Physics, 04.10.2019 21:00 rileydavidharless

Integrales inmediatas. desarrollar el ejercicio seleccionado utilizando el álgebra, la trigonometría y propiedades matemáticas para reducir las funciones a integrales inmediatas y compruebe su respuesta derivando el resultado.

Integrales inmediatas. desarrollar el ejercicio seleccionado utilizando el álgebra, la trigonometría

Answers: 1

Show answers

Answers

Answer from: ella3714

1. Resolver la integral:

Tenemos la siguiente función:

$f(x)=\frac{(x^2+1)^2}{3}+x-3$

De manera tal que necesitamos calcular la integral:

$\int f(x)dx$

Primero que nada, apliquemos un poco de álgebra a f(x):

$f(x)=\frac{x^4+2x^2+1}{3}+x-3 \\ \\ f(x)=\frac{x^4}{3}+\frac{2x^2}{3}+\frac{1}{3}+x-3 \\ \\ f(x)=\frac{x^4}{3}+\frac{2x^2}{3}+x-\frac{8}{3} \\ \\ Integrando: \\ \\ \int (\frac{x^4}{3}+\frac{2x^2}{3}+x-\frac{8}{3})dx \\ \\ Resolviendo: \\ \\ \frac{x^5}{3\times5}+\frac{2x^3}{3\times 3}+\frac{x^2}{2}-\frac{8}{3}x+C \\ \\ \boxed{\frac{x^5}{15}+\frac{2x^3}{9}+\frac{x^2}{2}-\frac{8}{3}x+C}$

2. Comprobando por derivada:

Necesitamos derivar la siguiente functión:

${\frac{x^5}{15}+\frac{2x^3}{9}+\frac{x^2}{2}-\frac{8}{3}x+C}$

Y comprobar que es igual al integrando:

$\frac{d}{dx}({\frac{x^5}{15}+\frac{2x^3}{9}+\frac{x^2}{2}-\frac{8}{3}x+C}) \\ \\ Derivando: \\ \\{\frac{5x^4}{15}+\frac{6x^2}{9}+\frac{2x}{2}-\frac{8}{3}+0$

Resolviendo:

$\boxed{\frac{x^4}{3}+\frac{2x^2}{3}+x-\frac{8}{3}}$

Como se observa, este resultado es el mismo que el integrando.

Answer from: Quest

a=d/t

a="triangle"v/t

explanation:

a = acceleration

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Another question on Physics

Physics, 22.06.2019 04:30

Asystem containing an ideal gas at a constant pressure of 1.22×10^5 pa gains 2140 j of heat. during the process, the internal energy of the system increases by 2320 j. what is the change in volume of the gas?

Answers: 3

Answer

Physics, 22.06.2019 04:30

Work out sian speed for the first 30 minutes of her journey. give your answer in km/h.

Answers: 1

Answer

Physics, 22.06.2019 09:50

If an athlete runs once around a track, back to the starting line, her average velocity is zero. true or false?

Answers: 3

Answer

Physics, 22.06.2019 23:00

According to the image below, which season is about to begin in the southern hemisphere? a. summer b. autumn c. winter d. spring