The value of _ x 0 1 x^3 ₀^x (1+ ^2 t) - ^2 t t^2 dt is

Step-by-step solution

The given limit is of the form 0 0 . Applying L'Hospital's rule and the Newton-Leibniz formula: L = _ x 0 (1+ ^2 x) - ^2 x x^2 3x^2 = _ x 0 (1+ ^2 x) - ^2 x 3x^4 Using Maclaurin series expansions: x = x - x^3 6 + ^2 x = (x - x^3 6 )^2 = x^2 - x^4 3 + (1+u) = u - u^2 2 + (1+ ^2 x) = (x^2 - x^4 3 ) - 1 2 (x^2 - x^4 3 )^2 + = x^2 - x^4 3 - x^4 2 + = x^2 - 5x^4 6 + Also, x = x + x^3 3 + ^2 x = (x + x^3 3 )^2 = x^2 + 2x^4 3 + Substituting these into the limit: L = _ x 0 (x^2 - 5x^4 6 ) - (x^2 + 2x^4 3 ) 3x^4 = _ x 0 - 5