suppose f has absolute minimum value m and absolute maximum value m. between what two values must ∫₂⁶f(x) dx…

suppose f has absolute minimum value m and absolute maximum value m. between what two values must ∫₂⁶f(x) dx lie? (enter your answers from smallest to largest.) which property of integrals allows you to make your conclusion? o if f(x) ≥ 0 for a ≤ x ≤ b, then ∫ₐᵇf(x) dx ≥ 0. o ∫ₐᵃf(x) dx = 0 o ∫ₐᶜf(x) dx + ∫ᶜᵇf(x) dx = ∫ₐᵇf(x) dx o ∫ₐᵇf(x) dx = -∫ᵇₐf(x) dx o if m ≤ f(x) ≤ m for a ≤ x ≤ b, then m(b - a) ≤ ∫ₐᵇf(x) dx ≤ m(b - a).

suppose f has absolute minimum value m and absolute maximum value m. between what two values must ∫₂⁶f(x) dx lie? (enter your answers from smallest to largest.) which property of integrals allows you to make your conclusion? o if f(x) ≥ 0 for a ≤ x ≤ b, then ∫ₐᵇf(x) dx ≥ 0. o ∫ₐᵃf(x) dx = 0 o ∫ₐᶜf(x) dx + ∫ᶜᵇf(x) dx = ∫ₐᵇf(x) dx o ∫ₐᵇf(x) dx = -∫ᵇₐf(x) dx o if m ≤ f(x) ≤ m for a ≤ x ≤ b, then m(b - a) ≤ ∫ₐᵇf(x) dx ≤ m(b - a).

Answer

Explanation:

Step1: Identificar los valores de a, b, m y M

Tenemos $a = 2$, $b = 6$, el mínimo $m$ y el máximo $M$ de $f(x)$.

Step2: Aplicar la propiedad de la integral

Utilizamos la propiedad: Si $m\leq f(x)\leq M$ para $a\leq x\leq b$, entonces $m(b - a)\leq\int_{a}^{b}f(x)dx\leq M(b - a)$. Sustituyendo $a = 2$ y $b = 6$, tenemos $b - a=6 - 2 = 4$. Entonces $4m\leq\int_{2}^{6}f(x)dx\leq4M$.

Answer:

$4m$ $4M$

Explanation (propiedad de integrales):

La propiedad que permite hacer esta conclusión es: Si $m\leq f(x)\leq M$ para $a\leq x\leq b$, entonces $m(b - a)\leq\int_{a}^{b}f(x)dx\leq M(b - a)$.

Answer (propiedad de integrales):

Si $m\leq f(x)\leq M$ para $a\leq x\leq b$, entonces $m(b - a)\leq\int_{a}^{b}f(x)dx\leq M(b - a)$.