What is the definite integral of 2x^2 + 4x + 1 with a lower limit of 3 and a higher limit of 6?

When integrating we add one to the power of x and divide the number in front of the x by the new power for each part of the function. So 2x^2 becomes 2/3 x^3, 4x becomes 4/2 x^2, 1 becomes 1/1 x^1. Then we need to put in the values of the upper and lower limits for x. So put the upper limit of 6 in first then subtract the value of putting the lower limit of 3 in. [(2/3 * 6^3) + (2 * 6^2) + (6)] - [ (2/3 * 3^3) + (2 * 3^2) + (3)] = (222) -(39) = 183 and that is the final answer.