# Use Riemann sums to evaluate? : #int_0^3 \ x^2-3x+2 \ dx #

# int_0^3 \ x^2-3x+2 \ dx = 3/2 #

We are asked to evaluate:

Using Riemann sums. By definition of an integral, then

That is

And so:

Using the standard summation formula:

we have:

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

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To use Riemann sums to evaluate the integral ( \int_0^3 (x^2 - 3x + 2) , dx ), you partition the interval ([0,3]) into subintervals and then approximate the area under the curve using rectangles.

Let's say we partition the interval into (n) subintervals of equal width, ( \Delta x = \frac{3}{n} ). Then, the (i)th subinterval is ( [x_{i-1}, x_i] ) where (x_i = \frac{3i}{n}) for (i = 1, 2, ..., n).

The Riemann sum for this integral is given by: [ \sum_{i=1}^{n} f(c_i) \Delta x ] where (c_i) is a point in the (i)th subinterval. We'll choose (c_i) to be the right endpoint of each subinterval.

[ c_i = x_i = \frac{3i}{n} ]

[ \Delta x = \frac{3}{n} ]

[ f(c_i) = (c_i)^2 - 3(c_i) + 2 ]

[ f(c_i) = \left(\frac{3i}{n}\right)^2 - 3\left(\frac{3i}{n}\right) + 2 ]

Now, we sum up all the rectangles and take the limit as (n) approaches infinity to get the integral:

[ \lim_{n \to \infty} \sum_{i=1}^{n} \left(\left(\frac{3i}{n}\right)^2 - 3\left(\frac{3i}{n}\right) + 2\right) \cdot \frac{3}{n} ]

[ = \lim_{n \to \infty} \frac{9}{n^3} \sum_{i=1}^{n} \left( i^2 - 9i + 6n \right) ]

[ = \lim_{n \to \infty} \frac{9}{n^3} \left( \frac{n(n+1)(2n+1)}{6} - 9\frac{n(n+1)}{2} + 6n^2 \right) ]

[ = \lim_{n \to \infty} \frac{9}{n^3} \left( \frac{2n^3 + 3n^2 + n}{6} - \frac{9n^2 + 9n}{2} + 6n^2 \right) ]

[ = \lim_{n \to \infty} \frac{9}{n^3} \left( \frac{2n^3 + 3n^2 + n}{6} - \frac{27n^2 + 18n}{6} + 6n^2 \right) ]

[ = \lim_{n \to \infty} \frac{9}{n^3} \left( \frac{2n^3 + 3n^2 + n - 27n^2 - 18n + 36n^2}{6} \right) ]

[ = \lim_{n \to \infty} \frac{9}{n^3} \left( \frac{2n^3 + 12n^2 - 17n}{6} \right) ]

[ = \lim_{n \to \infty} \frac{9}{n^3} \left( \frac{n(2n^2 + 12n - 17)}{6} \right) ]

[ = \lim_{n \to \infty} \frac{9}{n^2} \left( \frac{2n^2 + 12n - 17}{6} \right) ]

[ = \lim_{n \to \infty} \frac{3}{2} + \frac{9}{n} - \frac{153}{2n^2} ]

[ = \frac{3}{2} ]

Therefore, the value of the integral ( \int_0^3 (x^2 - 3x + 2) , dx ) using Riemann sums is ( \frac{3}{2} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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