What's the Correct Way to Calculate the Centroid of the Given Region?

[ineedhelpnow]

$y=x^3, x+y=2, x=0$
$A=\int_{0}^{1} \ (2-x-x^3) dx = [2x-\frac{x^2}{2} - \frac{x^4}{4}]_{0}^{1} = 5/4$

I found x and it was correct but there's something wrong with my y. Here's my work.

$\frac{4}{5} \int_{0}^{1} \ \frac{1}{2}[(2-x)^2-(x^3)^2]dx = \frac{4}{10}\int_{0}^{1} \ (4-4x-x^2-x^6)dx=\frac{4}{10} [4x-2x^2-\frac{x^3}{3}-\frac{x^7}{7}]_{0}^{1}= 64/105$

When I put it into my calculator as $\frac{4}{10} \int_{0}^{1} \ [(2-x)^2-(x^3)^2]dx $ I get 92/105 as my answer but when I put it in as $ \frac{4}{10}\int_{0}^{1} \ (4-4x-x^2-x^6)dx$ I get 64/105. What am I doing wrong?

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Oh I multiplied it out wrong. Never mind

 

[jvicens]

!
Thank you for sharing your work and asking for help with your calculations. It seems like you have already found your mistake and corrected it, but I would like to provide some feedback on your work.

Firstly, you correctly found the value of x to be 0. However, when finding the value of y, you used the equation x+y=2 instead of y=x^3. This is why your answer did not match the expected result.

Additionally, when solving the integral, you made a mistake in the expansion of the squared terms. The correct expansion would be (2-x)^2 = 4-4x+x^2, and (x^3)^2 = x^6. This is why your initial answer did not match the expected result.

I am glad that you were able to catch your mistake and correct it. Remember to always double-check your work and equations to ensure accuracy. Good luck with your future calculations!