Question

Find the area between the curves. The graph is attached.

Answer 1

THIS IS THE EQUATIONS

Answer 2

THIS IS THE GRAPH

Answer 3

area is \[\int\limits (t0p)-(bottom) dx \]
can be dy or dx. depending on what you chose or makes easier

Answer 4

then. set both your equations equal to each other to find your bounds. you're already given one, x=16.
now u just need 2 figure out the other.
u can also use ur given graph to find ur other one.

Answer 5

actually you're going to have to break it up into two integrals, since they cross over at x=4, so the first integral will be frome 0 to 4 and the next one will be from 4 to 16, keeping in mind that the bigger area goes in front for each integral so you don't get a negative area.

Answer 6

Hermeezy is correct.

Answer 7

\[\int\limits_{0}^{4} \sqrt{x} - \frac{ 1 }{ 2x } + \int\limits_{4}^{16} \frac{ 1 }{ 2x }- \sqrt{x}\]

Answer 8

When i do this i get a negative number though...-86/3. Should i switch the equations around? I thought i put the upper equation subtract lower equation corretctly

Answer 9

Is it:
\[\frac{ 1 }{ 2 }x \]
or
\[\frac{ 1 }{ 2x }\]

Answer 10

\[\frac{ 1 }{ 2x }\]

Answer 11

\[\frac{ 1 }{ 2 } \int\limits \frac{ 1 }{ x }dx= \frac{ 1 }{ 2 }\ln(x)\]

\[\int\limits(\sqrt{x})dx= \int\limits(x^{\frac{ 1 }{ 2 }})dx = \frac{ 2 }{ 3 } x^{\frac{ 3 }{ 2 }} \]

Evaluate on the new found bounds. Just make sure your integrals are correct.