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CALCULUS 1 PROBLEMS & SOLUTIONS

6.1.3
Areas As Limits Of Riemann Sums


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Review


1. Areas As Limits At Infinity Of Riemann Sums

We wish to compute the area of the plane region under the graph of a function and over a closed
interval, the function being non-negative and continuous on the interval.

Fig. 1 · y = (1/2)x + 1 over [–1, 3]. · n = 8.

We take an example. Let's find the area A of the plane region under the line y = (1/2)x + 1 and over
the interval [–1, 3], which is a trapezoid. See Fig. 1.

Using geometry, it is A = (1/2)(b₁ + b₂)h, where b₁ and b₂ are the bases and h is the height. Now,
b₁ = 1/2, b₂ = 5/2, and h = 4. So A = (1/2)(1/2 + 5/2)(4) = 6 square units.

Employing calculus, we proceed as follows. Let n be a positive integer. As shown in Fig. 1 with n = 8,
we approximate A by the Riemann Sum R_n with the regular partition of order n and using the left
endpoints. The length of each subinterval is (3 – (–1))/n = 4/n. The left endpoints are x₀ = –1, x₁ =
– 1 + 4/n, x₂ = – 1 + 2(4/n), ..., x_n–1 = – 1 + (n – 1)(4/n). Thus:



Now, if we increase n, say double it, then 4/n becomes smaller, hence the approximation R_n improves,
ie, gets closer to A, as illustrated in Fig. 2.

Fig. 2 · y = (1/2)x + 1 over [–1, 3]. · n = 16.

If we increase n further, then R_n gets closer to A still. Ultimately, if we let n ® ¥, then we obtain:



Now, if the geometric approach works and is simpler, why bother with the calculus approach? Well,
because the geometric approach works only with polygonal regions, ie, regions bounded by straight line
segments. If a region has a curved boundary, then the calculus approach is required, as shown in the
following example.

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2. Regions With Curved Boundaries

Fig. 3 Compute area under y = x^2 and over [1, 4].

Let's compute the area A of the plane region under the curve y = x² and over the interval [1, 4]. See
Fig 3. We utilize the Riemann Sum R_n with the regular partition of order n and using the right
endpoints. The length of each subinterval is (4 – 1)/n = 3/n. The right endpoints are x₁ = 1 + 3/n, x₂ =
1 + 2(3/n), ..., x_n = 1 + n(3/n) (= 4). So:



Thus, the area is A = lim_n®¥ R_n = 21 square units.

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Problems & Solutions


1. Compute the area of the plane region under the curve y = x² – 3x + 3 and over the interval [–1, 3].

Solution



Let f(x) = x² – 3x + 3. Let's use the Riemann Sum R_n with the regular partition of order n, where n is
a positive integer, and the right endpoints of the subintervals. The length of each subinterval is (3 –
(–1))/n = 4/n. The right endpoints are x₁ = – 1 + 4/n, x₂ = – 1 + 2(4/n), ..., x_n = – 1 + n(4/n) (= 3).
So:



Hence, the area is A = lim_n®¥ R_n = 28/3 square units.

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2. Find the area of the plane region under the curve y = x³ + 1 and over the interval [–1, 1].

Solution

Let f(x) = x³ + 1. Refer to Fig. 5. Let's use the Riemann Sum R_n with the regular partition of order n,
where n is a positive integer, and using the left endpoints. The length of each subinterval is
(1 – (–1))/n = 2/n. The left endpoints are x₀ = –1, x₁ = – 1 + 2/n, x₂ = – 1 + 2(2/n), ..., x_n–1 = – 1 +
(n – 1)(2/n). Thus:



It follows that the area is A = lim_n®¥ R_n = 2 square units.

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3. Probably you've accepted on faith that the area A of a circle of radius r is A = p r² although you've
never seen how it's proved. This Problem & Solution shows a way to do a proof of it. Let C be the
circumference of the circle. The number p is defined to be the ratio of the circumference and the
diameter of the circle: p = C/(2r). Thus, the circumference C is given by our old friend the formula
C = 2p r. The radian measure of the entire circle therefore is C/r = 2p.
a. In a circle of radius r, inscribe a regular polygon of n sides, where n ³ 3 is an integer. Prove that
the area A_n of this polygon is given by:



b. Prove that the area A of the circle is given by A = p r², observing that A = lim_n®¥ A_n. (Hence, it's
interesting to notice that the ratios A/r² of the area to the square of the radius and C/2r of the
circumference to the double of the radius are the same number, p.)

Solution

a. The area A₁ of triangle AOB is A₁ = (1/2)h(OB) = (1/2)hr = (1/2)(r sin a) r = (1/2)r² sin (2p/n).
In the above figure, we take n = 8 as an example. So the area A_n of the polygon is:



b. We have:



Now, lim_n®¥ r² cos (p/n) = r². For x Î R we have:



Thus, lim_n®¥ n sin (p/n) = p. Consequently, A = r² . p = p r².

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