Integration to find area between curve and straight line

~Bitzchick~ · Joined: 05 Aug 2007 Posts: 18

Hi all

My straight line is y = x-4
My curve is y = x^(2/3) - 2*x^(-1/3) +1

The straight line intersects the curve at (8,4) and meets the x-axis at
(4,0)

The curve touches the x-axis at (1,0)

I need to find the area above the x-axis.
The area of the triangle that is formed by the straight line and the x-axis
is 8 and I know I need to subtract that from the area under the curve and
the x-axis to find the area I need

However when integrating between the limits of x=1 to x=8, to get the area
under the curve and the x-axis, I get a negative area, which I shouldn't as
the area is above the x-axis

my integration after subtracting the curve from the straight line is:

[1/2x^2 - 5x - (3/5)x^(5/3) + 3x^(2/3)] between limits of x=1 to 8

I calculate the area under the curve only, to be -173/10

This can't be right as it's negative and when subtracting 8, the area I
require is also negtive which is wrong as the area required is above the
x-axis

What am I doing wrong?

I cannot see where I am going wrong at all

I hope you can help

TIA

sheri

Archived from group: uk>education>maths

Jean-Marc Gulliet · Joined: 05 Aug 2007 Posts: 4

~Bitzchick~ wrote:

> My straight line is y = x-4
> My curve is y = x^(2/3) - 2*x^(-1/3) +1
>
> The straight line intersects the curve at (8,4) and meets the x-axis at
> (4,0)
>
> The curve touches the x-axis at (1,0)
>
> I need to find the area above the x-axis.
> The area of the triangle that is formed by the straight line and the x-axis
> is 8 and I know I need to subtract that from the area under the curve and
> the x-axis to find the area I need
>
> However when integrating between the limits of x=1 to x=8, to get the area
> under the curve and the x-axis, I get a negative area, which I shouldn't as
> the area is above the x-axis
>
> my integration after subtracting the curve from the straight line is:
>
> [1/2x^2 - 5x - (3/5)x^(5/3) + 3x^(2/3)] between limits of x=1 to 8
>
> I calculate the area under the curve only, to be -173/10
>
> This can't be right as it's negative and when subtracting 8, the area I
> require is also negtive which is wrong as the area required is above the
> x-axis
>
> What am I doing wrong?
>
> I cannot see where I am going wrong at all

Assuming I have correctly understood your post, you want to compute
the area delimited by the x-axis, a curve, and a straight line,
(assuming the curve is above the straight line) for x between 1 and 8.

Using Maple, let's call the straight line f

> f := x-4;
x - 4

and the curve g

> g := x^(2/3)-2/x^(1/3)+1;
(2/3) 2
x - ------ + 1
(1/3)
x

Plotting both expressions on the same graph, we check that our initial
statements make sense

> plot([f, g], x);

We can see that the curve g lies above the straight line f on the
whole interval [1, 8], that g intersects the z-axis at (1, 0) and f at
(8, 4), and that f intersects the x-axis at (4,0). (This just confirms
what you claimed in your post.)
Now, you suggest to compute the areas of f and g with the x-axis and
subtract the former from the latter to get the desired area. Let's do
that. Using Maple we have,

> int(f, x = 4 .. Cool

;
8

> int(g, x = 1 .. Cool

;
83
--
5

> %-`%%`;
43
--
5

> evalf(%);
8.600000000

So the required area is positive, as expected. I hope this clarifies
the issue.

Regards,
--
Jean-Marc

~Bitzchick~ · Joined: 05 Aug 2007 Posts: 18

Jean-Marc Gulliet wrote:
> We can see that the curve g lies above the straight line f on the
> whole interval [1, 8], that g intersects the z-axis at (1, 0) and f at
> (8, 4), and that f intersects the x-axis at (4,0). (This just confirms
> what you claimed in your post.)

