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

Question

Isabella Ackerman · Accepted Answer

Please see below.

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#

Ezekiel Alexander · Answer

undefined To find the area between the curve $y = x^2$ and the x-axis from $x = 0$ to $x = 5$, you can integrate the absolute value of the function $y = x^2$ over the interval $[0, 5]$. This is because the function $y = x^2$ is always non-negative over this interval, so the area will be the same as the integral of $|x^2|$ over the interval $[0, 5]$. Therefore, the area can be calculated as follows:

$$A = \int_{0}^{5} |x^2| \, dx$$

$$= \int_{0}^{5} x^2 \, dx$$

$$= \left[\frac{x^3}{3}\right]_{0}^{5}$$

$$= \frac{5^3}{3} - \frac{0^3}{3}$$

$$= \frac{125}{3}$$

So, the area between the curve $y = x^2$ and the x-axis from $x = 0$ to $x = 5$ is $ \frac{125}{3} $ square units.