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Matemáticas 5 · Tema 15

Integrales definidas

Apuntes — Regla de Barrow

∫abf(x) dx=F(b)−F(a)\int_{a}^{b} f(x) \, dx = F(b) - F(a)

La función FF se llama primitiva o antiderivada: su derivada es la función ff, es decir F′=fF' = f.

∫xn dx=xn+1n+1+C\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C

Ejemplo

∫−35(4x2−x) dx=[4x33−x22]−35=[4(5)33−(5)22]−[4(−3)33−(−3)22]=[4(125)3−252]−[4(−27)3−92]=5003−252+1083+92=6083−162=11686=5843≈194.67\begin{aligned} \int_{-3}^{5} (4x^2 - x) \, dx &= \left[ \frac{4x^3}{3} - \frac{x^2}{2} \right]_{-3}^{5} \\ &= \left[ \frac{4(5)^3}{3} - \frac{(5)^2}{2} \right] - \left[ \frac{4(-3)^3}{3} - \frac{(-3)^2}{2} \right] \\ &= \left[ \frac{4(125)}{3} - \frac{25}{2} \right] - \left[ \frac{4(-27)}{3} - \frac{9}{2} \right] \\ &= \frac{500}{3} - \frac{25}{2} + \frac{108}{3} + \frac{9}{2} \\ &= \frac{608}{3} - \frac{16}{2} = \frac{1168}{6} = \mathbf{\frac{584}{3}} \approx \mathbf{194.67} \end{aligned}

Ejercicios

Bloque 1
∫−24(5x2+6x) dx\displaystyle \int_{-2}^{4} (5x^2 + 6x) \, dx
∫15(4−5x2) dx\displaystyle \int_{1}^{5} (4 - 5x^2) \, dx
∫−23(−4x2+x) dx\displaystyle \int_{-2}^{3} (-4x^2 + x) \, dx
Bloque 2
∫34(−2x2−7) dx\displaystyle \int_{3}^{4} (-2x^2 - 7) \, dx
∫−15(−8+3x) dx\displaystyle \int_{-1}^{5} (-8 + 3x) \, dx
∫46(5x2−4x) dx\displaystyle \int_{4}^{6} (5x^2 - 4x) \, dx

