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Matemáticas 5 · Tema 19

Suma de Riemann

Apuntes

La integral definida de una función continua f(x)f(x) en el intervalo cerrado [a, b][a,\,b] es el límite del área bajo la curva calculada con rectángulos, cuando el número de divisiones tiende a infinito:

∫abf(x) dx=lim⁡n→∞∑i=1nf(xi) Δx\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \, \Delta x

donde:

Δx=b−anxi=a+Δx⋅i\Delta x = \frac{b - a}{n} \qquad x_i = a + \Delta x \cdot i
Sumatoria de i2i^2
∑i=1ni2=n33+n22+n6=n(n+1)(2n+1)6\displaystyle \sum_{i=1}^{n} i^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} = \frac{n(n+1)(2n+1)}{6}
Sumatoria de ii
∑i=1ni=n22+n2=n(n+1)2\displaystyle \sum_{i=1}^{n} i = \frac{n^2}{2} + \frac{n}{2} = \frac{n(n+1)}{2}
Sumatoria de 11
∑i=1n1=n\displaystyle \sum_{i=1}^{n} 1 = n

Ejemplo desarrollado

Calcular con sumas de Riemann: ∫−23(x2+4) dx\displaystyle \int_{-2}^{3} (x^2 + 4) \, dx

1. Los componentes
Δx=3−(−2)n=5nxi=−2+5ni\begin{aligned} \Delta x &= \frac{3 - (-2)}{n} = \frac{5}{n} \\ x_i &= -2 + \frac{5}{n}i \end{aligned}
2. Evaluar f(xi)f(x_i) y armar f(xi)Δxf(x_i)\Delta x
f(xi)=(−2+5ni)2+4=4−20ni+25n2i2+4=25n2i2−20ni+8f(xi)Δx=(25n2i2−20ni+8)5n=125n3i2−100n2i+40n\begin{aligned} f(x_i) &= \left( -2 + \frac{5}{n}i \right)^2 + 4 \\ &= 4 - \frac{20}{n}i + \frac{25}{n^2}i^2 + 4 \\ &= \frac{25}{n^2}i^2 - \frac{20}{n}i + 8 \\ f(x_i)\Delta x &= \left( \frac{25}{n^2}i^2 - \frac{20}{n}i + 8 \right) \frac{5}{n} \\ &= \frac{125}{n^3}i^2 - \frac{100}{n^2}i + \frac{40}{n} \end{aligned}
3. Sumatoria y límite al infinito
∑i=1nf(xi)Δx=125n3∑i=1ni2−100n2∑i=1ni+40n∑i=1n1=125n3(n33+n22+n6)−100n2(n22+n2)+40n(n)=1253+1252n+1256n2−50−50n+40=953+1256n2+252nlim⁡n→∞∑i=1nf(xi)Δx=953≈31.67\begin{aligned} \sum_{i=1}^{n} f(x_i)\Delta x &= \frac{125}{n^3}\sum_{i=1}^{n}i^2 - \frac{100}{n^2}\sum_{i=1}^{n}i + \frac{40}{n}\sum_{i=1}^{n}1 \\ &= \frac{125}{n^3}\left( \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right) - \frac{100}{n^2}\left( \frac{n^2}{2} + \frac{n}{2} \right) + \frac{40}{n}(n) \\ &= \frac{125}{3} + \frac{125}{2n} + \frac{125}{6n^2} - 50 - \frac{50}{n} + 40 \\ &= \frac{95}{3} + \frac{125}{6n^2} + \frac{25}{2n} \\ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x &= \mathbf{\frac{95}{3}} \approx \mathbf{31.67} \end{aligned}

Ejercicios

Bloque 1
∫−14(x2+5) dx\displaystyle \int_{-1}^{4} (x^2 + 5) \, dx
∫58(4+3x2) dx\displaystyle \int_{5}^{8} (4 + 3x^2) \, dx
∫−25(x2−6) dx\displaystyle \int_{-2}^{5} (x^2 - 6) \, dx
Bloque 2
∫26(3−x2) dx\displaystyle \int_{2}^{6} (3 - x^2) \, dx
∫03(9−x2) dx\displaystyle \int_{0}^{3} (9 - x^2) \, dx
∫14(5−2x2) dx\displaystyle \int_{1}^{4} (5 - 2x^2) \, dx

