Question

I'm trying to find the area between these two curves:

y = sqrt(x) and y = 1/4x

I set them equal to each other to get:
1/4x - sqrt(x) = 0

But I don't know how to find x doing this. Should I multiply by 4 first? I've tried that and still don't know how to mess with the radical.

tl;dr: I need help factoring this polynomial!

Answer 1

Have you learned about integrals yet?

Answer 2

Of course. Once you know the critical point (dunno if I said that right? where they are both equal) of the two functions you can integrate them to get the area between the two curves, subtracting the smaller function from the larger one (according to FTC) and adding other integrals if necessary (which might have different limits of integration and the functions could be switched so you subtract the function you subtracted from and vice versa)

Answer 3

That was a mouthful. Guess I scared you away lol

Answer 4

Did you try graphing the two equations? Find the one on top and bottom. Subtract the bottom function from the top. And then integrate the function. Where the functions touch is where the points are to find the area.

Answer 5

just plot both the functions and integrate (sqrtx-x/4) from 0 to 16 they meet at (0,0) and then at (16,4) right

Answer 6

Wait I forgot to add that another one of the bounds was x = 25 so that might change things a bit. Sorry! The functions cross at a certain point between 0 and 25, finding that point is important because the functions switch position, so instead of \[\sqrt{x}-\frac{ x }{ 4 }\] you'd have \[\frac{ x }{ 4 } - \sqrt{x}\]

Answer 7

\[\int\limits_{0}^{16}(\sqrt{x}-(x/4))\]\[[((2x/3) ^{3/2}) - (x ^{2}/8)] _{0}^{16}\]\[[((32/3)^{3/2}) - 144] - 0 \]

Answer 8

in that case find the area under the triangle (0,0) (16,0) and (16,4) which is 32 sq units and to this add the area bound by sqrtx with x axis from 16 to 25 i.e integrate sqrtx from 16 to 25

Answer 9

How did you find the 16 value?

Answer 10

\[\frac{1}{4}x=\sqrt x \qquad \rightarrow \qquad \frac{x}{\sqrt x}=4 \qquad \rightarrow \qquad \sqrt x = 4 \qquad \rightarrow \qquad x=16\]