Thread: Area between two curves: y=2x^(1/3), y=(1/8)x^2, 0<=x<=6

1. Originally Posted by JeffM
That is exactly your problem. You did not integrate your function properly because you did not know how to cope with the coefficient.

$\displaystyle \int \ (2x)^{1/3} \ dx = \int \ 2^{1/3} * x^{1/3} \ dx = \dfrac{3 * 2^{1/3}x^{4/3}}{4} + C \ne \dfrac{3(2x)^{4/3}}{4} + C.$
Thanks for actually looking at the work. I will now try it with that and see what I get.

Still did not get the right answer. Could you show me all of the work with that to get the answer?

2. Originally Posted by crybloodwing
Still did not get the right answer. Could you show me all of the work with that to get the answer?
We ask that you show your work, so that we can see what you've done, so far. Thank you.

3. Originally Posted by crybloodwing

[(6/4)(64/3The book says the answer is (47/3)-[(9/2)(121/3) which is between 5 and 6. However, when I plug the value above into a calculator, I get between 7 and 8. Where am I going wrong?
There have now been so many posts that I cannot remember what you have shown and what not. Let's take it in pieces.

$\displaystyle f(x) = (2x)^{1/3} = \sqrt[3]{2} * x^{1/3} \implies \int \ f(x) \ dx = \dfrac{3 \sqrt[3]{2} * x^{4/3}}{4} + C_1 = F(x).$

$\displaystyle g(x) = \dfrac{x^2}{8} \implies \int \ g(x) \ dx = \dfrac{x^3}{24} + C_2 = G(x).$

Obviously f(0) = 0 = g(0).

$x > 0 \implies f(x) \div g(x) = 8 \sqrt[3]{2} * \dfrac{1}{x^{5/3}}.$

Now it is obvious by inspection that

$0 < x \le 1 \implies f(x) \div g(x) > 1 \because \dfrac{1}{x^{5/3}} \ge 1 \text { and } 8\sqrt[3]{2} > 1.$

It is also obvious by inspection that

$f(x) \div g(x)$ is monotonically falling as x increases: the numerator is a constant whereas the denominator is increasing as x does.

So the question is where (if anywhere) does that quotient = 1.

$f(x) \div g(x) = 1 \implies x^{5/3} = 8\sqrt[3]{2} = x^{5/3} \implies x^5 = 1024 \implies x = 4.$

So the area sought =

$(\{F(4) - F(0)\} - \{G(4) - G(0)\}) + (\{G(6) - G(4)\} - \{F(6) - F(4)\}) = what?$

By the way, I get the same answer your book does. I have no idea where your algebra or arithmetic went astray.

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