Consider the function` f(x)= (2/x^2)-(6/x^7)`?

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Consider the function` f(x)= (2/x^2)-(6/x^7)`?

Postby barnamah » Sat 02 April, 2011 16:31

Consider the function` f(x)= (2/x^2)-(6/x^7)`?

Let F(x) be the anti-derivative of f(x) with F(1)=0

Hint: F(1)=0 just gives us that coordinate point we need to determine the constant, C.

Then F(x)=?
Explore and know. That is asked.
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Re: Consider the function` f(x)= (2/x^2)-(6/x^7)`?

Postby Integr88 » Sat 02 April, 2011 16:53

The question say for the function ` f(x)= (2/x^2)-(6/x^7)` and let `F(x)=int(2/x^2)-(6/x^7)dx` so find initial equation

let's integrate `int(2/x^2)-(6/x^7)dx` (integrate means take the anti-derivatives

first rewrite the equation so it looks easy to take the antiderivatives
`int(2/x^2)-(6/x^7)dx=int(2x^(-2))-(6x^(-7))dx`

Now take the anti-derivatives (integrate):
`int(2x^(-2))-(6x^(-7))dx=2*x^(-2+1)/(-2+1)-6*x^(-7+1)/(-7+1)+C=2*x^(-1)/(-1)-6*x^(-6)/(-6)+C`

simplify and we have: `-2*x^(-1)+*x^(-6)+C=2x^(-1)+*x^(-6)+C`
`-2/x+1/x^6+C`

Now have F(1)=0 meaning when x=1 the whole equation is 0. so let's sub in the x=1
`0=-2/1+1/1^6+C`
`0=-2+1/1+C`
`0=-2+1+C`
`0=-1+C` adding 1 to both side and we have: `1=-1+1+C` we it becomes `1=C`

Now we have C=1, our equation is `F(x)=-2/x+1/x^6+1`
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