MHB Find the area bounded by the curve #2

[shamieh]

Find the area bounded by the curve $$x = 6x - x^2$$ and the y axis.

So can I use the even function rule to get:

$$2 \int^2_0 6x - x^2 dx$$

I just need someone to check my work.

$$2 [ 3x^2 - \frac{1}{3}x^3 ] | 2, 0$$
$$
2 [ 12 - \frac{8}{3}]$$
$$2 [ \frac{36}{3} - \frac{8}{3} ]$$
$$2 * 28/3$$

= $$\frac{56}{3}$$

is this correct?

 

[MarkFL]

I assume the curve is $y=6x-x^2$ and you are to find the area bounded by it and the $x$-axis. Correct?

Where did you get the limits of integration? What are the roots of the function?

If I wanted to use the even function rule here, I would transpose the graph of the curve 3 units to the left.

 

[shamieh]

I knew I was doing something wrong...

So this is what I should have

6x - x^2 = 0

x (6 - x ) = 0

x = 0 x = 6
$$
2 \int^6_0 6x - x^2 dx$$

which equals 72

 

[Prove It]

Did you read what Mark said? \displaystyle \begin{align*} f(x) = 6x - x^2 \end{align*} is NOT an even function, as \displaystyle \begin{align*} f(-x) = -6x - x^2 \neq f(x) \end{align*}.

 

[MarkFL]

I knew I was doing something wrong...

So this is what I should have

6x - x^2 = 0

x (6 - x ) = 0

x = 0 x = 6
$$
2 \int^6_0 6x - x^2 dx$$

which equals 72

Your limits are correct, however as Prove It showed, the function is not even, so the factor of 2 you added has doubled the correct result. As I stated, if you want to use the even function rule, transpose the graph 3 units to the left:

$$y(x+3)=6(x+3)-(x+3)^2=6x+18-x^2-6x-9=9-x^2$$

Now you have an even function and you may write:

$$A=2\int_0^3 9-x^2\,dx=2\left[9x-\frac{1}{3}x^3 \right]_0^3=2\cdot18=36$$