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Find the area of the indicated region

shamieh

Active member
Sep 13, 2013
539
Find the area of the indicated region.

photo(3).JPG


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
Jan 26, 2012
1,404
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 [tex]\displaystyle \begin{align*} y = x - \frac{1}{2} \end{align*}[/tex] and you are definitely not integrating over the region [tex]\displaystyle \begin{align*} 2 \leq x \leq 6 \end{align*}[/tex]. What's the point in even having a sketch if you don't look at it?
 

shamieh

Active member
Sep 13, 2013
539
From your sketch there is no way the top function is [tex]\displaystyle \begin{align*} y = x - \frac{1}{2} \end{align*}[/tex] and you are definitely not integrating over the region [tex]\displaystyle \begin{align*} 2 \leq x \leq 6 \end{align*}[/tex]. 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
Sep 13, 2013
539
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

- - - Updated - - -

is this correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
Aug 30, 2012
1,140
\(\displaystyle x^2 - 8x + 12 = (x - 6) (x - 2) = x = 6 x = 2\)
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