- $\displaystyle\sum_{j=1}^n{f(x_j)(x_j-x_{j-1})}$
- $\displaystyle\sum_{j=1}^n{f(x_{j-1})(x_j-x_{j-1})}$
- $\displaystyle\sum_{j=1}^n{f\left(\frac{x_j+x_{j-1}}{2}\right)(x_j-x_{j-1})}$
- $\displaystyle\int_0^{10}{f(x)dx}$
- $0$

Solution :

Consider the function $f$ defined by $f(x)=e^{-x}$ on the interval [0, 10]. Let $n>1$ and let $x_0,x_1,\ldots,x_n$ be numbers such that: \[0= x_0 < x_1 < \cdots < x_n =10.\] Which of the following is greatest?

**Choice 2 is the answer.**

The first three choices are Riemann sums with choice 1 the right sum, choice 2 the left sum and choice 3 the midpoint sum. Choice 4 is the actual area of $f$ from 0 to 10.

Since $e^{-x}$ is a decreasing function, the left Riemann sum is the greatest out of all the choices (see diagram below).

Thus, choice 2 (the left Riemann sum) is the answer.

- $\displaystyle\sum_{j=1}^n{f(x_j)(x_j-x_{j-1})}$
- $\displaystyle\sum_{j=1}^n{f(x_{j-1})(x_j-x_{j-1})}$
- $\displaystyle\sum_{j=1}^n{f\left(\frac{x_j+x_{j-1}}{2}\right)(x_j-x_{j-1})}$
- $\displaystyle\int_0^{10}{f(x)dx}$
- $0$

Solution :

The first three choices are Riemann sums with choice 1 the right sum, choice 2 the left sum and choice 3 the midpoint sum. Choice 4 is the actual area of $f$ from 0 to 10.

Since $e^{-x}$ is a decreasing function, the left Riemann sum is the greatest out of all the choices (see diagram below).

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