Calculate the area bound by the curve

y=x^3-6x^2+11x-6 and the x-axis.

Question

Calculate the area bound by the curve

y=x^3-6x^2+11x-6 and the x-axis.

anonymous · Answer

find the roots of this equation and integrate it from one root to the other

anonymous · Answer

$$\int\limits_{2}^{1}x^3-6x^2+11x-6 dx +\int\limits_{2}^{3}-(x^3-6x^2+11x-6)dx$$

is this correct?

anonymous · Answer

this can be written (x-3)(x-2)(x-1)

integrate from 1 to 3 directly

anonymous · Answer

$$\int\limits_{1}^{3}x^3 -6x^2 +11x -6$$

anonymous · Answer

http://www.wolframalpha.com/input/?i=y%3Dx%5E3-6x%5E2%2B11x-6

anonymous · Answer

thats how the graph looks its the same on the paper it cuts the x-axis and 3 and 1

anonymous · Answer

or

$$\int\limits_{1}^{2}x^3-6x^2+11x-6      +   \int\limits_{2}^{3}x^3-6x^2+11x-6 $$

anonymous · Answer

$$\int\limits_{1}^{2}$$ yeah thats how its suppose to look i have it on paper so for the 2nd integral why'd you put it without a negative sign?

anonymous · Answer

u dont need a negative sign

anonymous · Answer

okay thanks him1618!

anonymous · Answer

if u put a negative sign itll give you the sum of areas and not the algebraic area

anonymous · Answer

and all i need to do is solve the integrations?

anonymous · Answer

yes

anonymous · Answer

or cancel out the 2's and say $$\int\limits_{1}^{3}x^3-6x^2+11x-6 dx$$

anonymous · Answer

u can do that as well

anonymous · Answer

and i would get the same answer? okay thanks so much!

anonymous · Answer

yes

anonymous · Answer

bt here we knew the nature and location of all the roots so we could do that...if your roots were big messy numbers it wldve been wise to just break the integral up

anonymous · Answer

okay thanks i know what your talking about thanks for the tip!

anonymous · Answer

no probsd

anonymous · Answer

lols what?

anonymous · Answer

u such a noobs him

when they say "area", they mean area

anonymous · Answer

is this wrong?

anonymous · Answer

u do need the negative for the 2 to 3 region

anonymous · Answer

or calculate it normally then take the absolute value of it

anonymous · Answer

is the area 0 because i got 0

anonymous · Answer

no, the reason you got zero was because you didnt take the absolute value of the stuff below the x axis, and it cancelled with the with the area above.

Nevertheless, how can it be zero :|, obvious from the picture that its not zero