Z x 4 3 x G(x) = t dt = +C. 0 4 Question: How does this help us? COPYRIGHTED FundamentalTheoremofCalculus(Part2) Note: Wehavejustseenthat G(x) = Z xt3dt = x4 +C 0 4 for some constant C ∈ R. However, we also know that Z 0 3 04 0 = t dt = G(0) = +C=C 0 4 so Z x 3 x4 G(x) = t dt = . 0 4 Z 2 4 Finally, 3 2 t dt = G(2) = =4. 0 4 Question: Did we really need to find C? COPYRIGHTED FundamentalTheoremofCalculus(Part2) KeyObservation: Let F and G be any two antiderivatives of the same function f. Then G(x) = F(x)+C. Let a,b ∈ R. Then G(b)−G(a) = (F(b)+C)−(F(a)+C) = F(b)−F(a). COPYRIGHTED
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