Question

consider the integral
(a) find the riemann sum for this integral using left endpoints and ( n = 3 ).
( l _ { 3 } = )
(b) find the riemann sum for this same integral using right endpoints and ( n = 3 ).
( r _ { 3 } = )
note: you can earn partial credit on this problem.
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Explanation:

Step1: Calculate the width of sub - intervals

The interval is \([a,b]=[3,9]\) and \(n = 3\). The width of each sub - interval \(\Delta x=\frac{b - a}{n}=\frac{9 - 3}{3}=2\).
The sub - intervals are \([3,5]\), \([5,7]\), \([7,9]\).

Step2: Calculate \(L_3\) (using left - endpoints)

For \(L_3\), the left - endpoints are \(x_1 = 3\), \(x_2=5\), \(x_3 = 7\).
The function \(f(x)=4x^{2}+3x + 6\).
\(f(x_1)=4\times3^{2}+3\times3 + 6=4\times9+9 + 6=36+9+6=51\)
\(f(x_2)=4\times5^{2}+3\times5 + 6=4\times25+15 + 6=100+15+6=121\)
\(f(x_3)=4\times7^{2}+3\times7 + 6=4\times49+21 + 6=196+21+6=223\)
\(L_3=\Delta x\sum_{i = 1}^{3}f(x_i)=2\times(51 + 121+223)=2\times395 = 790\)

Step3: Calculate \(R_3\) (using right - endpoints)

For \(R_3\), the right - endpoints are \(x_1 = 5\), \(x_2=7\), \(x_3 = 9\).
\(f(x_1)=4\times5^{2}+3\times5 + 6=121\)
\(f(x_2)=4\times7^{2}+3\times7 + 6=223\)
\(f(x_3)=4\times9^{2}+3\times9 + 6=4\times81+27 + 6=324+27+6=357\)
\(R_3=\Delta x\sum_{i = 1}^{3}f(x_i)=2\times(121 + 223+357)=2\times701=1402\)

Answer:

\(L_3 = 790\), \(R_3=1402\)