HORRIBLE integration question - for those with time to kill! (1 Viewer)

[LuthienAdrianna]

I couldn't do it, my class couldn't do it, my teacher couldn't do it. ><

Find the exact area bounded by the parabola y=x^2 and the line y=4-x.

Doesn't look so bad, does it?

HAVE FUN >=)

 

[acmilan]

find where they intersect, call them a and b (where b>a)

then integrate (from a to b) 4 - x - x² dx?

 

[LuthienAdrianna]

Did you actually see where they intersect?

 

[acmilan]

yeah,

x = [-1 +- sqrt(17)]/2

so?

 

[LuthienAdrianna]

... pages and pages of work? My friend did this question about four times and got a different answer everytime b/c of little mistakes which screwed the whole thing up. My teacher suggested moving the whole thing to the right and then integrating from 0 to sqrt(17). Then he said to forget it.

 

[acmilan]

Oh, well yeah that part you need to be careful, but in terms of the integral its not a hard question.

 

[pLuvia]

Just use simultaneous equations since both are equal where the intersection is then use quadratic formula to find the x values

then use the higher function and minus the lower function and integrate to get the area between a and b (your x values)

 

[acmilan]

Do you mind posting the answer, LuthienAdrianna?

 

[LuthienAdrianna]

I think the challenge in this question is to do all the surd work without making a mistake, which ruins your whole answer.

 

[LuthienAdrianna]

17sqrt(17)
------------- units sq
6

 

[acmilan]

I get (i think) 17sqrt(17)/6

 

[icycloud]

Very simple question,

Let z = (-1+√17)/2 and Conj(z) = Z = (-1-√17)/2

Now, work out z + Z, z - Z, zZ, z² - Z², z² + Z², z³ - Z³:

z + Z = (-1+√17)/2 + (-1-√17)/2 = -1
z - Z = (-1+√17)/2 - (-1-√17)/2 = √17
zZ = (-1+√17)/2 * (-1-√17)/2 = -4
z² + Z² = (z+Z)² - 2zZ = 9
z² - Z² = (z+Z)(z-Z) = -√17
z³ - Z³ = (z-Z)(z²+zZ+Z²) = 5*√17

Now, the integral becomes:

4z - z²/2 - z³/3 - 4Z + Z²/2 + Z³/3
= 4(z-Z) - (z²-Z²)/2 - (z³-Z³)/3
= 4(√17) - (-√17)/2 - (5 * √17)/3
= 17/6 * √17 #

Quite simple, really. Takes around two minutes if you use my method. You just need a systematic approach on questions like this.

Edit: Hehe been doing too much complex numbers, hence the "z" and "Z" and "Conj" .

 

[acmilan]

Yeah it actually works out nicely, you just need to make sure you group them the right way and use algebraic laws, dont try evaluate each individual term on its own

 

[icycloud]

Yeah it actually works out nicely, you just need to make sure you group them the right way and use algebraic laws, dont try evaluate each individual term on its own

That's right. Just like specialisation in the production line. Compartmentalisation is your friend .

 

[skepticality]

Very simple question,

Let z = (-1+√17)/2 and Conj(z) = Z = (-1-√17)/2

Now, work out z + Z, z - Z, zZ, z² - Z², z² + Z², z³ - Z³:

z + Z = (-1+√17)/2 + (-1-√17)/2 = -1
z - Z = (-1+√17)/2 - (-1-√17)/2 = √17
zZ = (-1+√17)/2 * (-1-√17)/2 = -4
z² + Z² = (z+Z)² - 2zZ = 9
z² - Z² = (z+Z)(z-Z) = -√17
z³ - Z³ = (z-Z)(z²+zZ+Z²) = 5*√17

Now, the integral becomes:

4z - z²/2 - z³/3 - 4Z + Z²/2 + Z³/3
= 4(z-Z) - (z²-Z²)/2 - (z³-Z³)/3
= 4(√17) - (-√17)/2 - (5 * √17)/3
= 17/6 * √17 #

Quite simple, really. Takes around two minutes if you use my method. You just need a systematic approach on questions like this.

Edit: Hehe been doing too much complex numbers, hence the "z" and "Z" and "Conj" .

LoL , hialrious ass --> "very simple question..quite simple really.. " "compartmentlisation is ur frend"?

 

[LuthienAdrianna]

icycloud I did the question your way and I GOT IT!!

(question... was the answer supposed to come out negative? It's positive anyway because it's area but still... it's not below the x-axis...)

 

[Stan..]

For a title like that I expected something hard.