Finding the Area of a Curve Limited by x-Axis

[chawki]

Homework Statement

y=(x+2)*(3x-3)

Homework Equations

Find the area of the region limited by the curve y=(x+2)*(3x-3) and the axis x.

The Attempt at a Solution

we should do:
(x+2)*(3x-3)=3x²+3x-6
now we should differentiate ?

 

[Mark44]

Homework Statement

y=(x+2)*(3x-3)

Homework Equations

Find the area of the region limited by the curve y=(x+2)*(3x-3) and the axis x.

The Attempt at a Solution

we should do:
(x+2)*(3x-3)=3x²+3x-6
now we should differentiate ?

No. To find the area beneath a curve, integrate. Your textbook should have numerous examples.

 

[chawki]

ok i know we should integrate...but i was looking for kind of (integration from a to b) and we don't know that a and b. we just know axis x is y=0
and that will make ((x+2)*(3x-3))-0
if we integrate without a and b, the integration gives us:
x²+(3/2)*x²-6x

 

[Mark44]

Have you drawn a sketch of the region? The region is between the graph of y = (x + 2)(3x - 3) and the x-axis. If you draw a reasonably accurate sketch, you will see what you need to use for a and b.

 

[chawki]

i draw the curve but didnt help or maybe i didn't know what to look after.
maybe we should solve the equation of y=0 ?

 

[Mark44]

i draw the curve but didnt help or maybe i didn't know what to look after.
maybe we should solve the equation of y=0 ?

That's an excellent place to start. That will give you the x values at the points where the quadratic crosses the x-axis.

 

[chawki]

ok i found -2 and 1
and after integration we get -27/2

 

[pergradus]

ok i found -2 and 1
and after integration we get -27/2

If the curve lies above the x-axis between -2 and 1 the area should be positive. Check your work for sign errors and make sure your bounds are in the correct places.

 

[chawki]

If the curve lies above the x-axis between -2 and 1 the area should be positive. Check your work for sign errors and make sure your bounds are in the correct places.

maybe i was wrong in the integration ?
isn't x³+(3/2)*x²-6x ?

 

[Dick]

maybe i was wrong in the integration ?
isn't x³+(3/2)*x²-6x ?

No, you were right. The integral of (x+2)*(3*x-3) between -2 and 1 is -27/2. That's fine. The curve is below the x-axis though. What is meant by 'area' is usually the absolute value of the integral, since you usually think of area as positive.

 

[jhae2.718]

The curve lies below the x-axis on [-2,1].

Edit: too late...

 

[Mark44]

That's the negative of the integrand. Since the region lies below the x-axis, the typical area element is \Delta A = [0 - (3x^2 + 3x - 6)]\Delta x

This makes the integral
\int_{-2}^1 (-3x^2 -3x + 6)dx

When you carry out the integration, you get +27/2.

 

[Dick]

The curve lies below the x-axis on [-2,1].

Edit: too late...

Nothing wrong with being too late by 1 minute. What you say is still true. It's not a contest :).

 

[chawki]

The curve lies below the x-axis on [-2,1].

Edit: too late...

how you know it is below and not above ?

 

[Dick]

how you know it is below and not above ?

Graph it. You should always do that. The curve intersects the x-axis at 1 and -2. You do need to know what it does in between.

 

[jhae2.718]

how you know it is below and not above ?

Here's a graph by Wolfram|Alpha: http://www.wolframalpha.com/input/?i=y+=+3x^2+6x-3x-6

 

[chawki]

That's the negative of the integrand. Since the region lies below the x-axis, the typical area element is \Delta A = [0 - (3x^2 + 3x - 6)]\Delta x

This makes the integral
\int_{-2}^1 (-3x^2 -3x + 6)dx

When you carry out the integration, you get +27/2.

how we find that 3x²+3x-6 < 0 ?

 

[jhae2.718]

When finding the area under a curve, if on some interval (a,b) f(x)<0 for all x in (a,b), the area of the curve is defined by A=-\int_a^b f(x) dx.

So, after finding the zeros, pick a test case in the interval you are interested in and evaluate f(x).

For this problem:
f(x)=(x+2)(3x-3) = 3x^2+3x-6.
f(x) = 0 \Rightarrow x \in \{-2,1\}
Thus, we are interested in [-2,1]. Zero is contained in this interval, so let's evaluate f(0)=-6. This indicates that f(x) < 0 on [-2,1]. Then, evaluate the integrand and take the opposite as stated above.

 

[chawki]

ok thank you everyone, i get it now :)

 

[Mark44]

how we find that 3x²+3x-6 < 0 ?

Look at your graph. For -2 < x < 1, 3x²+3x-6 < 0.

 

[chawki]

Look at your graph. For -2 < x < 1, 3x²+3x-6 < 0.

Yep. thank you Mark
thank you all again :)