# Thread: Area Bound between curves

1. ## Area Bound between curves

$x+ 5 y = 19 , x + 5 = y^2$

I have already determined that this can be calculated with:

$
\displaystyle \int_{-5}^4 2\sqrt{x+5},dx + \int_{4}^{59} \frac{(19-x)}{5}+\sqrt{x+5},dx
$

When I solve this I get 22.183

The second part of this question wants me to put this into the form of a SINGLE integral with just one upper and lower limit.

How do I do this?

I assume I just add the two integrals like so:

$
\displaystyle \int_A^B 3\sqrt{x+5} + \frac{19-x}{5} dx
$

Is this correct? How do I get the upper and lower limits? I assume its not just -5 & 59 right?

2. Originally Posted by Vamz
$x+ 5 y = 19 , x + 5 = y^2$

I have already determined that this can be calculated with:

$
\displaystyle \int_{-5}^4 2\sqrt{x+5},dx + \int_{4}^{59} \frac{(19-x)}{5}+\sqrt{x+5},dx
$

When I solve this I get 22.183

The second part of this question wants me to put this into the form of a SINGLE integral with just one upper and lower limit.

How do I do this?

I assume I just add the two integrals like so:

$
\displaystyle \int_A^B 3\sqrt{x+5} + \frac{19-x}{5} dx
$

Is this correct? How do I get the upper and lower limits? I assume its not just -5 & 59 right?
No you write it as:

$\displaystyle A=\int_{-8}^3 (19-5y)-(y^2-5)\ dy$

(I would suggest you sketch the curves/area to see why)

CB

3. Can you please explain how you got that expression?

4. Originally Posted by Vamz
Can you please explain how you got that expression?
I drew a picture, then saw that you could integrate with respect to y more conveniently.

CB