Question

(a) Determine Δx and xi. Δx= xi= (b) Using the definition mentioned above, evaluate the integral. Value of integral ?

(c) evaluate the integral. Value of integral ?

Answer 1

(a) `Delta x = 3/n`, and `x_i=1+(3i)/n`

(b) `int_(1)^4 \ x^2+2x-5 \ dx = 21`

Explanation:

Using Riemann sums. 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. The limit of which is known as the Riemann sum (or the limit definition of an integral)

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-a }`

Here we have `f(x)=x^2+2x-5` and so we partition the interval `[1,4]` using:

`Delta = {1, 1+1(3)/n, 1+2 (3)/n, 1+3 (3)/n, ..., 4 }`

And so:

`I = int_(1)^4 \ (x^2+2x-5) \ dx`

`\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(1+(3i)/n)`

Giving us:

`Delta x = 3/n`, and `x_i=1+(3i)/n`

Using the definition of f(x), we have:

`I = lim_(n rarr oo) 3/n sum_(i=1)^n \ {(1+(3i)/n)^2+2(1+(3i)/n)-5}`

`\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {1+(6i)/n+(9i^2)/n^2 +2+(6i)/n-5}`

`\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {(9i^2)/n^2 +(12i)/n-2}`

`\ \ = lim_(n rarr oo) 3/n {9/n^2 sum_(i=1)^n i^2+ 12/n sum_(i=1)^n i-sum_(i=1)^n2}`

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)`

So, we have:

`I = lim_(n rarr oo) 3/n {9/n^2 1/6n(n+1)(2n+1)+ 12/n 1/2n(n+1)-2n}`

`\ \ = lim_(n rarr oo) 3/n {3/(2n)(n+1)(2n+1)+ 6(n+1)-2n}`

`\ \ = lim_(n rarr oo) 3/(2n^2) {3(2n^2+3n+1)+ 12n(n+1)-4n^2}`

`\ \ = lim_(n rarr oo) 3/(2n^2) {6n^2+9n+3+ 12n^2+12n-4n^2}`

`\ \ = 3/2 lim_(n rarr oo) 1/n^2 {14n^2+21+3}`

`\ \ = 3/2 lim_(n rarr oo) {14+21/n+3/n^2}`

`\ \ = 3/2 {14+0+0}`

`\ \ = 21`

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

`I = int_(1)^4 \ x^2+2x-5 \ dx`
`\ \ = [x^3/3+x^2-5x]_(1)^4`
`\ \ = (64/3+16-20)-(1/3+1-5)`
`\ \ = 64/3-4 - 1/3+4`
`\ \ = 21`