Integration: Area enclosed by the x-axis (1 Viewer)

[Makro]

Find the are bounded by the curve y=x² - 9 and the x-axis.

I ended up getting |18| + (-18) = 0

Answers say 36.

Aren't you supposed to put the abs. value signs around the one that uses the negative part of the x-axis, in this case -3 to 0?

Thanks in advance.

 

[tommykins]

If it says find the area, use absolute values for both integrals.

If it says evaluate, then the answer is -36.

 

[Makro]

If it says find the area, use absolute values for both integrals.

If it says evaluate, then the answer is -36.

Oh, thank you. What I've got in my notes makes more sense.

Thanks again.

 

[Prashant Sallan]

Heres a faster way of doing these types of questions.. by using symmetry.. its really helpful for cubics.. hope it helps.. =D

 

[Makro]

Didn't get too far after that :/

Find the area enclosed between the curve y = x² + 5x + 4 and the x-axis.

I must be factorising wrong? I ended up getting -4 and -1 for the x values, while the y value was 4

 

[Makro]

Heres a faster way of doing these types of questions.. by using symmetry.. its really helpful for cubics.. hope it helps.. =D

Thank you, I'll keep that in mind

 

[Timothy.Siu]

Didn't get too far after that :/

Find the area enclosed between the curve y = x² + 5x + 4 and the x-axis.

I must be factorising wrong? I ended up getting -4 and -1 for the x values, while the y value was 4

y=(x+4)(x+1)

x=-4,-1

S (x^2+5x+4) dx = [x^3/3+5x^2/2+4x] from -1 to -4

=-1/3+5/2-4-(-64/3+40-16)=-46.5=46.5units squared

i think thats right

 

[Makro]

Answer was 4.5 units :/

 

[bored of sc]

Find the are bounded by the curve y=x² - 9 and the x-axis.

I ended up getting |18| + (-18) = 0

Answers say 36.

Aren't you supposed to put the abs. value signs around the one that uses the negative part of the x-axis, in this case -3 to 0?

Thanks in advance.

All the area is negative, so it all needs to be absolute value. You can actually do it as a single integration from 3 to -3, or use symmetry as said above.

 

[bored of sc]

Find the area enclosed between the curve y = x² + 5x + 4 and the x-axis.

_-4~^-1(x²+5x+4)dx
= [x³/3+5x²/2+4x]^-1_-4
= |(-1/3+5/2-4)-(-64/3+40-16)|
= 4.5 units²

 

[Makro]

_-4~^-1(x²+5x+4)dx
= [x³/3+5x²/2+4x]^-1_-4
= |(-1/3+5/2-4)-(-64/3+40-16)|
= 4.5 units²

Why do you subtract in between them?

EDIT: Oh wait, man I'm silly, totally forgot the basics.

 

[Makro]

same type of question:

y = x² - 2x - 3 and the x-axis.

My answer ended up being 5/1/3, answer in textbook is 10/2/3.

 

[tommykins]

 

[tacogym27101990]

 

[Makro]

Thanks to both of you.

My teacher was telling us that we need to do [ xxx ]^-1 ₀ + [ xxx ]⁰ ₃

How come they're different..?

 

[tacogym27101990]

Thanks to both of you.

My teacher was telling us that we need to do [ xxx ]^-1 ₀ + [ xxx ]⁰ ₃

How come they're different..?

well im not sure why your teacher told you to do it like that, but you also need absolute values around them because its below the axis.
also you needed to integrate [xxx]⁰_{-1[/sup], not [ xxx ]^-1 0, this will also change to answer.}

 

[Makro]

well im not sure why your teacher told you to do it like that, but you also need absolute values around them because its below the axis.
also you needed to integrate [xxx]⁰_{-1[/sup], not [ xxx ]^-1 0, this will also change to answer.}

_{Oh... :/ So which number needs to be on top of the bracket? Is it the higher or lower x value? Just to clarify. Thanks a bunch, my teacher rushed through 2 exercises so hmm.}

 

[tacogym27101990]

Oh... :/ So which number needs to be on top of the bracket? Is it the higher or lower x value? Just to clarify. Thanks a bunch, my teacher rushed through 2 exercises so hmm.

the highest ( or most positive), needs to be the top limit