Respuestas

▼
Bloque 1
Ejercicio 1 — ∫−14(x2+5) dx\displaystyle \int_{-1}^{4} (x^2 + 5) \, dx
Δx=4−(−1)n=5nxi=−1+5nif(xi)=(−1+5ni)2+5=25n2i2−10ni+6f(xi)Δx=(25n2i2−10ni+6)5n=125n3i2−50n2i+30n∑i=1nf(xi)Δx=125n3(n33+n22+n6)−50n2(n22+n2)+30n(n)=1253+1252n+1256n2−25−25n+30=1403+1256n2+752nlim⁡n→∞∑i=1nf(xi)Δx=1403≈46.67\begin{aligned} \Delta x &= \frac{4 - (-1)}{n} = \frac{5}{n} \\ x_i &= -1 + \frac{5}{n}i \\ f(x_i) &= \left( -1 + \frac{5}{n}i \right)^2 + 5 = \frac{25}{n^2}i^2 - \frac{10}{n}i + 6 \\ f(x_i)\Delta x &= \left( \frac{25}{n^2}i^2 - \frac{10}{n}i + 6 \right) \frac{5}{n} = \frac{125}{n^3}i^2 - \frac{50}{n^2}i + \frac{30}{n} \\ \sum_{i=1}^{n} f(x_i)\Delta x &= \frac{125}{n^3}\left( \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right) - \frac{50}{n^2}\left( \frac{n^2}{2} + \frac{n}{2} \right) + \frac{30}{n}(n) \\ &= \frac{125}{3} + \frac{125}{2n} + \frac{125}{6n^2} - 25 - \frac{25}{n} + 30 \\ &= \frac{140}{3} + \frac{125}{6n^2} + \frac{75}{2n} \\ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x &= \mathbf{\frac{140}{3}} \approx \mathbf{46.67} \end{aligned}
Ejercicio 2 — ∫58(4+3x2) dx\displaystyle \int_{5}^{8} (4 + 3x^2) \, dx
Δx=8−5n=3nxi=5+3nif(xi)=4+3(5+3ni)2=27n2i2+90ni+79f(xi)Δx=(27n2i2+90ni+79)3n=81n3i2+270n2i+237n∑i=1nf(xi)Δx=81n3(n33+n22+n6)+270n2(n22+n2)+237n(n)=27+812n+272n2+135+135n+237=399+272n2+3512nlim⁡n→∞∑i=1nf(xi)Δx=399\begin{aligned} \Delta x &= \frac{8 - 5}{n} = \frac{3}{n} \\ x_i &= 5 + \frac{3}{n}i \\ f(x_i) &= 4 + 3\left( 5 + \frac{3}{n}i \right)^2 = \frac{27}{n^2}i^2 + \frac{90}{n}i + 79 \\ f(x_i)\Delta x &= \left( \frac{27}{n^2}i^2 + \frac{90}{n}i + 79 \right) \frac{3}{n} = \frac{81}{n^3}i^2 + \frac{270}{n^2}i + \frac{237}{n} \\ \sum_{i=1}^{n} f(x_i)\Delta x &= \frac{81}{n^3}\left( \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right) + \frac{270}{n^2}\left( \frac{n^2}{2} + \frac{n}{2} \right) + \frac{237}{n}(n) \\ &= 27 + \frac{81}{2n} + \frac{27}{2n^2} + 135 + \frac{135}{n} + 237 \\ &= 399 + \frac{27}{2n^2} + \frac{351}{2n} \\ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x &= \mathbf{399} \end{aligned}
Ejercicio 3 — ∫−25(x2−6) dx\displaystyle \int_{-2}^{5} (x^2 - 6) \, dx
Δx=5−(−2)n=7nxi=−2+7nif(xi)=(−2+7ni)2−6=49n2i2−28ni−2f(xi)Δx=(49n2i2−28ni−2)7n=343n3i2−196n2i−14n∑i=1nf(xi)Δx=343n3(n33+n22+n6)−196n2(n22+n2)−14n(n)=3433+3432n+3436n2−98−98n−14=73+3436n2+1472nlim⁡n→∞∑i=1nf(xi)Δx=73≈2.33\begin{aligned} \Delta x &= \frac{5 - (-2)}{n} = \frac{7}{n} \\ x_i &= -2 + \frac{7}{n}i \\ f(x_i) &= \left( -2 + \frac{7}{n}i \right)^2 - 6 = \frac{49}{n^2}i^2 - \frac{28}{n}i - 2 \\ f(x_i)\Delta x &= \left( \frac{49}{n^2}i^2 - \frac{28}{n}i - 2 \right) \frac{7}{n} = \frac{343}{n^3}i^2 - \frac{196}{n^2}i - \frac{14}{n} \\ \sum_{i=1}^{n} f(x_i)\Delta x &= \frac{343}{n^3}\left( \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right) - \frac{196}{n^2}\left( \frac{n^2}{2} + \frac{n}{2} \right) - \frac{14}{n}(n) \\ &= \frac{343}{3} + \frac{343}{2n} + \frac{343}{6n^2} - 98 - \frac{98}{n} - 14 \\ &= \frac{7}{3} + \frac{343}{6n^2} + \frac{147}{2n} \\ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x &= \mathbf{\frac{7}{3}} \approx \mathbf{2.33} \end{aligned}
