Question

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 ] ?

Answer 1

`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 `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)`

Here we have `f(x)=16-x^2` and we partition the interval `[2,4]` using `Delta = {2, 2+2*1/n, 2+2*2/n, ..., 2+2*n/n }`

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`