test 4 in class sols

# test 4 in class sols - Name: _ Test #4 Integrals In-class...

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Name: _________________ Test #4 Integrals In-class portion No Calculators 1. The easiest thing to do here would be to split it up. The first part needs u -substitution, but not the second part. 0 4 t 2 t 2 1 2 1 dt = 0 4 t 2 t 2 1 2 dt 0 4 1 dt u = 2 t 2 1 ;du = 4 t dt 1 4 9 u 1 1 4 0 4 4 t 2 t 2 1 2 dt 0 4 1 dt = 1 4 t = 0 t = 4 1 u 2 du 0 4 1 dt = ¿ t = 0 t = 4 t 0 4 = 1 4 − 2 t 2 1 1 0 4 t 0 4 = 1 4 − 1 2 4 2 1 −− 1 2 0 2 1  4 0 = 1 4 − 1 33 1  4 = 8 33 4 = 4 8 33 Find the average value of this function over the interval [0, 4] Divide the definite intergral we just found by the length of the interval (4-0 = 4), and we get 1 2 33 2. Find the general solution to the equation dy dx = x 2 y Separate the variables: dy y = x 2 dx and integrate: dy y = x 2 dx ln y = 1 3 x 3 C Solve for y : y = e 1 3 x 3 C = e 1 3 x 3 e C = Ce 1 3 x 3 This is the general form. Find the particular solution to the equation that passes through the point (2, 5

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## test 4 in class sols - Name: _ Test #4 Integrals In-class...

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