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

2 Answers
Sep 13, 2017

Answer:

#1088#

Explanation:

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#

Sep 13, 2017

Answer:

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

Explanation:

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 #