# Math Help - Integration c of mass

1. ## Integration c of mass

The 'question' is in blue on my drawing, I have attempted it but I dont believe it is correct ,I made up my own value for variable c , as they didnt give values
(I dont really understand it)

2. Originally Posted by wolfhound
,I made up my own value for variable c , as they didnt give values
For definite integrals you can ignore the constant term c as it will cancel out after inputing the terminals.

3. $\bar{y} = \frac{\int_{-c}^c x^4 \, dx}{2\int_{-c}^c x^2 \, dx}
$

taking advantage of symmetry ...

$\bar{y} = \frac{\int_0^c x^4 \, dx}{2\int_0^c x^2 \, dx}
$

$\bar{y} = \frac{\frac{c^5}{5}}{\frac{2c^3}{3}} = \frac{3c^2}{10}
$

so, if c = 3 , then the y-coordinate of the region's centroid is $\bar{y}= \frac{27}{10}$ , and the x-coordinate is $\bar{x} = 0$ as evidenced by the function's symmetry to the y-axis.

4. Originally Posted by skeeter
$\bar{y} = \frac{\int_{-c}^c x^4 \, dx}{2\int_{-c}^c x^2 \, dx}
$

taking advantage of symmetry ...

$\bar{y} = \frac{\int_0^c x^4 \, dx}{2\int_0^c x^2 \, dx}
$

$\bar{y} = \frac{\frac{c^5}{5}}{\frac{2c^3}{3}} = \frac{3c^2}{10}
$

so, if c = 3 , then the y-coordinate of the region's centroid is $\bar{y}= \frac{27}{10}$ , and the x-coordinate is $\bar{x} = 0$ as evidenced by the function's symmetry to the y-axis.
Thanks!
I am just a bit puzzled why do they say x= -c and x=c and c is positive,
so if I put the value of c=3 why dont I need to give -c a value ?or how does -c become 0?

5. If $f(x)$ is even, then $\int_{-a}^a f(x) dx = 2 \int_{0}^a f(x) dx$.

Just made your integral a little bit less annoying.

6. Originally Posted by o_O
If $f(x)$ is even, then $\int_{-a}^a f(x) dx = 2 \int_{0}^a f(x) dx$.

Just made your integral a little bit less annoying.
Thank you but I dont understand,
Could you explain this please:I am just a bit puzzled why do they say x= -c and x=c and c is positive,
so if I put the value of c=3 why dont I need to give -c a value ?or how does -c become 0?

7. They are saying that your region is bounded by the parabola $y=x^2$ and the two vertical lines $x = -c$ and $x = c$ where $c$ is just some positive constant (the $c>0$ part).

So when you set up your integral to find $\bar{y}$, we have: $\bar{y} = \frac{\int_{-c}^c x^4 \, dx}{2\int_{-c}^c x^2 \, dx}$

Now, if $c=3$, then we would have: $\bar{y} = \frac{\int_{-3}^3 x^4 \, dx}{2\int_{-3}^3 x^2 \, dx}$

You can solve that directly if you want. But what we noted earlier is that both $y= x^4$ and $y = x^2$ are even functions, and using the fact I gave you in my last post, we can convert each integral so that we have:

$\bar{y} = \frac{\int_{-3}^3 x^4 \, dx}{2\int_{-3}^3 x^2 \, dx} = \frac{{\color{red}2}\int_{{\color{red}0}}^3 x^4 \, dx}{2 \cdot {\color{red}2}\int_{{\color{red}0}}^3 x^2 \, dx} = \frac{\int_{0}^3 x^4 \, dx}{2\int_{0}^3 x^2 \, dx}$

Doesn't really do anything too much but it makes evaluating the definite integral a little easier. We like zeroes .. kind of..

8. Originally Posted by wolfhound
Thank you but I dont understand,
Could you explain this please:I am just a bit puzzled why do they say x= -c and x=c and c is positive,
so if I put the value of c=3 why dont I need to give -c a value ?or how does -c become 0?
-c did not become zero.

Both functions, $x^4$ and $x^2$ are symmetrical to the y-axis ... as such, I took advantage of the symmetry by writing $\int_{-c}^c x^4 \, dx = 2\int_0^c x^4 \, dx$ and $\int_{-c}^c x^2 \, dx = 2\int_0^c x^2 \, dx$ .

9. I see thanks!