Area Under a Curve

[generico]

The problem states, "In each problem, find the area of the region."

First problem: f(x)=x^3 g(x) = 2x-x^2

Now, what I did is first integrate both to get (x^4)/4 - x^2 - (x^3)/3. I set the original equations equal to each other, and found x = -1, 2, and then plugged those back into integrated functions and solved 2 - (-1). (If that makes sense.) I ultimately got 3.74. Might have been algebraic error, but not sure where I went wrong. Since I missed that one, thought maybe someone can help with the second:

Second: f(x) = 6/x g(x) = -x-5


Thank you in advance.

 

[arthur ohlsten]

1)
eq 1.
f[x]=x^3

eq2.
g[x]=2x-x^2 rewrite
g[x]=-[x^2-2x] complets the square
g[x]=-[x^2-2x+1]+1
g[x]=-[x-1]^2+1 a parabola, open down, vertex at 1,1

plot the two curves to aid in "seeing" the area of interest

S is the integral sign

area= S S dy dx
area= S [ {-[x-1]^2+1} - x^3 ] dx integrated from x=0 to x=1
area= S [-x^2+2x-x^3] integrated from 0 to1
area=-x^3/3 +x^2-x^4/4 evaluated at 1 and 0
area=[-1/3+1-1/4]-0
area= [-4+12-3] / 12
area=5/12 answer

I am not sure this is the area wanted, but you can see the area of interest, and integrate accordingly

Arthur

 

[generico]

The problem says the graphs "enclose a region." So, it's the area of the region.

I somewhat realize where I went wrong in the first problem. The three areas of intersect are -2, 0, and 1. It says to find them make the two functions equal to each other and solve for 0, then integrate over those numbers. SO, one would solve it from -2 to 0, and from 0 to 1, and add those two together. However, still getting wrong answer. 5/12 isn't accepted in the system as right, either.

 

[pitoten]

Make sure your limits of integration are in the correct spots and not switched around. You must know which curve is "on top" in the intervals in question. To do this, pick an arbritrary number within each interval and test both equations. The larger result comes from the equation that is on top within that particular interval. This equation must be the UPPER limit of integration, while the other one is the lower limit. If you get them mixed up, the area for that particular inteval will come out negative and your answer will be wrong.

I tried it and got 8/3 + 5/12 = 27/12