# How do you calculate the left Riemann sum for the given function over the interval [1,7], using n=3 for (3 x^2+2 x +5) ?

Nov 26, 2017

588 (Using a right Riemann sum, oops)

#### Explanation:

I just realized the actual question was for a left Riemann sum, not a right one. Pardon my mistake. Below is the procedure for evaluating it with a right Riemann sum:

The general formula for a right-sided rectangle Riemann approximation on the interval $\left[a , b\right]$ using $n$ rectangles is:
${\sum}_{i = 1}^{n} f \left(a + i \Delta x\right) \Delta x$
where $\Delta x = \frac{b - a}{n}$

Plugging in the numbers, we get:
$\Delta x = \frac{7 - 1}{3} = \frac{6}{3} = 2$
${\sum}_{i = 1}^{3} 2 \left(3 {\left(1 + 2 \cdot i\right)}^{2} + 2 \left(1 + 2 \cdot i\right) + 5\right)$

We can multiply out the $2$:
${\sum}_{i = 1}^{3} 6 {\left(1 + 2 \cdot i\right)}^{2} + 4 \left(1 + 2 \cdot i\right) + 10$

Now, let's evaluate it:
$6 {\left(1 + 2 \cdot 1\right)}^{2} + 4 \left(1 + 2 \cdot 1\right) + 10 + 6 {\left(1 + 2 \cdot 2\right)}^{2} + 4 \left(1 + 2 \cdot 2\right) + 10 + 6 {\left(1 + 2 \cdot 3\right)}^{2} + 4 \left(1 + 2 \cdot 3\right) + 10 =$

$= 6 {\left(3\right)}^{2} + 4 \left(3\right) + 10 + 6 {\left(5\right)}^{2} + 4 \left(5\right) + 10 + 6 {\left(7\right)}^{2} + 4 \left(7\right) + 10$

$= 6 \cdot 9 + 12 + 10 + 6 \cdot 25 + 20 + 10 + 6 \cdot 49 + 28 + 10$

$= 588$

So, a $3$-part right-sided rectangle Riemann approximation on the interval $\left[1 , 7\right]$ of $3 {x}^{2} + 2 x + 5$ gives an area of $588$.

We can compare this to the actual answer, which would be computed using the anti-derivative:
${\int}_{1}^{7} 3 {x}^{2} + 2 x + 5 = 420$

If we look at the difference, we get $588 - 420 = 168$, which is an error of 40%.

Nov 26, 2017

$L R S = 276$

#### Explanation:

Let:

$f \left(x\right) = 3 {x}^{2} + 2 x + 5$

We want to estimate $\int \setminus f \left(x\right) \setminus \mathrm{dx}$ the interval $\left[1 , 7\right]$ with $3$ strips; thus:

$\Delta x = \frac{7 - 1}{3} = 2$

Note that we have a fixed interval (strictly speaking a Riemann sum can have a varying sized partition width). The values of the function are tabulated as follows;

Left Riemann Sum

$L R S = {\sum}_{r = 0}^{2} f \left({x}_{i}\right) \setminus \Delta {x}_{i}$
$\text{ } = 2 \cdot \left(10 + 38 + 90\right)$
$\text{ } = 2 \cdot \left(138\right)$
$\text{ } = 276$

Actual Value

For comparison of accuracy:

$A r e a = {\int}_{1}^{7} \setminus 3 {x}^{2} + 2 x + 5 \setminus \mathrm{dx}$
$\text{ } = {\left[{x}^{3} + {x}^{2} + 5 x\right]}_{1}^{7}$
$\text{ } = \left(343 + 49 + 35\right) - \left(1 + 1 + 5\right)$
$\text{ } = 427 - 7$
$\text{ } = 420$