How do you use the limit process to find the area of the region between the graph #y=2x-x^3# and the x-axis over the interval [0,1]?

1 Answer
Nov 22, 2016

Answer:

See below.

Explanation:

Here is a limit definition of the definite integral. (I'd guess it's the one you are using.)

.#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 (2x-x^3) 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) = 2(x_i) - (x_i)^3 = (2i)/n - (i/n)^3 = 2i/n - i^3/n^3#

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( 2i/n - i^3/n^3) 1/n#

# = sum_(i=1)^n( 2i/n^2-i^3/n^4)#

# =sum_(i=1)^n ( 2i/n^2) - sum_(i=1)^n(i^3/n^4) #

# = 2/n^2sum_(i=1)^n(i)-1/n^4sum_(i=1)^n(i^3) #

Evaluate the sums

# = 2/n^2((n(n+1))/2) - 1/n^4( (n^2(n+1)^2)/4) #

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

Rewrite before finding the limit

#sum_(i=1)^n f(x_i)Deltax = 2/n^2((n(n+1))/2) - 1/n^4( (n^2(n+1)^2)/4) #

# = 2/2 ((n(n+1))/n^2) - 1/4((n^2(n+1)^2)/n^4)#

# = ((n(n+1))/n^2) - 1/4((n^2(n+1)^2)/n^4)#

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

#lim(nrarroo) ((n(n+1))/n^2) = 1#

#lim(nrarroo) ((n^2(n+1)^2)/n^4) = 1#

To finish the calculation, we have

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

# = (1)-1/4(1) = 3/4#