# Find the area of the indicated region

#### shamieh

##### Active member
Find the area of the indicated region.

so right minus left so I know the first thing I need to do is

find the integrals

so

$$\displaystyle \frac{x^2}{8} + 1 = x - \frac{1}{2}$$

So I took a stab at this and got this..

$$\displaystyle \frac{x^2}{8} - \frac{8x}{8} + \frac{12}{8}$$

$$\displaystyle x^2 - 8x + 12 = (x - 6) (x - 2) = x = 6 x = 2$$

so
$$\displaystyle \int^6_2 \frac{x^2}{8} + 1 - x - \frac{1}{2} dx$$

and I got $$\displaystyle \frac{128}{24}$$ .. can someone check my work please? I'm skeptical of my process to get there.

#### Prove It

##### Well-known member
MHB Math Helper
Find the area of the indicated region.

View attachment 1684

so right minus left so I know the first thing I need to do is

find the integrals

so

$$\displaystyle \frac{x^2}{8} + 1 = x - \frac{1}{2}$$

So I took a stab at this and got this..

$$\displaystyle \frac{x^2}{8} - \frac{8x}{8} + \frac{12}{8}$$

$$\displaystyle x^2 - 8x + 12 = (x - 6) (x - 2) = x = 6 x = 2$$

so
$$\displaystyle \int^6_2 \frac{x^2}{8} + 1 - x - \frac{1}{2} dx$$

and I got $$\displaystyle \frac{128}{24}$$ .. can someone check my work please? I'm skeptical of my process to get there.
From your sketch there is no way the top function is \displaystyle \begin{align*} y = x - \frac{1}{2} \end{align*} and you are definitely not integrating over the region \displaystyle \begin{align*} 2 \leq x \leq 6 \end{align*}. What's the point in even having a sketch if you don't look at it?

#### shamieh

##### Active member
From your sketch there is no way the top function is \displaystyle \begin{align*} y = x - \frac{1}{2} \end{align*} and you are definitely not integrating over the region \displaystyle \begin{align*} 2 \leq x \leq 6 \end{align*}. What's the point in even having a sketch if you don't look at it?
Nope. The top function should be x + 1. Seems my teacher made a mistake. Attempting the problem again.

#### shamieh

##### Active member
re did the problem and ended up getting
$$\displaystyle \int^8_0 \frac{x^2}{8} + 1 - x + 1 = \int^8_0 \frac{x^2}{8} + 2 - x$$

which is

$$\displaystyle \frac{1}{24}x^3 + 2x - \frac{1}{2}x^2$$ | 8,0 = 28

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is this correct?

#### MarkFL

Staff member
No, for your integrand, you want the "top" function minus the "bottom" function:

$$\displaystyle A=\int_0^8 (x+1)-\left(\frac{x^2}{8}+1 \right)\,dx$$

#### topsquark

##### Well-known member
MHB Math Helper
$$\displaystyle x^2 - 8x + 12 = (x - 6) (x - 2) = x = 6 x = 2$$
Pet peeve time. You do realize this line makes absolutely no sense? Note that graders will mark lines like this incorrect simply for being unintelligible...it's a bad habit to write things this way.

-Dan