Bloque 2
Ejercicio 4 — ∫26(3−x2) dx\displaystyle \int_{2}^{6} (3 - x^2) \, dx
Δx=6−2n=4nxi=2+4nif(xi)=3−(2+4ni)2=−16n2i2−16ni−1f(xi)Δx=(−16n2i2−16ni−1)4n=−64n3i2−64n2i−4n∑i=1nf(xi)Δx=−64n3(n33+n22+n6)−64n2(n22+n2)−4n(n)=−643−32n−323n2−32−32n−4=−1723−323n2−64nlim⁡n→∞∑i=1nf(xi)Δx=−1723≈−57.33\begin{aligned} \Delta x &= \frac{6 - 2}{n} = \frac{4}{n} \\ x_i &= 2 + \frac{4}{n}i \\ f(x_i) &= 3 - \left( 2 + \frac{4}{n}i \right)^2 = -\frac{16}{n^2}i^2 - \frac{16}{n}i - 1 \\ f(x_i)\Delta x &= \left( -\frac{16}{n^2}i^2 - \frac{16}{n}i - 1 \right) \frac{4}{n} = -\frac{64}{n^3}i^2 - \frac{64}{n^2}i - \frac{4}{n} \\ \sum_{i=1}^{n} f(x_i)\Delta x &= -\frac{64}{n^3}\left( \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right) - \frac{64}{n^2}\left( \frac{n^2}{2} + \frac{n}{2} \right) - \frac{4}{n}(n) \\ &= -\frac{64}{3} - \frac{32}{n} - \frac{32}{3n^2} - 32 - \frac{32}{n} - 4 \\ &= -\frac{172}{3} - \frac{32}{3n^2} - \frac{64}{n} \\ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x &= \mathbf{-\frac{172}{3}} \approx \mathbf{-57.33} \end{aligned}
Ejercicio 5 — ∫03(9−x2) dx\displaystyle \int_{0}^{3} (9 - x^2) \, dx
Δx=3−0n=3nxi=3nif(xi)=9−(3ni)2=−9n2i2+9f(xi)Δx=(−9n2i2+9)3n=−27n3i2+27n∑i=1nf(xi)Δx=−27n3(n33+n22+n6)+27n(n)=−9−272n−276n2+27=18−92n2−272nlim⁡n→∞∑i=1nf(xi)Δx=18\begin{aligned} \Delta x &= \frac{3 - 0}{n} = \frac{3}{n} \\ x_i &= \frac{3}{n}i \\ f(x_i) &= 9 - \left( \frac{3}{n}i \right)^2 = -\frac{9}{n^2}i^2 + 9 \\ f(x_i)\Delta x &= \left( -\frac{9}{n^2}i^2 + 9 \right) \frac{3}{n} = -\frac{27}{n^3}i^2 + \frac{27}{n} \\ \sum_{i=1}^{n} f(x_i)\Delta x &= -\frac{27}{n^3}\left( \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right) + \frac{27}{n}(n) \\ &= -9 - \frac{27}{2n} - \frac{27}{6n^2} + 27 \\ &= 18 - \frac{9}{2n^2} - \frac{27}{2n} \\ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x &= \mathbf{18} \end{aligned}
Ejercicio 6 — ∫14(5−2x2) dx\displaystyle \int_{1}^{4} (5 - 2x^2) \, dx
Δx=4−1n=3nxi=1+3nif(xi)=5−2(1+3ni)2=−18n2i2−12ni+3f(xi)Δx=(−18n2i2−12ni+3)3n=−54n3i2−36n2i+9n∑i=1nf(xi)Δx=−54n3(n33+n22+n6)−36n2(n22+n2)+9n(n)=−18−27n−9n2−18−18n+9=−27−9n2−45nlim⁡n→∞∑i=1nf(xi)Δx=−27\begin{aligned} \Delta x &= \frac{4 - 1}{n} = \frac{3}{n} \\ x_i &= 1 + \frac{3}{n}i \\ f(x_i) &= 5 - 2\left( 1 + \frac{3}{n}i \right)^2 = -\frac{18}{n^2}i^2 - \frac{12}{n}i + 3 \\ f(x_i)\Delta x &= \left( -\frac{18}{n^2}i^2 - \frac{12}{n}i + 3 \right) \frac{3}{n} = -\frac{54}{n^3}i^2 - \frac{36}{n^2}i + \frac{9}{n} \\ \sum_{i=1}^{n} f(x_i)\Delta x &= -\frac{54}{n^3}\left( \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right) - \frac{36}{n^2}\left( \frac{n^2}{2} + \frac{n}{2} \right) + \frac{9}{n}(n) \\ &= -18 - \frac{27}{n} - \frac{9}{n^2} - 18 - \frac{18}{n} + 9 \\ &= -27 - \frac{9}{n^2} - \frac{45}{n} \\ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x &= \mathbf{-27} \end{aligned}

