If f: R rightarrow R is defined by f(x)=x+1,t | vedclass

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(B) The given expression is a Riemann sum of the form $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} f\left(\frac{5r}{n}\right)$.This can

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$\frac{7}{2}$

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(B) The given expression is a Riemann sum of the form $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} f\left(\frac{5r}{n}\right)$.This can be expressed as a definite integral: $I = \int_{0}^{1} f(5x) \,dx$.Substituting $f(x) = x+1$,we get $f(5x) = 5x+1$.Thus,$I = \int_{0}^{1} (5x+1) \,dx$.Evaluating the integral: $I = \left[ \frac{5x^2}{2} + x \right]_{0}^{1}$.$I = \left( \frac{5(1)^2}{2} + 1 \right) - (0) = \frac{5}{2} + 1 = \frac{7}{2}$.

If $f: R \rightarrow R$ is defined by $f(x)=x+1$,then the value of $\lim _{n \rightarrow \infty} \frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+\ldots+f\left(\frac{5(n-1)}{n}\right)\right]$ is:

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