How do you use limits to evaluate #int8x^3dx# from [3,5]?

Question

Charles Anctil · Accepted Answer

#1088#

we need to use the power rule

#intx^ndx =x^(n+1)/(n+1)+c, " " n!=-1#

then apply the limits given

#int_3^5 8x^3dx#

#=8[ x^4/4]_3^5#

#=2[x^4]_3^5#

#=2{[x^4]^5-[x^4]_3}#

#2(5^4-3^4)#

#=2(625-81)#

#=1088#

Aurora Alvarado · Answer

# int_3^5 \ 8x^3 \ dx = 1088 #

By definition of an integral, then # int_a^b \ f(x) \ dx # represents the area under the curve #y=f(x)# between #x=a# and #x=b#. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles. That is # int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)# And we partition the interval #[a,b]# equally spaced using: # Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) } # # \ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b } # Here we have #f(x)=8x^3# and so we partition the interval #[3,5]# using: # Delta = {3, 3+2/n, 3+2 2/n, 3+3 2/n, ..., 5 } # And so: # I = int_3^5 \ 8x^3 \ dx # # \ \ = lim_(n rarr oo) 8*2/n sum_(i=1)^n \ f(3+i*2/n)# # \ \ = lim_(n rarr oo) 16/n sum_(i=1)^n \ (3+(2i)/n)^3 # # \ \ = lim_(n rarr oo) 16/n sum_(i=1)^n \ {(3)^3+3(3)^2((2i)/n)^1 + 3(3)((2i)/n)^2 + ((2i)/n)^3# # \ \ = lim_(n rarr oo) 16/n sum_(i=1)^n \ {27+54i/n\ + 36i^2/n^2 + 8i^3/n^3 }# # \ \ = lim_(n rarr oo) 16/n { sum_(i=1)^n 27 + sum_(i=1)^n 54i/n + sum_(i=1)^n36i^2/n^2 + sum_(i=1)^n8i^3/n^3 }# # \ \ = lim_(n rarr oo) 16/n { 27sum_(i=1)^n 1 + 54/n sum_(i=1)^n i + 36/n^2 sum_(i=1)^n i^2 + 8/n^3 sum_(i=1)^n i^3 }# Using the standard summation formula: # sum_(r=1)^n a \ = an # # sum_(r=1)^n r \ = 1/2n(n+1) # # sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) # # sum_(r=1)^n r^3 = 1/4n^2(n+1)^2 # we have: # I = 16 lim_(n rarr oo) 1/n { 27n + 54/n 1/2n(n+1) + 36/n^2 1/6n(n+1)(2n+1) + 8/n^3 1/4n^2(n+1)^2 # # \ \ = lim_(n rarr oo) 16/n { 27n + 27(n+1) + 6/n(n+1)(2n+1) + 2/n(n+1)^2 } # # \ \ = lim_(n rarr oo) 16/n^2 { 27n^2 + 27n(n+1) + 6(n+1)(2n+1) + 2(n+1)^2 } # # \ \ = lim_(n rarr oo) 16/n^2 { 27n^2 + 27n^2+27n +6(2n^2+3n+1) + 2(n^2+2n+1) } # # \ \ = lim_(n rarr oo) 16/n^2 { 27n^2 + 27n^2+27n +12n^2+18n+6 + 2n^2+4n+2 } # # \ \ = lim_(n rarr oo) 16/n^2 { 68n^2 + 49n+8 } # # \ \ = lim_(n rarr oo) { 1088 +784/n +8/n^2} # # \ \ = 1088 +0+0 # # \ \ = 1088 # Using Calculus If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get: # int_3^5 \ 8x^3 \ dx = [ 8x^4/4 ]_3^5 # # " " = 2(5^4-3^4) # # " " = 2(625-81 # # " " = 2 * 544 # # " " = 1088 #

Samantha Adkison · Answer

undefined To evaluate $ \int_{3}^{5} x^3 \, dx $ using limits, follow these steps:

1. Find the antiderivative of $ x^3 $, which is $ \frac{x^4}{4} $.
2. Evaluate the antiderivative at the upper and lower limits of integration.
3. Subtract the result of the antiderivative evaluated at the lower limit from the result of the antiderivative evaluated at the upper limit.
4. This will give you the value of the definite integral.