How do you integrate #(2x+1)dx# with b=3,a=1 using reimann sums?

Answer 1

I'm going to assume that you want to evaluate the integral, not simply get an approximation.

You didn't specify, so I'll use right endpoint in my calculations.

Also, you may be required to write equalities all the way through, for students learning this, I think it's more clear to evaluate the pieces, then the sum and finally evaluate the limit.

#int_1^3 (2x+1)dx = lim_(nrarroo) sum_(i=1)^n f(x_i) Delta x#
For each #n#, we have:
#Delta x = (b-a)/n = 2/n#
#x_i = a+iDeltax=1+(2i)/n #
#f(x_i) = 2x_i+1 = 2(1+(2i)/n)+1=(4i)/n+3#
#sum_(i=1)^n f(x_i) Delta x = sum_(i=1)^n ((4i)/n+3) 2/n = sum_(i=1)^n ((8i)/n^2+6/n)#

So the sum is:

#sum_(i=1)^n f(x_i) Delta x = sum_(i=1)^n (8i)/n^2+sum_(i=1)^n 6/n#.

And finally:

#sum_(i=1)^n f(x_i) Delta x = 8/n^2sum_(i=1)^n i + 6/n sum_(i=1)^n 1#

Apply the formulas for these sums to get:

#sum_(i=1)^n f(x_i) Delta x = 8/n^2 (n(n+1))/2 + 6/n (n)#.
In preparation for evaluating the limit as #nrarroo#, rewrite this as:
#sum_(i=1)^n f(x_i) Delta x = (4 n)/n (n+1)/n + (6n)/n = 4(1+1/n)+6#
Evaluate the limit as #nrarroo#:
#int_1^3 (2x+1)dx = lim_(nrarroo) sum_(i=1)^n f(x_i) Delta x#
#color(white)"ssssssssssssssss"# # = lim_(nrarroo) sum_(i=1)^n ((4i)/n+3) 2/n#
#color(white)"ssssssssssssssss"# # = lim_(nrarroo) (4(1+1/n)+6)#
#color(white)"ssssssssssssssss"# # = 4+6 = 10#
#int_1^3 (2x+1)dx = 10#
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Answer 2

To integrate ( (2x+1) , dx ) with ( a = 1 ) and ( b = 3 ) using Riemann sums, you can approximate the integral by partitioning the interval ([1,3]) into ( n ) subintervals of equal width ( \Delta x ). Then, choose sample points within each subinterval and evaluate the function at those points. Finally, sum up the products of the function values and the subinterval widths, and take the limit as ( n ) approaches infinity to obtain the integral.

The Riemann sum for the integral ( \int_{1}^{3} (2x+1) , dx ) is given by:

[ \sum_{i=1}^{n} f(x_i) \Delta x ]

Where ( f(x) = 2x+1 ), ( \Delta x = \frac{b-a}{n} ), and ( x_i ) are the sample points in the ( i )th subinterval.

Substituting the values ( a = 1 ), ( b = 3 ), and ( f(x) = 2x+1 ), we get:

[ \Delta x = \frac{3-1}{n} = \frac{2}{n} ]

[ x_i = 1 + i \cdot \Delta x = 1 + i \cdot \frac{2}{n} ]

Thus, the Riemann sum becomes:

[ \sum_{i=1}^{n} \left(2\left(1 + i \cdot \frac{2}{n}\right) + 1\right) \cdot \frac{2}{n} ]

Now, you can simplify and compute this sum, and then take the limit as ( n ) approaches infinity to find the value of the integral.

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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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