Ejercicios extra 1

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Bloque A
∫13(x2+2) dx\displaystyle \int_{1}^{3} (x^2 + 2) \, dx
∫04(2x2+1) dx\displaystyle \int_{0}^{4} (2x^2 + 1) \, dx
∫−22(3x2−5) dx\displaystyle \int_{-2}^{2} (3x^2 - 5) \, dx
Bloque B
∫−12(4−x2) dx\displaystyle \int_{-1}^{2} (4 - x^2) \, dx
∫15(10−2x2) dx\displaystyle \int_{1}^{5} (10 - 2x^2) \, dx
∫03(6−3x2) dx\displaystyle \int_{0}^{3} (6 - 3x^2) \, dx

Ejercicios extra 2

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Bloque C
∫−31(x2+3) dx\displaystyle \int_{-3}^{1} (x^2 + 3) \, dx
∫25(5+2x2) dx\displaystyle \int_{2}^{5} (5 + 2x^2) \, dx
∫−13(x2−4) dx\displaystyle \int_{-1}^{3} (x^2 - 4) \, dx
Bloque D
∫−40(2x2+3) dx\displaystyle \int_{-4}^{0} (2x^2 + 3) \, dx
∫36(x2−1) dx\displaystyle \int_{3}^{6} (x^2 - 1) \, dx
∫−21(8−x2) dx\displaystyle \int_{-2}^{1} (8 - x^2) \, dx
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