Please help me with this math problem lease I’ll mark brainlist

“encontrar la integral indefinida y verificar el resultado mediante derivación”

Oliga [24]

$I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx$

Haz la sustitución:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C$

Para confirmar el resultado:

$\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}$

$I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx$

Sustituye:

$y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx$

$\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C$

(Te dejaré confirmar por ti mismo.)

$I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx$

Sustituye:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C$

$I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}$

Sustituye:

$u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}$

$\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C$

Podemos hacer que esto se vea un poco mejor:

$\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}$

$\implies I=-\dfrac{(t+1)^4}{4t^4}+C$

4 0

3 years ago

Which graph best represents the feasibility region for the system shown above? ( pics are in this)

zalisa [80]

Consider the system of inequalities

$\left\{ \begin{array}{l} y\ge 2 \\ x\le 6 \\ y\le 3x+2 \\ y\le -x+10. \end{array}\right.$

1. Plot all lines that are determined by equalities (see attached diagram)

$\left\{ \begin{array}{l} y=2 \text{ (red line)} \\ x= 6 \text{ (blue line)}\\ y=3x+2 \text{ (green line)} \\ y= -x+10 \text{ (orange line)}. \end{array}\right.$

2. Determine which bounded part of the plane you should select:

$y\ge 2$ means that you should take points with y-coordinates greater than or equal to 2 (top part of the coordinate plane that was formed by the red line);
$x\le 6$ means that you should take points with x-coordinates less than or equal to 6 (left part of the coordinate plane that was formed by the blue line);
for $y\le 3x+2$ you can check where the origin is placed. Since $0\le 3\cdot 0+2$ , the origin belongs to the needed part and you have to take the right part of the coordinate plane that was formed by green line.
for $y\le -x+10$ you can check where the origin is placed. Since $0\le -1\cdot 0+10$ , the origin belongs to the needed part and you have to take the bottom part of the coordinate plane that was formed by orange line.