Gardner Reese
01/07/2024 · Senior High School
\( \int_{3}^{1} f(x) d x=5 \) y \( \int_{-2}^{3} f(x) d x=2 \), el valor de la integral \( \int_{-2}^{1} f(x) d x \) es: ccione una: 7 3 -3 -7
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Para resolver el problema, utilizaremos la propiedad de la aditividad de las integrales definidas. Sabemos que:
\[
\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx
\]
En este caso, queremos encontrar \( \int_{-2}^{1} f(x) \, dx \). Podemos dividir esta integral en dos partes:
\[
\int_{-2}^{1} f(x) \, dx = \int_{-2}^{3} f(x) \, dx + \int_{3}^{1} f(x) \, dx
\]
Ya tenemos los valores de las integrales:
1. \( \int_{3}^{1} f(x) \, dx = 5 \)
2. \( \int_{-2}^{3} f(x) \, dx = 2 \)
Sin embargo, notamos que \( \int_{3}^{1} f(x) \, dx \) se puede reescribir como:
\[
\int_{3}^{1} f(x) \, dx = -\int_{1}^{3} f(x) \, dx
\]
Por lo tanto, podemos decir que:
\[
\int_{3}^{1} f(x) \, dx = -5
\]
Ahora, sustituimos estos valores en la ecuación:
\[
\int_{-2}^{1} f(x) \, dx = \int_{-2}^{3} f(x) \, dx + \int_{3}^{1} f(x) \, dx
\]
Sustituyendo los valores:
\[
\int_{-2}^{1} f(x) \, dx = 2 + (-5) = 2 - 5 = -3
\]
Por lo tanto, el valor de la integral \( \int_{-2}^{1} f(x) \, dx \) es:
\[
\boxed{-3}
\]
Quick Answer
El valor de la integral \( \int_{-2}^{1} f(x) \, dx \) es -3.
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