1. ## Estimate the area

The problem asks: Estimate the area under the graph of f(x0 = 4x^2 + 4x + 4 from x=5 to x= 8.
using 3 approximating rectangles and right endpoints.
R3=?
Using 3 approximating rectangles and left endpoints.
L3=?
I under stand up to here:
4x^3/3-2x^2-4x
I know I am supposed to evaluate from 5 to 8 but that is as far as it goes.
Thank You,
Keith

2. Well, first find the co-ordinatdes where the graph crosses the x-axis.
y = 4x^2 + 4x + 4
4x^2 + 4x + 4 = 0 (0 is the y co-ordinate where the graph crosses the x-axis)
Using the discriminant b²-4ac = 4²-(4*4*4)
= -48 < 0 therefore the graph doesn't cross the x-axis at all. So to do this i would suggest drawing the graph of 4x^2 + 4x + 4. But it is still possible to do it without drawing the graph it just takes a little bit more thought of which ill divulge into tomorrow. It's late now .

3. See here.

4. Originally Posted by keith
The problem asks: Estimate the area under the graph of f(x0 = 4x^2 + 4x + 4 from x=5 to x= 8.
using 3 approximating rectangles and right endpoints.
R3=?
Using 3 approximating rectangles and left endpoints.
L3=?
I under stand up to here:
4x^3/3-2x^2-4x
I know I am supposed to evaluate from 5 to 8 but that is as far as it goes.
Thank You,
Keith
I don't know what the problem means but I can guess. It is about finding the approximate area under the parabola from x=5 to x=8 using 3 approximating rectangles. It is not through the usual Integral, from 5 to 8, of (4x^2 +4x +4)dx.

The approximating "dx" here is actually 1 unit wide. The 3 approximating rectangles are:
a) For R3 or right endpoints, f(6)high by 1 wide, f(7)high by 1 wide, and f(8) high by 1 wide.
b) For L3 or left endpoints, f(5)high by 1 wide, f(6)high by 1 wide, and f(7)high by 1 wide.

So,
f(x) = 4x^2 +4x +4
f(x) = 4(x^2 +x +1) --------***

R3 = 4[6^2 +6 +1]*1 +4[7^2 +7 +1]*1 +4[8^2 +8 +1]*1 = 692 sq.units. ----answer

L3 = 4[5^2 +5 +1]*1 +4[6^2 +6 +1]*1 +4[7^2 +7 +1]*1 = 524 sq.units. ----answer.

--------------------------
So, the approximate area under the curve from x=5 to x=8 is the average of R3 and L3,
= (R3 +L3)/2 = (692 +524)/2 = 608 sq.units ------***

If we use integration to find the exact area,
A = INT.(5-->8)[4x^2 +4x +4]dx
A = 4*INT.(5-->8)[x^2 +x +1] dx
A = 4[(x^3)/3 +(x^2)/2 +x]|(5-->8)
A = 4[{(8^3)/3 +(8^2)/2 +8} -{(5^3)/3 +(5^2)/2 +5}]
A = 4[151.5]
A = 606 sq.units exactly. ------------***

So the approximate 608 sq.units above is not bad.