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.

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.

Tutors, please sign in to answer this question.

Aliquippa, PA

∫ 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)**

Westford, MA

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

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