Use a power series to approximate the definite integral,I, to six decimal places. Solution (x^3)/(1 + x^4) = (x^3)*(1 + x^4)^-1 = (x^3)*(1 - x^4 + x^8 + . . . using the binomial formula. (I doubt whether any more is necessary.) INT [0, 0.2] (x^3)/(1 + x^4) dx = INT [0, 0.2] x^3 - x^7 + x^11 - . . . = (1/4)x^4 - (1/8)x^8 + (1/12)x^12 - . . . = (1/4)*0.2^4 - (1/8)*0.2^8 + (1/12)*0.2^12 - . . . = 0.000399680 =0.000400 (6 dec. pl.).