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Calculus I, Section 5.3, #12 The Fundamental Theorem of Calculus 1 Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. R(y) = Z 2t3sin(t) dt y Let’s remind ourselves of the Fundamental Theorem of Calculus, Part 1: The Fundamental Theorem of Calculus, Part 1 If f is continuous on [a;b], then the function g defined by g(x) = Z xf(t) dt a ≤ x ≤ b a is continuous on [a;b] and differentiable on (a;b) and g′(x) = f(x). First, we’ll use properties of the definite integral to make the integral match the form in the Fundamental Theorem. Z 2t3sin(t) dt = −Z y t3sin(t) dt y 2 so we have R(y) = −Z yt3sin(t) dt 2 The minus sign is just a constant factor, so d [R(y)] = −1· d Z yt3sin(t) dt dy dy 2 R′(y) = −1·y3sin(y) Thus, R′(y) = −y3sin(y) 1Stewart, Calculus, Early Transcendentals, p. 399, #12.
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