Question

Find the area of the region bounded by y= x(x-1)(x-2) and the x-axis.

Answer 1

I can see on my calculator that it has a region from 0 to 1 which is positive and 1 to 2 that is negative, do I add both of the areas together?

Answer 2

What are the intervals?

Answer 3

\[\int\limits_{a}^{b} x(x-1)(x-2) ~dx\]
need the values of b and a to calculate the area.

Answer 4

I hate questions like this.
There are two regions, I wish they would make questions like this more clear..

I dunno what they want here..

Answer 5

So you either want:\[\Large\bf\sf\int\limits\limits_{0}^{1} x(x-1)(x-2) \;dx\]
or:
\[\Large\bf\sf\int\limits\limits\limits_{0}^{1} x(x-1)(x-2) \;dx\quad-\quad \int\limits\limits\limits_{1}^{2} x(x-1)(x-2) \;dx\]

Answer 6

I'm not sure which :(
Do you have an answer key you can check?

Answer 7

i do not have an answer key. but i also think that you have to add the two regions

Answer 8

\[\large \int\limits_{0}^{1} x(x-1)(x-2) dx + \int\limits_{1}^{2} x(x-1)(x-2) dx\]

Do you know the fundamental theorem of calculus, or are you supposed to construct a graph and use shapes to approximate the area?

Answer 9

if they are after the total region, the area below the axis will be negative and above will be positive

so for mine its

\[\int\limits_{0}^{1} f(x) dx + | \int\limits_{1}^{2} f(x) dx|\]

hope it helps

Answer 10

so for negative area use have the integral inside absolute value symbols... that way you will be summing 2 positives..