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**Vamz** $\displaystyle x+ 5 y = 19 , x + 5 = y^2$

I have already determined that this can be calculated with:

$\displaystyle

\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

\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?