The value of the integral _ 1 3 ^1 (x-x^3 )^ 1 3 x^4 d x is

Step-by-step solution

Let I = _ 1 3 ¹ (x-x^3)^ 1 3 x^4 dx . Factor out x^3 from the term inside the cube root: I = _ 1 3 ¹ (x^3(x⁻²-1))^ 1 3 x^4 dx = _ 1 3 ¹ x(x⁻²-1)^ 1 3 x^4 dx = _ 1 3 ¹ (x⁻²-1)^ 1 3 x^3 dx . Let u = x⁻² - 1 . Then du = -2x⁻³ dx , which implies x⁻³ dx = - 1 2 du . Change the limits of integration: When x = 1 3 , u = ( 1 3 )⁻² - 1 = 9 - 1 = 8 . When x = 1 , u = 1⁻² - 1 = 0 . Substituting these into the integral: I = ₈⁰ u^ 1 3 (- 1 2 ) du = 1 2 ₀⁸ u^ 1 3 du . Evaluating the integral: I = 1 2 [ u^ 4 3 4 3 ]₀⁸ = 1 2 3 4 [