How do you use limits to find the area between the curve #y=x^2# and the x axis from [0,5]?

1 Answer
Oct 22, 2017

Answer:

Please see below.

Explanation:

Here is a limit definition of the area (actually for the definite integral). I will use what I think is frequently used 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.)
We'll do one small step at a time.

#int_0^5 x^2 dx#.

Find #Delta x#

For each #n#, we get

#Deltax = (b-a)/n = (5-0)/n = 5/n#

Find #x_i#

And #x_i = a+iDeltax = 0+i5/n = (5i)/n#

Find #f(x_i)#

#f(x_i) = {:x_i:}^2 = ((5i)/n)^2#

# = (25i^2)/n^2#

Find and simplify #sum_(i=1)^n f(x_i)Deltax # in order to evaluate the sum.

#sum_(i=1)^n f(x_i)Deltat = sum_(i=1)^n ((25i^2)/n^2) 5/n#

# = sum_(i=1)^n( (125i^2)/n^3)#

# = 125/n^2 sum_(i=1)^n(i^2)#

Evaluate the sum

# = 125/n^3((n(n+1)(2n+1))/6)#

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

Rewrite before finding the limit

#sum_(i=1)^n f(x_i)Deltax = 125/6((n(n+1)(2n+1))/n^3)#

# = 125/6((n(n+1)(2n+1))/n^3)#

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

#lim_(nrarroo) ((n(n+1)(2n+1))/n^3)= lim_(nrarroo) (n/n (n+1)/n (2n+1)/n)#

# = (1)(1)(2) = 2#

To finish the calculation, we have

#int_0^5 x^2dx = lim_(nrarroo)(125/6((n(n+1)(2n+1))/n^3))#

# = 125/6(2) = 125/3#