5 Steps to a 5 AP Calculus BC 2019

Areas, Volumes, and Arc Lengths 265

Or, using

∑

notation:

∑^5

i= 1

f(xi)Δxi=

∑^5

i= 1

f(4+i)(1)=

∑^5

i= 1

√

4 + 1.

Enter
∑(√
( 4 +x),x,1,5

)

and obtain 13.160.

Thus the area under the curve is approximately 13.160.

Example 3

The functionfis continuous on [1, 9] and f >0. Selected values offare given below:

x 1 2 3 4 5 6 7 8 9

f(x) 1 1.41 1.73 2 2.37 2.45 2.65 2.83 3

Using 4 midpoint rectangles, approximate the area under the curve off forx=1tox=9.
(See Figure 12.2-4.)

I

II

III IV

10

1

2

3

2 3 45 67 89

y

x

f

Figure 12.2-4

LetΔxibe the length ofith rectangle. The lengthΔxi=

9 − 1

4

= 2.

Area of RectI= f(2)(2)=(1.41)2= 2. 82.

Area of RectII= f(4)(2)=(2)2= 4.

Area of RectIII= f(6)(2)=(2.45)2= 4. 90.

Area of RectIV= f(8)(2)=(2.83)2= 5. 66.

Area of (RectI+RectII+RectIII+RectIV)= 2. 82 + 4 + 4. 90 + 5. 66 = 17 .38.

Thus the area under the curve is approximately 17.38.