How do you use the limit process to find the area of the region between the graph #y=x^2+2# and the x-axis over the interval [0,1]?

1 Answer
Nov 13, 2016

Answer:

Please see the explanation section below.

Explanation:

Here is a limit definition of the definite integral. (I'd guess it's the one you are using.) I will use what I think is somewhat standard notation in US textbooks.

.#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_0^1 (x^2+2) dx#.

Find #Delta x#

For each #n#, we get

#Deltax = (b-a)/n = (1-0)/n = 1/n#

Find #x_i#

And #x_i = a+iDeltax = 0+i1/n = i/n#

Find #f(x_i)#

#f(x_i) = (x_i)^2+2 = (i/n)^2+2#

# = i^2/n^2+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)Deltax = sum_(i=1)^n( i^2/n^2+2) 1/n#

# = sum_(i=1)^n(i^2/n^3+2/n)#

# =sum_(i=1)^n ( i^2)+sum_(i=1)^n(2/n) #

# =1/n^3 sum_(i=1)^n ( i^2/n^3)+2/nsum_(i=1)^n(1) #

Evaluate the sums

# = 1/n^3((n(n+1)(2n+1))/6) + 2/n (n) #

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

Rewrite before finding the limit

#sum_(i=1)^n f(x_i)Deltax = 1/6((n(n+1)(2n+1))/n^3) + 2#

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

Now we need to evaluate the limit as #nrarroo#.

#lim_(nrarroo) (n/n) = 1#

#lim_(nrarroo) ((n+1)/n) = 1#

#lim_(nrarroo) (2n+1)/n = 2#

To finish the calculation, we have

#int_0^1 (x^2 +2) dx = lim_(nrarroo) (1/6(n/n*(n+1)/n*(2n+1)/n) +2)#

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

# = 1/3 + 2 = 7/3#.