Yes. The first thing I did was sketch the curve and straight line and
determine all intersection points

> Now, you suggest to compute the areas of f and g with the x-axis and
> subtract the former from the latter to get the desired area. Let's do
> that. Using Maple we have,
>
> > int(f, x = 4 .. Cool

;
> 8
>
> > int(g, x = 1 .. Cool

;
> 83
> --
> 5
>
> > %-`%%`;
> 43
> --
> 5

> So the required area is positive, as expected. I hope this clarifies
> the issue.
>
Thank you Jean-Marc, but I do not know what Maple is (I assume it's a
software program?)

I really want to know how to do it manually and where I have gone wrong with
my integration

I am going wrong with my int(g, x = 1 .. Cool

; but I don't know where? It's
driving me bonkers. Please could you help further?

TIA

sheri

--
Life might not be the party we hoped for but whilst we are here we may
as well dance

~Bitzchick~ · Joined: 05 Aug 2007 Posts: 18

~Bitzchick~ wrote:
> Jean-Marc Gulliet wrote:
> I really want to know how to do it manually and where I have gone
> wrong with my integration

Hi all

I've realised where I've gone wrong!

I was trying to integrate to find an area between the straight line boundry
and the curve boundary and then subtracting the area between the straight
line and the x-axis, rather than integrating to find the area between the
x-axis and the curve boundary and THEN subtracting the area between the
straight line and the x-axis

When I found the area between the x-axis and the curve boundary (which was a
positive value) and then subtracted the triangular area between the straight
line and the x-axis I got 43/5

So it must be correct

sheri

--
Life might not be the party we hoped for but whilst we are here we may
as well dance

Frederick Williams · Joined: 05 Aug 2007 Posts: 20

~Bitzchick~ wrote:

> My curve is y = x^(2/3) - 2*x^(-1/3) +1

You're a funny shape for a girl.

--
How unlike the home life of our own dear Queen.
Remove "antispam" and ".invalid" for e-mail address.

~Bitzchick~ · Joined: 05 Aug 2007 Posts: 18

Frederick Williams" <"Frederick Williams wrote:
> ~Bitzchick~ wrote:
>
>> My curve is y = x^(2/3) - 2*x^(-1/3) +1
>
> You're a funny shape for a girl.

Heh heh
Your humour is incalculable

sheri

--
Life might not be the party we hoped for but whilst we are here we may
as well dance

Nicolas Jeannequin · Joined: 19 Jan 2008 Posts: 1

Let's call f1(x) = x-4 and f2(x) = x^{2/3}-2*x^{-1/3}+1
you want to compute:
\int_{1}^{8} f2(x) - f1(x) \, dx = I

I = \int_{1}^{8} x^{2/3}-2*x^{-1/3} -x +5 \, dx
= [3/5 * x^{5/3} - 3*x^{2/3} - 1/2*x^{2} + 5*x]_{1}^{8}
= 131/10

Cheers,
Nicolas

On Mon, 26 Nov 2007, ~Bitzchick~ wrote:

> Hi all
>
> My straight line is y = x-4
> My curve is y = x^(2/3) - 2*x^(-1/3) +1
>
> The straight line intersects the curve at (8,4) and meets the x-axis at
> (4,0)
>
> The curve touches the x-axis at (1,0)
>
> I need to find the area above the x-axis.
> The area of the triangle that is formed by the straight line and the x-axis
> is 8 and I know I need to subtract that from the area under the curve and
> the x-axis to find the area I need
>
> However when integrating between the limits of x=1 to x=8, to get the area
> under the curve and the x-axis, I get a negative area, which I shouldn't as
> the area is above the x-axis
>
> my integration after subtracting the curve from the straight line is:
>
> [1/2x^2 - 5x - (3/5)x^(5/3) + 3x^(2/3)] between limits of x=1 to 8
>
> I calculate the area under the curve only, to be -173/10
>
> This can't be right as it's negative and when subtracting 8, the area I
> require is also negtive which is wrong as the area required is above the
> x-axis
>
> What am I doing wrong?
>
> I cannot see where I am going wrong at all
>
> I hope you can help
>
> TIA
>
> sheri
>
>
>
>

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