# Approximate $\int\limits_{1}^{7} f{\left(x \right)}\, dx$ with the Riemann sum using the table $\left[\begin{array}{ccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7\\4 & -2 & 3 & 1 & 0 & 5 & 9\end{array}\right]$

The calculator will approximate the integral $\int\limits_{1}^{7} f{\left(x \right)}\, dx$ with the Riemann sum using the table $\left[\begin{array}{ccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7\\4 & -2 & 3 & 1 & 0 & 5 & 9\end{array}\right]$, with steps shown.

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 $x$ $f{\left(x \right)}$

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Approximate the integral $\int\limits_{1}^{7} f{\left(x \right)}\, dx$ with the left Riemann sum using the table below:

 $x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $f{\left(x \right)}$ $4$ $-2$ $3$ $1$ $0$ $5$ $9$

### Solution

The left Riemann sum approximates the integral using left endpoints: $\int\limits_{a}^{b} f{\left(x \right)}\, dx\approx \sum_{i=1}^{n - 1} \left(x_{i+1} - x_{i}\right) f{\left(x_{i} \right)}$, where $n$ is the number of points.

Therefore, $\int\limits_{1}^{7} f{\left(x \right)}\, dx\approx \left(2 - 1\right) 4 + \left(3 - 2\right) \left(-2\right) + \left(4 - 3\right) 3 + \left(5 - 4\right) 1 + \left(6 - 5\right) 0 + \left(7 - 6\right) 5 = 11.$

$\int\limits_{1}^{7} f{\left(x \right)}\, dx\approx 11$A