Question

How do you use the limit process to find the area of the region between the graph `y=-2x+3` and the x-axis over the interval [0,1]?

Answer 1

`int_0^1 \ -2x+3 \ dx = 2`

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)=-2x+3` and we partition the interval `[0,1]` using:

`Delta = {0, 0+1*1/n, 0+1*2/n, ..., 0+1*n/n }`
`\ \ \ = {0, 1/n, 2/n, ..., n/n }`

And so:

`I = int_0^1 \ (-2x+3) \ dx`
`\ \ = lim_(n rarr oo) 1/n sum_(i=1)^n \ f(0+1*i/n)`
`\ \ = lim_(n rarr oo) 1/n sum_(i=1)^n \ f(i/n)`
`\ \ = lim_(n rarr oo) 1/n sum_(i=1)^n \ -2(i/n)+3`
`\ \ = lim_(n rarr oo) 1/n sum_(i=1)^n \ (-2i)/n+3`
`\ \ = lim_(n rarr oo) 1/n {-2/n sum_(i=1)^ni +sum_(i=1)^n3}`

Using the standard summation formula:

`sum_(r=1)^n r = 1/2n(n+1)`

we have:

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

Using Calculus

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

`int_0^1 \ -2x+3 \ dx = [ -x^2+3x ]_0^1`
`" " = (-1+3)-(0+0)`
`" " = 2`