Find the integral of x(x^2+2)^2 dx from 0 to 1. u=x^2+2 du=2x dx 1/2 du=x dx 1/2 ∫ u^2 du What do I do next? The answer is 19/6. Please show all your work. 1/29/2014 | Sun from Los Angeles, CA | 2 Answers | 0 Votes Mark favorite Subscribe Comment
∫ x(x^{2} + 2)^{2}dx Let u = x^{2} + 2 du = 2xdx ∫ x(x^{2} + 2)2dx = (1/2)∫u^{2}du = (1/2)*(u^{3}/3) = (1/6)*(x^{2} + 2)^{3 }(evaluated between 0 and 1) = (1/6)[(1)^{2} + 2]^{3} - (1/6)[(0)^{2} + 2]^{3} = (1/6)*(27 - 8) = (19/6) 1/29/2014 | William S. Comment Comments Thanks. 1/29/2014 | Sun from Los Angeles, CA Comment
Alternatively or as a check: f(x) = x(x^2+2)^2 f(x) = x(x^4+4x^2+4) f(x) = x^5+4x^3+4x One of f's antiderivatives is F(x) = x^6/6+4x^4/4+4x^2/2 F(x) = x^6/6+x^4+2x^2 F(1)-F(0) = 1/6+1+2 = 19/6 1/30/2014 | Steve S. Comment
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