Area under the curve of x^3 from -1 to 3 with Summation?

I have already done the problem using the Fundamental Theorem of Calculus and I got an answer of 20. Now how would I solve it using Sigma Notation. Thanks in advance to anyone who responds

1 Answer
Jim H
Nov 27, 2017

Please see below.

Explanation:

#Deltax = 4/n#

#x_i = -1+iDelta x = -1+(4i)/n#

#f(x_i) = (-1+(4i)/n)^3 = -1+3((4i)/n)-3((4i)/n)^2+((4i)/n)^3#

# = (64i^3)/n^3-(48i^2)/n^2+(12i)/n-1#

#sum_(i=1)^n f(x_i)Deltax = sum_(i=1)^n ((64i^3)/n^3-(48i^2)/n^2+(12i)/n-1)(4/n)#

# = sum_(i=1)^n((256i^3)/n^4-(192i^2)/n^3+(48i)/n^2-(4)/n)#

# = 256/n^4 sum_(i=1)^n i^3 - 192/n^3 sum_(i=1)^n i^2 + 48/n^2 sum_(i=1)^n i - 4/n sum_(i=1)^n 1#

# = 256/n^4 (n^2(n+1)^2)/4 - 192/n^3 ((n(n+1)(2n+1))/6) + 48/n^2 ((n(n+1))/2) - 4/n (n)#

# = 256/4 (n^2(n+1)^2)/n^4 - 192/6 ((n(n+1)(2n+1))/n^3) + 48/2 ((n(n+1))/n^2) - 4#

#lim_(nrarroo)sum_(i=1)^n f(x_i)Deltax = 256/4(1)-192/6(2)+48/2(1) -4#

# = 64-64+24-4=20#

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