2 Multiple choice (1) Suppose that f(x) is continuous at [0,1] int _(0)^1f(x)dx= A lim _(narrow infty )sum _(k=1)^nf((2k-1)/(2n))(1)/(2n) B lim _(narrow infty )sum _(k=1)^nf((2k-1)/(2n))(1)/(n) C. lim _(narrow infty )sum _(k=1)^2nf((k-1)/(2n))(1)/(n) D. lim _(narrow infty )sum _(k=1)^2nf((k)/(2n))(2)/(n) (2) If lim _(narrow infty )((n^2+1)/(n+1)-an-b)=0 then A. a=1,b=0 B. a=1,b=1 C. a=1,b=-1 D. a=0,b=1

(1) The correct answer is A. $\lim _{n\rightarrow \infty }\sum _{k=1}^{n}f(\frac {2k-1}{2n})\frac {1}{2n}$<br /><br />Explanation: The integral $\int _{0}^{1}f(x)dx$ can be approximated using a Riemann sum. In this case, the Riemann sum is given by $\lim _{n\rightarrow \infty }\sum _{k=1}^{n}f(\frac {2k-1}{2n})\frac {1}{2n}$, where $n$ is the number of subintervals, $\frac {2k-1}{2n}$ represents the $k$th subinterval endpoint, and $\frac {1}{2n}$ is the width of each subinterval.<br /><br />(2) The correct answer is C. $a=1,b=-1$<br /><br />Explanation: To find the values of $a$ and $b$, we need to simplify the expression $\lim _{n\rightarrow \infty }(\frac {n^{2}+1}{n+1}-an-b)$. By dividing the numerator and denominator of the fraction by $n$, we get $\lim _{n\rightarrow \infty }(\frac {n+1}{1+\frac {1}{n}}-an-b)$. As $n$ approaches infinity, $\frac {1}{n}$ approaches 0, so the expression simplifies to $\lim _{n\rightarrow \infty }(1-an-b)$. For this limit to be equal to 0, we must have $a=1$ and $b=-1$.