Integration Question (1 Viewer)

[jash1228]

If the gradient of a curve is given by dy/dx= x/(x^2-1)^2 and the curve passes through (2,-1), find the equation of the curve.

Answer: y= -1/2(x^2-1) -5/6

I got part of the answer correct but don't know where the -5/6 came from. Can someone please explain?

 

[InteGrand]

If the gradient of a curve is given by dy/dx= x/(x^2-1)^2 and the curve passes through (2,-1), find the equation of the curve.

Answer: y= -1/2(x^2-1) -5/6

I got part of the answer correct but don't know where the -5/6 came from. Can someone please explain?

It's because the curve has to pass through (2, -1). When you integrated you probably ended up with a y = something "+C". You should then sub. in x = 2 and y = -1 to find what the constant C must be in order for (2, -1) to lie on the curve.

 

[jash1228]

I sort of understand that process. But where do I sub the point (2,-1) into to find the '+c' ?

 

[InteGrand]

I sort of understand that process. But where do I sub the point (2,-1) into to find the '+c' ?

 

[jash1228]

Thank you!

 

[tlv6554]

Can someone please put in the steps for the integration process. I dont get how you do it if theres an x at the front

 

[InteGrand]

Can someone please put in the steps for the integration process. I dont get how you do it if theres an x at the front