# Create a grid to provide upper and lower bounds on the area under the curve y=4x-x^2 using 8 rectangles of equal width for each bound

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Expert Answers

rcmath | Certified Educator

Let's start by finding the x-intercepts.

`4x-x^2=0=>x(4-x)=0=>x=0 or x=4`

If we graph the function we will notice that the area under the curve and above the x-axis is between the two intercept. Hence our interval is [0,4]. We need to divide that into 8 equal parts, thus `Deltax=4/8=0.5`

Let's find the area by using f(0) for the height of the first rectangle, etc.

`A_1=0.5(f(0)+f(.5)+f(1)+f(1.5)+f(2)+f(2.5)+f(3)+f(3.5))=>`

`A_1=0.5(0+1.75+3+3.75+4+3.75+3+1.75)=>`

`A_1=0.5*21=10.5`

`A_2=0.5*(f(0.5)+f(1)+f(1.5)+f(2)+f(2.5)+f(3)+f(3.5)+f(4))=>`

`A_2=0.5(21+0)=10.5`

Given the symetric nature of the curve over the interval [0,4], we ended up with the lower and upper bound equal.