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]?

1 Answer
May 19, 2017

Answer:

# 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 #