Integration (1 Viewer)

[FDownes]

I'm having some trouble with this question, and I'm not sure if it's because the textbook has the wrong answer or if it's because I'm going completely off the track with my working. Could someone walk me through the problem, so I can see if and/or where I'm making a mistake? Here's the question

a) Find the area enclosed between the curves y = x³ and y = x²

b) Find the volume of the solid formed if this area is rotated about the x-axis

∫

 

[cwag]

x^2 is a little higher than x^3.

therefore your area becomes.
¹₀∫(x² - x³)dx
= [x³/3 - x⁴/4]¹₀
= 1/3 - 1/4
=1/12 units²

..for volume. y² = x⁴ for parabola
y²=x⁶ for cubic

V = pi ¹₀∫(x⁴ - x⁶) dx
= pi [ x⁵/5 - x⁷/7]¹₀
=pi (1/5 - 1/7)
=2pi/35 units³

sorry...migth be wrong

 

[FDownes]

No, you're right. Thanks.

EDIT: Ugh... I see my mistake now. Stupid error, for some reason (in the second question) I made one of the integrals 1⁶/6 instead of 1⁷/7...

 

[FDownes]

Question time again;

Find the area enclosed between the curve y = 3/(x - 2), the y-axis and the lines y = 1 and y = 3

My working;

y = 3/(x - 2)
y(x - 2) = 3
x - 2 = 3/y
x = 3/y + 2
x = 3y^-1 + 2

₁∫³ 3y^-1 + 2 dy
[3 + 2y]₁³
[3 + 2(3)] - [3 + 2(1)]
[9] - [5]
4

However, the answer should be 7.35. What am I doing wrong?

 

[Slidey]

Sorry. Anyway, you forgot that Int f'(x)/f(x) dx = ln(f(x))

More specifically: Int 1/y dy = lny

 

[FDownes]

So I did. Thanks.

 

[FDownes]

Two questions this time. The first one is;

Find the exact volume of the solid formed by rotating the curve y = 1/x about the x-axis between x = 1 and x = 3.

The answer should be 2pi/3, but I end up with -2pi/3.

The second question is;

Find the total area enclosed between the curve y = x(x - 1)(x + 2) and the x-axis.

The answer should be about 3.08 units², but I end up with 5.41.

Show your working if possible, please.

 

[tommykins]

1) Volume cannot be negative so you're right.
2) y = x(x²+x-2) = x³+x²-2x

Drawing the graph, cuts the x axis at -2,0,1

so the total area would be

int x³+x²-2x dx -2->0 + int x³+x²-2x 0->1

 

[FDownes]

1) Volume cannot be negative so you're right.

D'oh... That was stupid of me. I really have to concentrate more.

EDIT: Ah, I see what my problem was with the second question now. Thanks.

 

[FDownes]

Here we go again;

Find the exact area bounded by the parabola y = x² and the line y = 4 - x.

It's not so much that I don't know how to tackle this kind of question, it's more that I've made a mistake somewhere in the mass of equations that results, and I'm not sure where. Although, to be honest, I wouldn't blame anyone for not being bothered typing their working up, it's a pretty massive slab of numbers (unless I'm making a mistake, which is entirely possible). If my working is correct, the two points at which the graphs intersect are x = (-1 + √17)/2 and
x = (-1 - √17)/2

 

[cwag]

yea they are right......

area = int: 4-x-x² dx

= [4x - x²/2 - x³/3] between those gay numbers

i saved these numbers in the memory of my calculator and came up with

4.50647....+ 7.9243997.......

approx = 12.43 units ²

is this right?

 

[FDownes]

The answer is apparently (17√17)/6, I'll check if it's the same.

EDIT: Not quite. Anyways, I'm just gonna skip that question because it's kind of pointless; I'm not going to learn anything from it.

 

[cwag]

The answer is apparently (17√17)/6, I'll check if it's the same.

EDIT: Not quite. Anyways, I'm just gonna skip that question because it's kind of pointless; I'm not going to learn anything from it.

mmm..yea...i dunno buddy....stumped me

 

[cwag]

The answer is apparently (17√17)/6, I'll check if it's the same.

EDIT: Not quite. Anyways, I'm just gonna skip that question because it's kind of pointless; I'm not going to learn anything from it.

actually sorry mate...after redoing it in my calculator i got

3.75773....+7.92439......

which approx = 11.68

which is the same as yours....but it would seriosuly be an effort to left it all in surds...

 

[FDownes]

Yeah, I agree. It's probably not worth it.

Anyways, time for another question;

Find the volume of the solid formed when the curve y = (x + 5)² is rotated about the y-axis from y = 1 to y = 4.

My main problem is figuring out exactly what x² is equal to.

 

[tommykins]

y = (x+5)²
x+5 = +- sqrt y
x = +-sqrt y - 5
x² = (+-sqrt y - 5)²

I guess the issue is using + or - with sqrt y * 5, try using both and see how it goes.

ie. 5sqrty becomes 5y^(1/2) and so forth.

 

[FDownes]

Ah... My problem was that I was expanding the brackets before trying to find x². Thanks.

 

[FDownes]

Nope, no matter how many times I go through this question I just can't seem to get the right answer. I don't suppose anyone could show me their working?

 

[cwag]

Nope, no matter how many times I go through this question I just can't seem to get the right answer. I don't suppose anyone could show me their working?

hmm..it seems like your going to get two answers....what is the answer?

 

[FDownes]

215pi/6 or 112.6 units²