1. ## Curve area

I got a curve y = n ( 16-x2 )
Now the question is to find 'n' when the area between the curve and x axis is 33.

how would i go about it?

2. Originally Posted by taurus
I got a curve y = n ( 16-x2 )
Now the question is to find 'n' when the area between the curve and x axis is 33.

how would i go about it?
think basic. how do we find the area between the curve and the axis?

$y = n \left( 16 - x^2 \right)$

where does y meet the x-axis? that is, when is y zero?

$\Rightarrow 0 = n \left( 16 - x^2 \right)$

$\Rightarrow 0 = 16 - x^2$

$\Rightarrow 0 = (4 + x)(4 - x)$

$\Rightarrow x = \pm 4$

So we want: $A = n \int_{-4}^{4} \left( 16 - x^2 \right)~dx = 2n \int_{0}^{4} \left( 16 - x^2 \right)~dx = 33$

now continue...

3. just a quick question, wat topic is that rule you are using there?

4. Originally Posted by taurus
just a quick question, wat topic is that rule you are using there?
definite integrals

5. hmm ok so

=> 2n[(16^2/2)-(4^3/3)]-[(16^2/2)-(0^3/3)] = 33
=> 2n(106.6666667 - 128) = 33
=> 2n(21.33333333) = 33

But i think im goinf wrong?

6. Originally Posted by taurus
hmm ok so

=> 2n[(16^2/2)-(4^3/3)]-[(16^2/2)-(0^3/3)] = 33
=> 2n(106.6666667 - 128) = 33
=> 2n(21.33333333) = 33

But i think im goinf wrong?
yes, you are going wrong. find the indefinite integral first, and then plug in the limits using the fundamental theorem of calculus. what is the indefinite integral?

7. ya the indefinate integral is 2n [(16^2/2)-(x^3/3)]?

8. Originally Posted by taurus
ya the indefinate integral is 2n [(16^2/2)-(x^3/3)]?
16 is a constant! what is the integral of any constant c?

9. oh right 1?. so

ya the indefinate integral is 2n [1-(x^3/3)]

10. Originally Posted by taurus
oh right 1?. so

ya the indefinate integral is 2n [1-(x^3/3)]
no. the integral of 16 with repsect to x is 16x.

remember, $\int cx^n~dx = \frac {cx^{n + 1}}{n + 1} + C$

we can think of $16$ as $16x^0$

so to integrate a constant, we just attach the variable we are integrating with respect to

anyway, now continue

11. $2n \int_{0}^{4} \left( 16 - x^2 \right)~dx = 2n \left[ 16x - \frac 13x^3 \right]_{0}^{4} = 2n \cdot \frac {128}3 = 33$

now solve for n