# area between curves.

• Jan 11th 2010, 10:19 AM
Tweety
area between curves.

http://www.mathhelpforum.com/math-he...3&d=1263236914

I have attached the sketch of the curves,

I am not sure how to integrate, because one part is below the x-axis and the other is above.

Appreciate if someone can show me how to solve these problems.

• Jan 11th 2010, 10:26 AM
If you subtract \$\displaystyle x^2-4x\$ from \$\displaystyle 4x-x^2,\$

then you will always have the difference between the curves.
Integrating this difference gives the area between the points of intersection.

Or, since the region is symmetrical about the x-axis,
integrate the top curve and multiply the answer by 2,
or integrate the region under the x-axis and multiply by -2.
• Jan 11th 2010, 10:34 AM
Remember....

integration sums f(x),
so if f(x) is negative, integration sums negative values.
If you want area, you just change the sign,
though you must be aware of where the graph crosses the x-axis
to isolate where f(x) is positive or negative.
• Jan 11th 2010, 10:53 AM
Tweety
Quote:

If you subtract \$\displaystyle x^2-4x\$ from \$\displaystyle 4x-x^2,\$

then you will always have the difference between the curves.
Integrating this difference gives the area between the points of intersection.

Or, since the region is symmetrical about the x-axis,
integrate the top curve and multiply the answer by 2,
or integrate the region under the x-axis and multiply by -2.

Thank you, so should I integrate each area from 4 to 0 separately and than subtract?
• Jan 11th 2010, 11:12 AM
No, Tweety,

When you are dealing with the area between curves as in your case,
with one below the other, you can subtract the functions first,
then integrate.
Subtract the lower curve from the upper curve.

\$\displaystyle 4x-x^2-(x^2-4x)=8x-2x^2.\$

now you can integrate the difference between them
and that gives the area.

Notice that \$\displaystyle 4x-x^2\$ is \$\displaystyle x^2-4x\$ upside down.

Hence the area above the x axis between the points of intersection,
is always positive there.

Hence integrating that will give a positive answer.

Or, integrate the lower curve, but since that one is negative, your answer
will be negative.
So just change the sign and multiply by 2.

If you do all 3 ways, it would be excellent practice.
• Jan 11th 2010, 11:26 AM
Tweety
Quote:

No, Tweety,

When you are dealing with the area between curves as in your case,
with one below the other, you can subtract the functions first,
then integrate.
Subtract the lower curve from the upper curve.

\$\displaystyle 4x-x^2-(x^2-4x)=8x-2x^2.\$

now you can integrate the difference between them
and that gives the area.

Notice that \$\displaystyle 4x-x^2\$ is \$\displaystyle x^2-4x\$ upside down.

Hence the area above the x axis between the points of intersection,
is always positive there.

Hence integrating that will give a positive answer.

Or, integrate the lower curve, but since that one is negative, your answer
will be negative.
So just change the sign and multiply by 2.

If you do all 3 ways, it would be excellent practice.

Okay I understand now thanks, I will give all three methods ago.
• Jan 11th 2010, 11:39 AM
if you subtract the lower function from the upper one,
the difference between them is always positive.
Therefore, integrating the difference gives the area,