for the following problems, the units of a variable are given along with the units for a function (or two…

for the following problems, the units of a variable are given along with the units for a function (or two functions). give the units of each definite integral. (a) if x is in “seconds” and f(x) is in “feet/second” then ∫a to b f(x) dx is in \no feet/sec²\no feet/sec\no feet\no feet·sec\n(b) if t is in “seconds” and g(t) is in “feet/seconds²” then ∫a to b g(t) dt is in \no feet/sec²\no feet/sec\no feet\no feet·sec

for the following problems, the units of a variable are given along with the units for a function (or two functions). give the units of each definite integral. (a) if x is in “seconds” and f(x) is in “feet/second” then ∫a to b f(x) dx is in \no feet/sec²\no feet/sec\no feet\no feet·sec\n(b) if t is in “seconds” and g(t) is in “feet/seconds²” then ∫a to b g(t) dt is in \no feet/sec²\no feet/sec\no feet\no feet·sec

Answer

Explanation:

Step1: Record la regla de unidades de integración

La integral definida $\int_{a}^{b}y(x)dx$ tiene unidades que son el producto de las unidades de $y(x)$ y las unidades de $x$.

Step2: Resolver (a)

Dado que $x$ está en segundos y $f(x)$ está en pies/segundo, entonces las unidades de $\int_{a}^{b}f(x)dx$ son (pies/segundo) $\times$ segundos = pies.

Step3: Resolver (b)

Dado que $t$ está en segundos y $g(t)$ está en pies/segundo$^{2}$, entonces las unidades de $\int_{a}^{b}g(t)dt$ son (pies/segundo$^{2}$) $\times$ segundos = pies/segundo.

Answer:

(a) feet (b) feet/sec