# week10 - c = 2000 or directly c(0 = 2000 Solution We have c...

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Integration with initial conditions Example We know that an antiderivative of x 2 is any function whose derivative is x 2 . All antiderivatives of x 2 form a “family” of functions: y ( x ) = x 3 3 + C, where C can be chosen any constant (for instance C could be the number 1 but also the number ( - 5) , or 10000 . If an additional condition is given regarding y ( x ) , then we may determine the constant that fulﬁlls this additional condition. For example, say we are given y (1) = 5 . Then we have y (1) = 1 3 3 + C and y (1) = 5 and so: 1 3 3 + C = 5 C = 5 - 1 3 C = 14 3 In conclusion, the antiderivative y ( x ) of x 2 such that y (1) = 5 is y ( x ) = x 3 3 + 14 3 Example (Ex. 14/Sect 14.3) If dc dq is a marginal-cost function and the ﬁxed costs is indicated in braces, ﬁnd the total-cost function. dc dq = 2 q + 75 [2000] Note The ﬁxed costs are constant regardless the output. So here, when q = 0 we have

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Unformatted text preview: c = 2000 , or directly c (0) = 2000 Solution We have c ( q ) = Z (2 q + 75) dq = 2 Z q dq + Z 75 dq = q 2 + 75 q + C Since c (0) = 0 2 + 75 Â· 0 + C and c (0) = 2000 , we have C = 2000 . So c ( q ) = q 2 + 75 q + 2000 . More Integration Formulas Integration by substitution Z f ( u ( x )) Â· u ( x ) dx = Z f ( u ) du where du = u ( x ) dx Example Z ( x + 5) 11 dx = Z u 11 du where u = x + 5 and du = u ( x ) dx = 1 dx that is du = dx So Z ( x + 5) 11 dx = Z u 11 du = u 12 12 + C = ( x + 5) 12 12 + C Example Find: Z x 3 e x 4 dx Solution Denote u = x 4 â‡’ du = u ( x ) dx that is du = 4 x 3 dx So Z x 3 e x 4 dx = Z e x 4 x 3 dx = 1 4 Z e x 4 4 x 3 dx = 1 4 Z e u du = 1 4 e u + C = 1 4 e x 4 + C...
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week10 - c = 2000 or directly c(0 = 2000 Solution We have c...

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