Use the limit process to find the area of the region between the graph of f(x) = 16 – x^2 and the x-axis over the interval [ 2,4 ] ?
1 Answer
int_2^4 16 - x^2 = 40/3
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_2^4 \ (16-x^2) \ dx
\ \ = lim_(n rarr oo) 2/n sum_(i=1)^n \ f(2+2*i/n)
\ \ = lim_(n rarr oo) 2/n sum_(i=1)^n \ f(2+(2i)/n)
\ \ = lim_(n rarr oo) 2/n sum_(i=1)^n \ (16-(2+(2i)/n)^2)
\ \ = lim_(n rarr oo) 2/n sum_(i=1)^n \ (16-(4+(8i)/n+(4i^2)/n^2))
\ \ = lim_(n rarr oo) 2/n sum_(i=1)^n \ (12-(8i)/n-(4i^2)/n^2)
\ \ = lim_(n rarr oo) 2/n {sum_(i=1)^n12 - sum_(i=1)^n(8i)/n - sum_(i=1)^n(4i^2)/n^2}
\ \ = lim_(n rarr oo) 2/n {sum_(i=1)^n12 - 8/nsum_(i=1)^n i - 4/n^2sum_(i=1)^n i^2}
Using the standard summation formula:
sum_(r=1)^n r \ = 1/2n(n+1)
sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1)
we have:
I = lim_(n rarr oo) 2/n {12n - 8/n 1/2n(n+1) - 4/n^2 1/6n(n+1)(2n+1)}
\ \ = lim_(n rarr oo) 2/n {n(12 - 4/n (n+1) - 2/3 1/n^2 (n+1)(2n+1))}
\ \ = lim_(n rarr oo) 2 {1/(3n^2)(36n^2 - 12n (n+1) - 2 (n+1)(2n+1))}
\ \ = lim_(n rarr oo) 2/(3n^2) {36n^2 - 12n^2-12 - 2 (2n^2+3n+1)}
\ \ = lim_(n rarr oo) 2/(3n^2) {36n^2 - 12n^2-12 - 4n^2-6n-2)}
\ \ = 2/3lim_(n rarr oo) 1/n^2{20n^2 - 6n-14)}
\ \ = 2/3lim_(n rarr oo) {20 - 6/n-14/n^2}
\ \ = 40/3
We can compare this result using calculus to evaluate the integral
int_2^4 16 - x^2 = [16x-x^3/3]_2^4
" " = (64-64/3)-(32-8/3)
" " = 64-64/3-32+8/3
" " = 40/3