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# Approximating the Area Under a Curve for AP Calculus

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By — McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these concepts can be found at: Areas and Volumes Practice Problems for AP Calculus

### Rectangular Approximations

If f ≥ 0, the area under the curve of f can be approximated using three common types of rectangles: left-endpoint rectangles, right-endpoint rectangles, or midpoint rectangles. (See Figure 12.2-1.)

The area under the curve using n rectangles of equal length is approximately:

If f is increasing on [a, b], then left-endpoint rectangles are inscribed rectangles and the right-endpoint rectangles are circumscribed rectangles. If f is decreasing on [a, b], then left-endpoint rectangles are circumscribed rectangles and the right-endpoint rectangles are inscribed. Furthermore,

inscribed rectangle ≤ area under the curve ≤ circumscribed rectangle.

#### Example 1

Find the approximate area under the curve of f (x ) = x2 +1 from x =0 to x =2, using 4 left-endpoint rectangles of equal length. (See Figure 12.2-2.)

Let Δxi be the length of ith rectangle. The length .

Area under the curve .

Enter Σ (((0.5(x ≤ 1))2 +1) * 0.5, x, 1, 4) and obtain 3.75.

Or, find the area of each rectangle:

Area of RectI = (f(0))Δx1 = (1) = .
Area of RectII = (f(0.5))Δx2 = ((0.5)2 + 1) = 0.625.
Area of RectIII = (f(1))Δx3 = ((12 + 1) = 1.
Area of RectIV = (f(1.5))Δx4 =((1.5)2 + 1) = 1.625.

Area of (RectI + RectII + RectIII + RectIV) = 3.75.

Thus, the approximate area under the curve of f(x ) is 3.75.

#### Example 2

Find the approximate area under the curve of f (x = from x = 4 to x = 9 using 5 rightendpoint rectangles. (See Figure 12.2-3.)

Let Δxi be the length of ith rectangle. The length Δxi = = 1; xi = 4 + (1)i = 4 + i.

Or, using Σ notation:

Thus the area under the curve is approximately 13.160.

#### Example 3

The function f is continuous on [1, 9] and f > 0. Selected values of f are given below:

Using 4 midpoint rectangles, approximate the area under the curve of f for x =1 to x =9. (See Figure 12.2-4.)

Let Δxi be the length of ith rectangle. The length Δxi = 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.

### Trapezoidal Approximations

Another method of approximating the area under a curve is to use trapezoids. See Figure 12.2-5.

#### Formula for Trapezoidal Approximation

If f is continuous, the area under the curve of f from x = a to x = b is:

#### Example 1

Find the approximate area under the curve of from x = 0 to x = π, using 4 trapezoids. (See Figure 12.2-6.)

Since n = 4,

Area under the curve:

Practice problems for these concepts can be found at: Areas and Volumes Practice Problems for AP Calculus

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