Write integral subscript 1 superscript 6 left parenthesis 7 x plus 2 right parenthesis d x as the limit of a Riemann sum and evaluate. a.) limit as n rightwards arrow infinity of space sum from k equals 1 to n of open parentheses 1 plus fraction numerator 35 k over denominator n end fraction close parentheses open parentheses 5 over n close parentheses equals 85 over 2 b.) limit as n rightwards arrow infinity of space sum from k equals 1 to n of open parentheses 1 plus fraction numerator 5 k over denominator n end fraction close parentheses open parentheses 5 over n close parentheses equals 35 over 2 c.) limit as n rightwards arrow infinity of space sum from k equals 1 to n of open parentheses fraction numerator 35 k over denominator n end fraction close parentheses open parentheses 5 over n close parentheses equals 175 over 2 d.) limit as n rightwards arrow infinity of sum from k equals 1 to n of open parentheses 9 plus fraction numerator 35 k over denominator n end fraction close parentheses open parentheses 5 over n close parentheses equals 265 over 2