Approximating Areas

Question

Estimate the area under the graph of f(x) = 25 - x^2 over the interval [-2,5] using five approximating rectangles and right endpoints.

Rn = ?

Repeat the approximation using left endpoints.

Ln = ?

*Report answers accurate to 4 places. Remember not to round too early in your calculations.

Repeat this with a graph of f(x) = 2x^2+8x+13 Over the interval [0,3] using four approximating rectangles and right endpoints. Repeat the approximation using left endpoints.

Answer 1

integral of 25-x^2 = 25x-x^3/3

evaluated from -2 to 5

= 25(5) - 125/3 - (-50 +8/3)= = 250/3+142/3

= 392/3 = 130 2/3 = Area under the curve

which the rectangles approximate

the integral is the exact area under the curve

which should be in between the right & left rectangle approximations

right end points of the 5 rectangles are 7/5-2 = -.6, .8, 2.2, 3.6, 5

rectangle heights are f(-.6), f(.8), f(2.2), f(3.6), f(5)

= 25-.36, 25-.64, 25- 2.2^2, 25-3.6^2, 25-25

= 24.64, 24.36, 20.16, 12.04.0

base of each of the 5 rectangles = 7/5 = 1.4

areas are: (24.64+24.36+20.16+12.04+0)(1.4)=85.2(1.4)

=119.28

left end points are 3.6, 2.2, .8, -.6 and -2

rectangle heights are f(-2), f(-.6), f(.8), f(2.2), f(3.6)

= 21, 24.64, 24.36, 20.16, 12.04

areas sum to (21+24.64+24.36+20.16+12.04)(1.4)

=106.2(1.4)

= 148.68

2x^2+8x+13 from 0 to 3

(2/3)x^3 + 4x^2 +13x evaluate from 0 to 3

=18 +36+39 -0 = 93

base of each of 4 rectangles = 3/4=.75

left ends are 0, 3/4, 3/2, 9/4

heights are f(0), f(3/4), f(3/2), f(9/4)

areas sum to (f(0)+f(3/4)+f(3/2)+f(9/4))(3/4)

=(13+ 2(3/4)^2+6+13+ 2(3/2)^2+12+13 + 2(9/4)^2+18+13)(3/4)

=(88+ 2(9/16 + 9/4+ 81/16)(3/4)

=(88 +2(126/16))(3/4)

= (88+126/8)(3/4)

= (22+63/16)(3)

= 66+ 189/16

=66+ 11 13/16

= 77 13/16

= 77.815

right end areas sum to (f(3/4)+f(3/2)+f(9/4)+f(3))(3/4)

=(88-13 + 126/8 + 2(9)+8(3)+13)(3/4)

= (88+63/4 + 18+24)(3/4)

=(130+63/4)(3/4)

= (65+ 63/8)(3/2)

=195/2 + 189/16

= 97.5 + 11 13/16

= 108.5 + 13/16

= 109.25 +.065

= 109.315