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

Answer 1

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

We are asked to evaluate:

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

Using Riemann sums. By definition of an integral, then

# int_a^b \ f(x) \ dx #
represents the area under the curve #y=f(x)# between #x=a# and #x=b#. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

# int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)#
Here we have #f(x)=x^2-3x+2# and we partition the interval #[0,3]# using:
# Delta = {0, 1*3/n, 2*3/n, ..., n*3/n } #

And so:

# I = int_0^3 \ (x^2-3x+2) \ dx # # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(0+i*(3-0)/n)# # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f((3i)/n)# # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {((3i)/n)^2-3((3i)/n)+2}# # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {(9i^2)/n^2-(9i)/n+2}#
# \ \ = lim_(n rarr oo) 3/n {sum_(i=1)^n (9i^2)/n^2 - sum_(i=1)^n (9i)/n + sum_(i=1)^n 2}#
# \ \ = lim_(n rarr oo) 3/n {9/n^2 sum_(i=1)^n i^2 - 9/n sum_(i=1)^n i + 2 sum_(i=1)^n 1}#

Using the standard summation formula:

# sum_(r=1)^n r \ = 1/2n(n+1) # # sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #

we have:

# I = lim_(n rarr oo) 3/n {9/n^2 1/6n(n+1)(2n+1) - 9/n 1/2n(n+1) + 2 n}#
# \ \ = lim_(n rarr oo) 3/n {3/(2n) (n+1)(2n+1) - 9/2 (n+1) + 2 n}#
# \ \ = lim_(n rarr oo) 3/n 1/(2n) {3 (n+1)(2n+1) - 9n (n+1) + 4n^2}#
# \ \ = lim_(n rarr oo) 3/(2n^2) {3 (2n^2+3n+1) - 9(n^2+1) + 4n^2}#
# \ \ = lim_(n rarr oo) 3/(2n^2) {6n^2+9n+3 -9n^2-9 + 4n^2}#
# \ \ = 3/2 lim_(n rarr oo) 1/(n^2) {n^2+9n -6}#
# \ \ = 3/2 lim_(n rarr oo) {1+9/n -6/n^2}#
# \ \ = 3/2 {1+0 -0}#
# \ \ = 3/2#

Using Calculus

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

# I = int_0^3 \ x^2-3x+2 \ dx # # \ \ = [x^3/3-(3x^2)/2+2x]_0^3 # # \ \ = (9-27/2+6) - (0) # # \ \ = 3/2 #
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Answer 2

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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Answer from HIX Tutor

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.

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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