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Find the midpoint rule approximations to the following integral. 12 S x x3dx using n= 1, 2, and 4 subintervals. 4 M(1) = (Sim

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0 12 x² da n=1 bra a24, b=12,122 Axa bis 1948 8 0228 Sub Internals [4, 12] mid point: ue2-8 3 f(x)= x f (8) 3 2512 midpoint snziu Ana 12-4 8 4 4 -२ =2=6222 Sub Polerrals: [4, 6], [68],[8,10],[10,123 midpointsi ut 6 6-18 8710 a ) 10612 5, 7, 9, 10 f(5

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