How do you evaluate the definite integral by the limit definition given #int x^2+1dx# from [1,2]?

Answer 1

Please see the explanation section below.

Here is a limit definition of the definite integral.

.#int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax#.
Where, for each positive integer #n#, we let #Deltax = (b-a)/n#
And for #i=1,2,3, . . . ,n#, we let #x_i = a+iDeltax#. (These #x_i# are the right endpoints of the subintervals.)

I prefer to do this type of problem one small step at a time.

#int_1^2 (x^2+1) dx#.
Find #Delta x#
For each #n#, we get
#Deltax = (b-a)/n = (2-1)/n = 1/n#
Find #x_i#
And #x_i = a+iDeltax = 1+i1/n = 1+i/n#
Find #f(x_i)#
#f(x_i) = x_icolor(white)()^2 =(1+i/n)^2+1#
# = 2+(2i)/n+i^2/n^2#
Find and simplify #sum_(i=1)^n f(x_i)Deltax # in order to evaluate the sums.
#sum_(i=1)^n f(x_i)Deltat = sum_(i=1)^n ( 2+(2i)/n+i^2/n^2) 1/n#
# = sum_(i=1)^n( 2/n+(2i)/n^2+i^2/n^3)#
# = 2/nsum_(i=1)^n (1) +2/n^2sum_(i=1)^n (i)+1/n^3 sum_(i=1)^n (i^2)#

Evaluate the sums

# = 2/n(n)+2/n^2((n(n+1))/2) + 1/n ((n(n+1)(2n+1))/6)#

(We used a summation formula for the sums in the previous step.)

Rewrite before finding the limit

#sum_(i=1)^n f(x_i)Deltax = 2/n(n)+2/n^2((n(n+1))/2) + 1/n ((n(n+1)(2n+1))/6)#
# = 2+ ((n(n+1))/n^2) + 1/6 ((n(n+1)(2n+1))/n^3) #
Now we need to evaluate the limit as #nrarroo#.
#lim_(nrarroo) ((n(n+1))/n^2) = lim_(nrarroo) (n/n*(n+1)/n) = 1#
and #lim_(nrarroo) ((n(n+1)(2n+1))/n^3) = lim_(nrarroo) (n/n*(n+1)/n (2n+1)/n) = 2#

To finish the calculation, we have

#int_0^4 (x^2+1) dx = lim_(nrarroo) ( 2+ ((n(n+1))/n^2) + 1/6 ((n(n+1)(2n+1))/n^3))#
# = 2+(1)+1/6(2) = 10/3#
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Answer 2

To evaluate the definite integral (\int_{1}^{2} (x^2 + 1) , dx) using the limit definition:

  1. Start with the definite integral: (\int_{1}^{2} (x^2 + 1) , dx).
  2. Break the integral into smaller intervals: (\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x), where (f(x_i) = x_i^2 + 1) and (\Delta x = \frac{b-a}{n}), with (a = 1) and (b = 2).
  3. Compute (\Delta x = \frac{2-1}{n} = \frac{1}{n}).
  4. Choose sample points within each interval. One common approach is to use the right endpoint of each interval. So, (x_i = 1 + i \Delta x) for (i = 0, 1, 2, ..., n-1).
  5. Substitute the expression for (f(x_i)) into the sum: (\lim_{n \to \infty} \sum_{i=1}^{n} (1 + (1 + i\Delta x)^2)\Delta x).
  6. Simplify the expression inside the sum: (\lim_{n \to \infty} \sum_{i=1}^{n} (1 + 1 + 2i\Delta x + i^2(\Delta x)^2)\Delta x).
  7. Expand and compute the sum: (\lim_{n \to \infty} \sum_{i=1}^{n} (1 + 1 + 2i\frac{1}{n} + i^2(\frac{1}{n})^2)\frac{1}{n}).
  8. Simplify and evaluate the sum: (\lim_{n \to \infty} \sum_{i=1}^{n} (1 + \frac{2i}{n} + \frac{i^2}{n^2})\frac{1}{n}).
  9. Evaluate the limit of the sum as (n) approaches infinity.
  10. The result of the limit is the value of the definite integral (\int_{1}^{2} (x^2 + 1) , dx).
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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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