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.

Related calculator: Riemann Sum Calculator for a Function

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

Answer

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

Approximate ∫17f(x) dx\int\limits_{1}^{7} f{\left(x \right)}\, dx1∫7​f(x)dx with the Riemann sum using the table [12345674−231059]\left[\begin{array}{ccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7\\4 & -2 & 3 & 1 & 0 & 5 & 9\end{array}\right][14​2−2​33​41​50​65​79​]

Your Input

Solution

Answer

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