Respuestas

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Bloque 1
Ejercicio 1
∫−24(5x2+6x) dx=[5x33+3x2]−24=[5(4)33+3(4)2]−[5(−2)33+3(−2)2]=[5(64)3+3(16)]−[5(−8)3+3(4)]=[3203+48]−[−403+12]=3203+48+403−12=3603+36=120+36=156\begin{aligned} \int_{-2}^{4} (5x^2 + 6x) \, dx &= \left[ \frac{5x^3}{3} + 3x^2 \right]_{-2}^{4} \\ &= \left[ \frac{5(4)^3}{3} + 3(4)^2 \right] - \left[ \frac{5(-2)^3}{3} + 3(-2)^2 \right] \\ &= \left[ \frac{5(64)}{3} + 3(16) \right] - \left[ \frac{5(-8)}{3} + 3(4) \right] \\ &= \left[ \frac{320}{3} + 48 \right] - \left[ \frac{-40}{3} + 12 \right] \\ &= \frac{320}{3} + 48 + \frac{40}{3} - 12 \\ &= \frac{360}{3} + 36 = 120 + 36 = \mathbf{156} \end{aligned}
Ejercicio 2
∫15(4−5x2) dx=[4x−5x33]15=[4(5)−5(5)33]−[4(1)−5(1)33]=[20−5(125)3]−[4−5(1)3]=20−6253−4+53=16−6203=48−6203=−5723≈−190.67\begin{aligned} \int_{1}^{5} (4 - 5x^2) \, dx &= \left[ 4x - \frac{5x^3}{3} \right]_{1}^{5} \\ &= \left[ 4(5) - \frac{5(5)^3}{3} \right] - \left[ 4(1) - \frac{5(1)^3}{3} \right] \\ &= \left[ 20 - \frac{5(125)}{3} \right] - \left[ 4 - \frac{5(1)}{3} \right] \\ &= 20 - \frac{625}{3} - 4 + \frac{5}{3} \\ &= 16 - \frac{620}{3} \\ &= \frac{48 - 620}{3} = \mathbf{-\frac{572}{3}} \approx \mathbf{-190.67} \end{aligned}
Ejercicio 3
∫−23(−4x2+x) dx=[−4x33+x22]−23=[−4(3)33+(3)22]−[−4(−2)33+(−2)22]=[−4(27)3+92]−[323+42]=−1083+92−323−2=−1403+52=−280+156=−2656≈−44.17\begin{aligned} \int_{-2}^{3} (-4x^2 + x) \, dx &= \left[ -\frac{4x^3}{3} + \frac{x^2}{2} \right]_{-2}^{3} \\ &= \left[ -\frac{4(3)^3}{3} + \frac{(3)^2}{2} \right] - \left[ -\frac{4(-2)^3}{3} + \frac{(-2)^2}{2} \right] \\ &= \left[ -\frac{4(27)}{3} + \frac{9}{2} \right] - \left[ \frac{32}{3} + \frac{4}{2} \right] \\ &= -\frac{108}{3} + \frac{9}{2} - \frac{32}{3} - 2 \\ &= -\frac{140}{3} + \frac{5}{2} \\ &= \frac{-280 + 15}{6} = \mathbf{-\frac{265}{6}} \approx \mathbf{-44.17} \end{aligned}
Bloque 2
Ejercicio 4
∫34(−2x2−7) dx=[−2x33−7x]34=[−2(4)33−7(4)]−[−2(3)33−7(3)]=[−1283−28]−[−543−21]=−1283−28+543+21=−743−7=−74−213=−953≈−31.67\begin{aligned} \int_{3}^{4} (-2x^2 - 7) \, dx &= \left[ -\frac{2x^3}{3} - 7x \right]_{3}^{4} \\ &= \left[ -\frac{2(4)^3}{3} - 7(4) \right] - \left[ -\frac{2(3)^3}{3} - 7(3) \right] \\ &= \left[ -\frac{128}{3} - 28 \right] - \left[ -\frac{54}{3} - 21 \right] \\ &= -\frac{128}{3} - 28 + \frac{54}{3} + 21 \\ &= -\frac{74}{3} - 7 \\ &= \frac{-74 - 21}{3} = \mathbf{-\frac{95}{3}} \approx \mathbf{-31.67} \end{aligned}
Ejercicio 5
∫−15(−8+3x) dx=[−8x+3x22]−15=[−8(5)+3(5)22]−[−8(−1)+3(−1)22]=[−40+752]−[8+32]=−40+752−8−32=−48+722=−48+36=−12\begin{aligned} \int_{-1}^{5} (-8 + 3x) \, dx &= \left[ -8x + \frac{3x^2}{2} \right]_{-1}^{5} \\ &= \left[ -8(5) + \frac{3(5)^2}{2} \right] - \left[ -8(-1) + \frac{3(-1)^2}{2} \right] \\ &= \left[ -40 + \frac{75}{2} \right] - \left[ 8 + \frac{3}{2} \right] \\ &= -40 + \frac{75}{2} - 8 - \frac{3}{2} \\ &= -48 + \frac{72}{2} = -48 + 36 = \mathbf{-12} \end{aligned}
Ejercicio 6
∫46(5x2−4x) dx=[5x33−2x2]46=[5(6)33−2(6)2]−[5(4)33−2(4)2]=[10803−72]−[3203−32]=10803−72−3203+32=7603−40=760−1203=6403≈213.33\begin{aligned} \int_{4}^{6} (5x^2 - 4x) \, dx &= \left[ \frac{5x^3}{3} - 2x^2 \right]_{4}^{6} \\ &= \left[ \frac{5(6)^3}{3} - 2(6)^2 \right] - \left[ \frac{5(4)^3}{3} - 2(4)^2 \right] \\ &= \left[ \frac{1080}{3} - 72 \right] - \left[ \frac{320}{3} - 32 \right] \\ &= \frac{1080}{3} - 72 - \frac{320}{3} + 32 \\ &= \frac{760}{3} - 40 \\ &= \frac{760 - 120}{3} = \mathbf{\frac{640}{3}} \approx \mathbf{213.33} \end{aligned}

Ejercicios extra 1

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Bloque A
∫13(2x2+5x) dx\displaystyle \int_{1}^{3} (2x^2 + 5x) \, dx
∫−22(6−3x2) dx\displaystyle \int_{-2}^{2} (6 - 3x^2) \, dx
∫04(−x2+4x) dx\displaystyle \int_{0}^{4} (-x^2 + 4x) \, dx
∫−13(3x2−2x) dx\displaystyle \int_{-1}^{3} (3x^2 - 2x) \, dx
∫25(−5x2+8) dx\displaystyle \int_{2}^{5} (-5x^2 + 8) \, dx
∫−31(4−6x) dx\displaystyle \int_{-3}^{1} (4 - 6x) \, dx
Bloque B
∫02(5x2−3x) dx\displaystyle \int_{0}^{2} (5x^2 - 3x) \, dx
∫−24(2−x2) dx\displaystyle \int_{-2}^{4} (2 - x^2) \, dx
∫14(−3x2+2x) dx\displaystyle \int_{1}^{4} (-3x^2 + 2x) \, dx
∫−12(6x2+4x) dx\displaystyle \int_{-1}^{2} (6x^2 + 4x) \, dx
∫36(8−2x2) dx\displaystyle \int_{3}^{6} (8 - 2x^2) \, dx
∫−30(−4x2−5) dx\displaystyle \int_{-3}^{0} (-4x^2 - 5) \, dx
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