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Calculus, Riemann sum pic below
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largely arithmetic right hand endpoints means \(f(3),f(7),f(8),f(10)\)
on the interval \([1,3]\) the interval is of length 2 and \(f(3)=0\) from the table, so that give \(0\times 2=0\) for that part
on the interval \([3,7]\) the length is 4 and \(f(7)=3\) again from the table, so that gives \[3\times 4=12\]
now lets cut to the chase it is \[0\times 2+3\times 4+3\times 1+(-4)\times 2\] whatever that is
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7 :) yay thank you
yw easy right?
yes! you explained it very well
thanks
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