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# area under curve watch

1. a curve is given by y=x^2-6x+13
the line AB is the tangent at A and meets y-axis at B. A=(4,5) B=(0,-3) and equation of line AB is y=2x-3
find the area bounded by the curve, the line AB and the y-axis
2. (Original post by evariste)
a curve is given by y=x^2-6x+13
the line AB is the tangent at A and meets y-axis at B. A=(4,5) B=(0,-3) and equation of line AB is y=2x-3
find the area bounded by the curve, the line AB and the y-axis
Tip; Draw a sketch. That should help you see your limits and whether you need to subtract any areas due to the line.
3. (Original post by evariste)
a curve is given by y=x^2-6x+13
the line AB is the tangent at A and meets y-axis at B. A=(4,5) B=(0,-3) and equation of line AB is y=2x-3
find the area bounded by the curve, the line AB and the y-axis
I did it! we havent covered this in class but i just taught myself. Wot u need to do is draw a sketch showing the line AB and the curve that touches point A and that it is U shaped because x^2 is positive. So wot you do then is you use integration to find the area under the curve.
To start with we want to find the area between the cureve and the point x = o and x= 4. But at this point anything below the x axis is not being considered.
So you need to integrate the equation of the curve. so y = x^2 - 6x +13 becomes
x^3/3 - 6x^2/2 +13 x which goes to x^3/3 - 3x^2 +13x

Then to find the area you subtract the function of 0 from the function of 4 (so just find the function of 4)
F(4) = 84 - 48 +52 = 88

But that area includes the triangle that is between the line x = 4 and the line AB. To work out the triangles area you need to find the co ordinate where the line AB intersects the x axis. So when y = 0 0 = 2x - 3 so x = 1.5

Therefore we want to find the line at the base of this small triangle so its length is 4 - 1.5 = 2.5. This is then multiplied by its height which is 5 and then divided by 2 to give an area of 6.25.
Subtract this from 88 to give 81.75.

Now we need to add the triangle's area that is below the x axis. So knowing where AB intersects the x axis we can determine the length base of the triangle (its upside down tho) so that is just from 0 to 1.5. then multiply this by 3 (from 0 to -3) to give 4.5 and then divide by 2. = 2.25.
Add this to the 81.75 to give 84!
4. The area under a curve (between it and the x-axis) is given by the intgral,

integral[f(x) {x2,x1}] dx

The area between any two curves (this ignores the axes) is given by,

integral[f_1(x) - f_2(x) {x2,x1}] dx

we have,

f_1(x) = y1 = x² - 6x + 13
f_2(x) = y2 = 2x - 3
x1 = 0, x2 = 4

A = integral[y1 - y2 {4, 0}] dx
A = integral[x² - 6x + 13 - 2x + 3 {4, 0}] dx
A = integral[x² - 8x + 16 {4, 0}] dx
A = [x^3 / 3 - 4x² + 16x]{4, 0}
A = 64/3 - 64 + 64
A = 64/3
=======
5. im afraid that newton's and Fermat's solutions are flawed. Newton has misread the question - it is the area bound by the curve and the line (which is a tangent to the curve) and the y axis. Newton assumed that it was 2 curves but in fact it was a curve and its tangent. So he wasted much time working out their intercept but as this was the tangent to the curve which met at A we knew the co - ordinates of their "intercept" = 4,5.
From the information in the equations we can also see that the curve has no real roots because the value of (b^2 - 4 ac)^1/2 (the discriminant) is below zero. This is helpful to know when sketching the graph but not essential.

I can see what fermat has done but his calculations are emitting crucial details that don't compensate for the areas under the curve which are not in the area bound by the straight line. Also there is the area below the x - axis which is not added on.

I thought my answer out very methodically and my calculations show all steps clearly. They might be wrong but i am certainly closer than newton or fermat. I am surprised with names like that aswell. i hope someone can verify my answer.
6. (Original post by ryan750)
So you need to integrate the equation of the curve. so y = x^2 - 6x +13 becomes
x^3/3 - 6x^2/2 +13 x which goes to x^3/3 - 3x^2 +13x

Then to find the area you subtract the function of 0 from the function of 4 (so just find the function of 4)
F(4) = 84 - 48 +52 = 88

Therefore we want to find the line at the base of this small triangle so its length is 4 - 1.5 = 2.5. This is then multiplied by its height which is 5 and then divided by 2 to give an area of 6.25.

Now we need to add the triangle's area that is below the x axis. So knowing where AB intersects the x axis we can determine the length base of the triangle (its upside down tho) so that is just from 0 to 1.5. then multiply this by 3 (from 0 to -3) to give 4.5 and then divide by 2. = 2.25.
Add this to the 81.75 to give 84!
The error is in F(4) it is 64/3-48+52=64/3+4=76/3
area of triangle is ok=25/4
area of other triangle ok=9/4
so total area is 76/3+9/4-25/4=76/3-16/4=256/12=64/3 which is in agreement with Fermat.
7. (Original post by evariste)
The error is in F(4) it is 64/3-48+52=64/3+4=76/3
area of triangle is ok=25/4
area of other triangle ok=9/4
so total area is 76/3+9/4-25/4=76/3-16/4=256/12=64/3 which is in agreement with Fermat.
Oh yes. Fermat was right - i apologise for my lack of understanding and premature righteousness. I was confused because fermat said the area bound by any two curves and i thought he was assuming wrong as newton was. But does this mean that the method (much quicker than mine) he used can assume the regions bound by 2 curves, a curve and a line or even 2 lines?
I wasn't to far off!
8. A straight line is also a curve and is sometimes referred to as a straight-line curve.
The method I used can be used to find the area between any two curves, as described by yourself, and bounded by two x-ordinates, or vertical lines.
An alternative version is to find the area of a region which is bounded by two curves and two y-ordinates, or horizontal lines.
Then you would have,

A = integral[g_1(y) - g_2(y) {y2, y1}] dy

where g_1(y) and g_2(y) are the curves expressed as functions of y.
9. Tangent => ((b^2)-4ac)=0 => One distinct root => Touching point which can be called an intersection.

Newton.
10. (Original post by Newton)
Tangent => ((b^2)-4ac)=0 => One distinct root => Touching point which can be called an intersection.

Newton.
((b^2) - 4ac) = (-6^2) - 4 x 1 x 13
=36-52
=-16

Therefore the discriminant is below zero - which means it has no real roots. So the curve does not intersect with the x - axis. This information doesn't matter anyway because if it did have roots, they would occur outside the area we are considering.

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