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MATH-112-301-001 Unproctored Midcourse Exam 4 Practice Exam 1

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Given that ๐บ ( ๐‘ฅ ) = โˆซ 3 ๐‘ฅ cos โก ( ๐œ‹ ๐‘ก / 6 ) ๐‘ก 2 ๐‘‘ ๐‘ก , what is ๐บ โ€ฒ ( 2 ) ?

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The question as provided defines G(x) as an integral: G(x) = โˆซ 3 x cos(ฯ€ t/6) t^2 dt, but it does not specify the limits of integration. Without limits, the integral is not a well-defined function of x, so we cannot proceed to differentiate with respect to x or evaluate G'(2). First, note a key principle: if you have a parameter x inside an integrand and want d/dx G(x) where G(x) = โˆซ f(x, t) dt with fixed ......Login to view full explanation

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