Suppose that the n-th Riemann sum (with n sub-intervals of equal length), for some function f(x) on the intrval [0, 2], is Sn=2n2 × ∑k=1n k. What is the value of ∫02 f(x) dx?

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Suppose that the n-th Riemann sum (with n sub-intervals of equal length), for some function f(x) on the intrval [0, 2], is Sn=2n2 × ∑k=1n k.  What is the value of ∫02 f(x) dx?

saiyansruleA · Accepted Answer

Step 1 Sn=2n2 ∑k=1n k Sn=2n2 [1 + 2 + 3 + ⋯ + n] Sn=2n2n(n + 1)2 Sn=n + 1n f(x)=x + 1x Step 2 Now to find =∫02 f (x) dx =∫02[xx + 1x]dx =∫02(1 + 1x) dx =[x + ln⁡ x]02 =(2 + ln⁡ 2) − (0 + ln⁡ 0) =2 + ln⁡ 2

Suppose that the n-th Riemann sum (with n sub-intervals of equal length), for some function...

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