Area of 2 curves -Integration

I'm getting really frustrated after having spent 5 hours on this question. I get a feeling the entire answer sheet itself is off.

Find the area of the region trapped between the 2 given curves and the indicated lines.

f(x) = 3x

g(x) = x^2

x = -1, x = -2

int[3x-x^2] = 3x^2/2 - x^3/3 ----> call this function z(x)

Take z(-2) - z(-1) = results

What I also did was to integrate each function separately and sub in the the range of x and find the difference.

Apparently, the results I obtain does not tally with the answer sheet that says 31/6

I really don't wish to go through the hassle of typing the entire working. Pretty much in a foul mood now and I need a break. I would appreciate if someone could help.

My answer works out to be 3/2.

Re: Area of 2 curves -Integration

First of all, when is and when is ? To figure this out, find out when they are equal.

when .

So, at . Hence on the interval .

How about on ? Well, , so on the interval .

So, the correct set-up for this would be:

.

1 Attachment(s)

Re: Area of 2 curves -Integration

Quote:

Originally Posted by

**Darrylcwc** I'm getting really frustrated after having spent 5 hours on this question. I get a feeling the entire answer sheet itself is off.

Find the area of the region trapped between the 2 given curves and the indicated lines.

f(x) = 3x

g(x) = x^2

x = -1, x = -2

int[3x-x^2] = 3x^2/2 - x^3/3 ----> call this function z(x)

Take z(-2) - z(-1) = results

What I also did was to integrate each function separately and sub in the the range of x and find the difference.

Apparently, the results I obtain does not tally with the answer sheet that says 31/6

I really don't wish to go through the hassle of typing the entire working. Pretty much in a foul mood now and I need a break. I would appreciate if someone could help.

My answer works out to be 3/2.

Graph the functions to visualzse what you are doing

Attachment 29415

Do you see what you can do now: integrate under the parabola and subtruct the area of the triangle

Re: Area of 2 curves -Integration

Where did the negative one came from?

Equation is equal only at 0 ll -3.

Re: Area of 2 curves -Integration

Quote:

Originally Posted by

**Darrylcwc** Where did the negative one came from?

Equation is equal only at 0 ll -3.

You are told to bound the functions between the lines x=-1 and x=2. Finding where the equations are equal only tells you the points where it "may" switch which one is "above" the other.

Re: Area of 2 curves -Integration

Quote:

Originally Posted by

**Darrylcwc** Where did the negative one came from?

Equation is equal only at 0 ll -3.

The integration in the orginal post is area from -2 to -1.

if your question is about the line y = 3x, at x=-2 y=-6, at x=-1 y=3*(-1) = -3

Re: Area of 2 curves -Integration

Quote:

Originally Posted by

**votan** The integration in the orginal post is area from -2 to -1.

if your question is about the line y = 3x, at x=-2 y=-6, at x=-1 y=3*(-1) = -3

Oh, oops! I thought it was -1 and 2. Good catch! So then the integral should be

Re: Area of 2 curves -Integration

It just doesn't make sense.

Since x = 0 and x =3 is where the entire curve intersect, then why can't I use x = [0,3] as the range?

graph y=x^2 && y=3x - Wolfram|Alpha=

It just doesn't make sense mathematically to cover the domain outside of 0 and 3 because the area is not enclosed.

I'm solving this, otherwise I'm not sleeping.

Re: Area of 2 curves -Integration

Re: Area of 2 curves -Integration

Hang on. Isn't x the domain? Why is it a function?

Re: Area of 2 curves -Integration

Quote:

Originally Posted by

**Darrylcwc** Hang on. Isn't x the domain? Why is it a function?

Because you said it was:

Quote:

Originally Posted by

**Darrylcwc** Find the area of the region trapped between the 2 given curves and **the indicated lines**.

f(x) = 3x

g(x) = x^2

x = -1, x = -2

Re: Area of 2 curves -Integration

In this case, shouldn't the set up thus be

[-1,-2]int (x^2-3x) + [0,3]int (x^2 - 3x) ?

1 Attachment(s)

Re: Area of 2 curves -Integration

Quote:

Originally Posted by

**Darrylcwc** In this case, shouldn't the set up thus be

[-1,-2]int (x^2-3x) + [0,3]int (x^2 - 3x) ?

notice athe -2 < -1, you could integrate from -1 to -2 if you place a negative sign infront of the integral.

I couldn't post a new reply, I append this post by the image.

Attachment 29422

Re: Area of 2 curves -Integration

I understand that. This set up I gave should be correct. But it still doesn't tally.

Answer sheet states 31/6.

Is something wrong? It's can't be the case. This is supposed to be a straight forward and simple abstract question.

Re: Area of 2 curves -Integration

Quote:

Originally Posted by

**SlipEternal** First of all, when is

and when is

? To figure this out, find out when they are equal.

when

.

So, at

. Hence

on the interval

.

How about on

? Well,

, so

on the interval

.

So, the correct set-up for this would be:

.

I contrast my set up with yours and I don't understand why the second set up ranges from 0,2. The point of intersection is at 0,3.