Question

Area under a curve stuff.

Find the area bounded by the curves y^2 = 2x + 6 and x = y + 1. Your work must include an integral in one variable.

Answer 1

so far i've found that the curves in terms of x are

y = sqrt(2x + 6)

and

y = x-1

Answer 2

@Hero do you know how to do these ?

Answer 3

oh okay thanks for stopping by either way :)

Answer 4

Graph it and find the bounded region in order to set up the integral.

Answer 5

The way you got the y by itself is correct.

Answer 6

yep i already graphed it

Answer 7

which function is on top of the bounded region?

Answer 8

um the y = sqrt(2x + 6) one right?

Answer 9

Does the problem also say bounded by x=0?

Answer 10

no copyed and pasted exactly what was given to me .

Answer 11

i'd assume it is a given though right?

Answer 12

It should be or the answer is infinite from just looking at the graph.

So the bottom function is y=x-1

\[\int\limits_{0}^{5}\sqrt{2x+6}-(x-1)dx\]

Answer 13

since x-1 is cut off by the x=0, the square root of (2x+6) remains

you add the first integral I gave you to this
\[\int\limits_{-3}^{0}\sqrt{2x+6}\]

Answer 14

Know how I got that?

Answer 15

yes , why dont you just use the \[\int\limits_{-3}^{5}\sqrt{2x+6}-(x-1)\]

Answer 16

Because the x=0 cuts the x-1 out.

You can see that the bounded region does not touch the x-1 from x=-3 to x=0

Do you see the graph?

Answer 17

yes i do
is this what that integral would find?