Question

ind the area of the region enclosed by these graphs and the vertical lines x = -1 and x = 2
f(x)=x+2 g(x)=x^2-4

Answer 1

@jim_thompson5910

Answer 2

Interesting question. Could you show us what you've done so far please?

Answer 3

i have to figure out which one is the upper and which one is the lower but I am not entire sure but I did find the graph

Answer 4

Okay first you have to combine the two equations together.
f(x)=x+2 g(x)=x^2-4
\[x+2=x^2-4\]
\[x^2-x-4-2=0\]
And you would get this right?
\[x^2-x-6=0\]

Answer 5

yes that is what I got

Answer 6

Then you would put your end points.
\[\int\limits_{-1}^{2}(x^2-x-6)dx\]

Answer 7

Basically if you end up with a negative, just put absolute value signs. It won't matter because the negative number comes up due to the combinging process at the very beginning.

Answer 8

So could you integrate that for me please?

Answer 9

just a moment, workign it

Answer 10

@RoseRoque Have you gotten an answer?

Answer 11

hang on

Answer 12

Is something wrong? Because this integral only requires anti-differentiation. Are you anti-differentiating or using another method because it's much easier to anti-differentiate integrals such as these than using other methods.

Answer 13

-33/2

Answer 14

look it takes me a minute because I do it all piece by piece by piece okay

Answer 15

Okay, well in actual fact, it actually took 13-14 minutes.
And unfortunately the answer is not -33/2. You must of done a mistake somewhere in your working.

Answer 16

What did you get when you integrated the integral?

Answer 17

Did you get this?
\[=\frac{1}{3}x^3-\frac{1}{2}x^2-6x\]

Answer 18

And it's good to have a calculator with you if you're not that good at adding/subtracting fractions. Makes everything easier for you and you don't need to work it out piece by piece, etc.

Answer 19

yes

Answer 20

Okay, well you should of gotten this.
\[\left|=(\frac{8}{3}-2-12)-(-\frac{1}{3}-\frac{1}{2}+6)\right|\]
\[\left|=\frac{9}{3}-8-\frac{1}{2}\right|\]
\[\left| =-\frac{11}{2} \right|\]

Answer 21

And then what would be your final answer. the absolute value signs weren't meant toinclude the equal signs inside. Sorry about that, butthe main quesion I want to ask is what would the final answer be?

Answer 22

-33/2

Answer 23

to include*

Answer 24

but the*

Answer 25

Using what I have as my working, what would be the final answer?

Answer 26

i got the final answer to be -33/2

Answer 27

-33/2 is incorrect. Are you sure you didn't make a mistake somewhere, because I double checked mine and I couldn't find a mistake.

Answer 28

is it not supposed to be negative?

Answer 29

Ah wait. Yes that's correct!

Answer 30

Sorry about that. And yes, you're meant to put absoulte value signs around yourlines of working out.

Answer 31

It doesn't really matter whether it's positive or negative, because if it is negative which is true in this case, you jsut put absolute value signs.

Answer 32

so it is -33/2

Answer 33

No, the answer has to be positive. But what you're jsut doing is making the number positive by putting absolute value signs around your number and then on the next line you write the number without the minus sign in front.

Answer 34

Here is how I would approach it.
The top line, is the actual line, y=x+2.
The bottom is the curve, y=x^2-4.
So I would then set up the integral like this.
You take the integral from -1 to 2. And you set it up, top minus bottom.

\[\int\limits_{-1}^{2}(x-2)-(x^2-4) dx\]

Which, when solved is 9/2.
Did this help?

Answer 35

I apologize, I made a typo. It is supposed to be x+2 in the integral.

Answer 36

The correct answer is 33/2