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MATH 151 Engineering Math I, Spring 2014 JD Kim Week14 Section 6.2, 6.3 Section 6.2 Area Using Rectangles to Approximate the Area Under a Curve Let f (x) be a function defined on the interval [a, b]. We wish to approximate the area bounded by the curve f (x), the x-axis, x = a and x = b. We begin by partitioning the interval [a, b] into n smaller subintervals. We call P = {a = x0 , x1 , x2 , · · · , xn−1 , xn = b}, where a = x0 < x1 < x2 < · · · < xn−1 < xn = b, the partition points. For each subinterval [xi−1 , xi ], choose a representative point x∗i , that is x∗i is any point on the interval [xi−1 , xi ]. For each subinterval [xi−1 , xi ], we will construct a rectangle under the curve and above the x-axis, where the height of this rectangel is f (x∗i ) and the width is ∆xi = xi − xi−1 . The sum of the approximating rectangles gives an approximation under the graph n P f (x∗i )∆xi , of f (x) from x = a to x = b. Moreover, the area under the curve ≈ i=1 where n is the number of rectangles constructed. We call this sum a Riemann Sum. 1 Summary Subdividing the interval [a, b] into n smaller subintervals a = x0 < x1 < x2 < · · · < xn−1 < xn = b. Then the n subintervals are [x0 , x1 ], [x1 , x2 ], · · · , [xn−1 , xn ]. This subdivision is called a Partition of [a, b]. ∆xi = xi − xi−1 is the length of the ith subinterval [xi−1 , xi ]. The length of the longest subinterval is denoted by ||p|| and is called the norm of p, ||p|| = max{∆x1 , ∆x2 , · · · , ∆xn }. We define the area A as the limiting values (if it exists) of the areas of the approximating polygons, n X A = lim f (x∗i )∆xi ||p||→0 i=1 2 x∗i Ex1) A function f , an interval, partition points, and a description of the point within the ith subinterval are given, 1. Find ||p||. 2. Sketch the graph of f and the approximating rectangles. 3. Find the sum of the approximating rectangles. 1-1) f (x) = 16 − x2 , 1-2) f (x) = 4 cos x, [0, 4], p = {0, 1, 2, 3, 4}, h πi 0, , 2 x∗i = left endpoint. n π π π πo p = 0, , , , , 6 4 3 2 3 x∗i = right endpoint Ex2) For the following functions set up the limit of a Riemann Sum that represents the area under the graph of f (x) on the given interval. Do not evaluate the limit!. 2-1) f (x) = x2 + 3x − 2 on the interval [1, 4] using right endpoints. 2-2) f (x) = √ x2 + 1 on the interval [0, 5] using right endpoints. 4 Ex3) The following limits represent the area under the graph of f (x) from x = a to x = b. Identify f (x), a, and b. r n 3 3P 1+ i 3-1) lim n→∞ n i=1 n n 10 P n→∞ n i=1 3-2) lim 1 3 10 1+ 7+ i n 5 Section 6.3 The Definite Integral If f (x) ≥ 0 on the interval [a, b], the area under the curve of f (x), above the n P x-axis, from x = a to x = b is ≈ f (x∗i )∆xi . We call this sum a Riemann Sum. i=1 Ex4) If f (x) = x + x2 , the interval [0, 4], partition points p = {0, 1, 2, 3, 4}, and x∗i is left endpoint. Find the Riemann sum. If f (x) ≥ 0 on the interval [a, b], then the true area under the graph of f (x) from n P b−a and x∗i is any point on f (x∗i )∆xi , where ∆xi = x = a to x = b is A = lim n→∞ i=1 n the ith subinterval. We would like to define the limit of a Riemann Sum irregardless of whether the function is positive. To that end, we will define the definite integral. Definition The Definite Integral The Definite Integral of f (x) from x = a to x = b is Z a b f (x)dx = lim x→∞ n X f (x∗i )∆x i=1 b−a and x∗i is any point on the ith subinterval. In the event f (x) n is positive on the interval [a, b], then the definite integral is the same as the area bounded by f (x), the x-axis, x = a and x = b. If f (x) is not always positive on the interval [a, b], then the definite integral is the net area. where ∆x = 6 Remark Any Riemann Sum can approximate a definite integral. Specifically, the Midpoint Rule can be used to approximate the definite integral. Midpoint Rule Z a where ∆x = b f (x)dx ≈ n X f (x̄i )∆x, i=1 b−a and x̄i is the midpoint of the ith subinterval. n Ex5) Use the Midpoint Rule with n = 4 to approximate Ex6) Use Geometry to evaluate the following integrals. R3 6-1) 0 (1 − 2x)dx 7 R5√ 1 x2 + 1dx. 6-2) 6-3) R3 −1 |x − 2|dx R0 √ −2 4 − x2 dx 8 Theorem 1. Z b c dx = c(b − a) a 2. Z b cf (x) dx = c a 3. Z b f (x) dx a b a Z (f (x) ± g(x)) dx = 4. Z Z b a f (x) dx ± Z b g(x) dx a a f (x) dx = 0 a 5. 6. Z Z b a f (x) dx = − b f (x) dx = a Z Z a f (x) dx b c f (x) dx + a Z b f (x) dx c 7. If m ≤ f (x) ≤ M for all x in the interval [a, b], then m(b − a) ≤ Z a b f (x) dx ≤ M(b − a) 9 Ex7) Find Ex8) If R3 1 R1√ 1 x5 + x2 + 1 dx f (x) dx = 4 and Ex9) Write R5 f (x) dx − −3 R3 1 g(x) dx = −3, find R0 f (x) dx + −3 R6 5 Ex10) Find an upper and lower bound on 10 R3 1 (f (x) + 2g(x)) dx. f (x) dx as a single integral. R2√ 0 x3 + 1 dx Ex11) Express the following limits as a definite integral n 1P n→∞ n i=1 11-1) lim 1 2 i 1+ n ! 5 n 2i 2P 3 1+ −6 11-2) lim n→∞ n i=1 n 11