How do you use limits to evaluate #int (x+2)dx# from [1,4]?
2 Answers
Perform the integration as if the integral were indefinite but without a constant of integration.
Subtract the expression evaluated at the lower limit from the expression evaluated at the upper limit.
Explanation:
Given:
Perform the integration as if the integral were indefinite but without a constant of integration:
Subtract the resulting expression evaluated at the lower limit from the expression evaluated at the upper limit:
# int_1^4 \ (x+2) \ dx = 13.5#
Explanation:
By definition of an integral, then
# int_a^b \ f(x) \ dx #
represents the area under the curve
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)#
Here we have
And so:
# I = int_1^4 \ (x+2) \ dx #
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(1+3*i/n)#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ ((1+3*i/n)+2) #
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ (3+(3i)/n) #
# \ \ = lim_(n rarr oo) 3/n {sum_(i=1)^n 3+ sum_(i=1)^n (3i)/n }#
# \ \ = lim_(n rarr oo) 3/n {3n+ 3/nsum_(i=1)^n i }#
Using the standard summation formula:
# sum_(r=1)^n r = 1/2n(n+1) #
we have:
# I = lim_(n rarr oo) 3/n {3n+ 3/n*1/2n(n+1) }#
# \ \ = lim_(n rarr oo) 3/n {3n+ 3/2(n+1) }#
# \ \ = lim_(n rarr oo) 3/n*3/2 {2n+ (n+1) }#
# \ \ = lim_(n rarr oo) 9/(2n) (3n+1)#
# \ \ = lim_(n rarr oo) (27/2+9/(2n)) #
# \ \ = 27/2#
# \ \ = 13.5#