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frez [133]
3 years ago
5

2)simplify: y^6 / y^2 a: y^8 b: y^4 c: y^3 ( my answer) d: y^1

Mathematics
1 answer:
pogonyaev3 years ago
3 0
Y^6 / y^2 = y^(6 - 2) = y^4
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Find the distance between two points. (1,3)(-2,8)
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Answer:

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Step-by-step explanation:

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“encontrar la integral indefinida y verificar el resultado mediante derivación”
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I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx

Haz la sustitución:

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

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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

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(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

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Can yall please help me
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