Did I do this calc problem right?

Question

I had a final in calc BCAP today and it was terrible. I keep running over the problems in my head.

Okay, it was something like "get the area between y=3x and y=(x^3)- (2x^2)"

I set 3x equal to [(x^3)- (2x^2)] I ended up with x=-1, x=0 and x=3. Then I was confused, thinking I'm only supposed to get two numbers. So I just used -1 for the lower limit and 3 for the upper. So it was (bear with me, no idea how to type this) a definite intergral with -1 and 3 as the limits and [(x^3)- (2x^2)-3x) dx.

thus, plugging the upper and lower limits in: (27-18-9)- (-1-2-3)= 6

Answer 1

You made several mistakes.

Mistake #1: Misunderstanding the geometry of the situation

There are in fact two bounded regions which are between y=3x and y=x³ - 2x² -- there is the region between x=-1 and x=0 that is below y=x³ - 2x² and above y=3x, and then a second region between x=0 and x=3 which is above y=x³ - 2x² and below y=3x. Since they did not specify which one of the two regions they wanted you to find the area of, presumably they wanted you to find the total area of both. So the correct setup would be:

[-1, 0]∫x³ - 2x² - 3x dx + [0, 3]∫3x - x³ + 2x² dx

Mistake #2: You forgot to integrate. In order to find the integral of a function over an interval, you evaluate an _antiderivative_ of the function at the endpoints. You evaluated the function itself at the endpoints. Which, had you done your arithmetic correctly, would have given you a value of 0. Which brings me to my next point:

Mistake #3: Arithmetic error - evaluating x³ - 2x² - 3x at -1 yields -1-2+3 = 0, not -1-2-3 = -6.

The correct solution to his problem is as follows:

[-1, 0]∫x³ - 2x² - 3x dx + [0, 3]∫3x - x³ + 2x² dx
(1/4 x⁴ - 2/3 x³ - 3/2 x²)|[-1, 0] + (3/2 x² - 1/4 x⁴ + 2/3 x³)|[0, 3]
- (1/4 + 2/3 - 3/2) + (3/2 * 9 - 1/4 * 81 + 2/3 * 27)
71/6

And we are done.

Answer 2

no this is not correct
get the aboslute value of definite integral from -1 ---> 0
then the absolute value of definite integral from 0 -->3
then add them to get the area
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also u plugged into the function not he intergration
integrate 1st then plug the values
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[-1-->0]∫x^3-2x^2-3x .dx = [-1-->0] ( x^4/4 - 2x^3 /3 - 3x^2 /2 )
= 0 - (1/4 +2/3 -3/2) = - ( 7/12) ---->asboute value= 7/12
the 2nd area
[0-->3] ( x^4/4 - 2x^3 /3 - 3x^2 /2 ) = 81/4 -2*9 -27/2 = -11.25
abs. value = 11.25

so the total area = 7/12 + 11.25 =11.8333