1 answer
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since the interval width is 3, the left sum is
3(f(0)+f(3)+f(6)+f(9)+f(12)) = 3(63+62+59+52+38) = 822
the right sum is
3(f(3)+f(6)+f(9)+f(12)+f(15)) = 3(62+59+52+38+10) = 663