Integrate y= x^3+3x^2-4x-7 between x values 1 and 3

Firstly, integrate y with respect to dx. Increase the powers of x by 1 and then divide the coefficient of x by the new power of x. I.e.: x^3 becomes 1/4x^4. The power increases from 3 to 4 and the coefficient, 1, is divided but the new power 4 to give a new coefficient of a quarter. Integrating the full expression gives: = 1/4x^4+x^3-2x^2-7x+c. C is the constant but at the next stage of the question will become irrelevant.
now the x values need to be added into this new integral and subtracted from one another as follows:[1/4(3)^4+(3)^3-2(3)^2-7(3)]-[1/4(1)^4+(1)^3-2(1)^2-7(1)]=[81/4+27-18-21]-[1/4+1-2-7]=[20-12+8]=16