1. ## [SOLVED] Basic Integration

Here's a question from a past paper

I have attempted the first part and my values are:

K = +8 or -8 C = 2

Whereas the answer in the marking scheme states K = 32 and C = 2
Is there a mistake in the marking scheme because the value of K can't be 32 if the value of C is 2 ... right?

Thanks!

P.S

My equation of the curve comes out to be

$\displaystyle y=k^2/4x^2+c$

2. Originally Posted by unstopabl3
Here's a question from a past paper

I have attempted the first part and my values are:

K = +8 or -8 C = 2

Whereas the answer in the marking scheme states K = 32 and C = 2
Is there a mistake in the marking scheme because the value of K can't be 32 if the value of C is 2 ... right?

Thanks!

P.S

My equation of the curve comes out to be

$\displaystyle y=k^2/4x^2+c$
$\displaystyle -\frac{k}{x^3} = -kx^{-3}$

$\displaystyle \int -kx^{-3} \rightarrow \frac{-k}{-2}x^{-2} +c$

then we know it goes through two points.

sub it in the equation. $\displaystyle y= \frac{-k}{-2}x^{-2} +c$

we have $\displaystyle 18= \frac{-k}{-2}1^{-2} +c$

and $\displaystyle 3= \frac{-k}{-2}4^{-2} +c$ .

now we have 2 simultaneous equations.

solve to find k and c. which comes to your mark scheme answer.

3. By the way, I'm not sure why you thought this question belonged in the Pre-Algebra and Algebra subforum since it is clearly a calculus problem.

4. Thanks, sorry for the mix up, as the question was so basic I thought it belonged in the basic forums. Nevertheless, thanks for moving it, although I don't think this belongs in the university level calculus, maybe in the high-school calculus forum.

Thanks anyways!