Integration C4

Why is it that when I integrate this curve and plug in A and B values, I get exactly half the area of the enclosed loop (that I am meant to work out). The integration and everything has been done correctly when I look at the mark scheme, but at the very final step they multiply the area that I worked out by 2. My foundation on integration isn't very good so I'm assuming that when integrating a loop like this , only half the area can be worked out. But then what if the loop was non-uniform and bulging out more on one side?

Reply 1

TenOfThem

Original post by scientific222

When you say that you plug in the A and B values - do you mean the x-coordinates or the t-values

Reply 2

davros

Original post by scientific222

Probably the values you chose for the parameter limits correspond to a traversal of the upper half of the curve (for example) so your "area" is the area between the upper curve and the x-axis. If the loop is symmetric about the x-axis, then by symmetry you have to double "your" area to get the total area.

If you had some weird curve without symmetry then a different approach would be needed, but I don't think you'll get one of those at A level.

Reply 3

scientific222

Original post by TenOfThem

When you say that you plug in the A and B values - do you mean the x-coordinates or the t-values

I plugged in t-values into the integrated "t" form because I used integration by substitution and they gave x and y in terms of t initially

Reply 4

TenOfThem

Original post by scientific222

I plugged in t-values into the integrated "t" form because I used integration by substitution and they gave x and y in terms of t initially

So you should have used the 2 t values at B

If I am recognising the question I think they were 1 and -1

Reply 5

scientific222

Original post by TenOfThem

So you should have used the 2 t values at B

If I am recognising the question I think they were 1 and -1

Initially they gave the equation y = t (9 - t^2)

So as y=0 where the curve crosses the X-axis, t=0, t=+ or - 3 are the solutions. How do I use the two t values at B?

Reply 6

TenOfThem

Original post by scientific222

Initially they gave the equation y = t (9 - t^2)

So t=0, t=+ or - 3 are the solutions. How do I use the two t values at B?

Oh, yes

You use 3 and -3

I assume you either used 0 and 3 or -3 and 0

Reply 7

scientific222

Original post by TenOfThem

Oh, yes

You use 3 and -3

I assume you either used 0 and 3 or -3 and 0

I used 0 and 3 as the limits of the integral, but I don't understand how I am meant to use 0 3 and -3. I evaluated the integral at the upper limit (3) then subtracted it from the lower limit(0) how I usually would. But how do I use 3 and -3 because then I ignore 0?

Reply 8

TenOfThem

Original post by scientific222

You are not ignoring the 0, it is between the limits

That is like asking Why do I ignore 2 if you are integrating between 1 and 3

When you use a parameter you have to think differently as t moves from -3 to 0 it sweeps out half of the shape as it continues to +3 it sweeps out the other half

Reply 9

scientific222

Original post by TenOfThem

I see. Thanks a lot for your help

Reply 10

VannR

The definite integral of a function in the interval (A, B) will give the area bounded by the lines x = A, x = B, the curve and the x-axis. If you wish to find the area beneath the x-axis, you need to either translate the graph by some amount k before integration, or use geometric reasoning to find the total area.

In this case, it is probably evident from given function that the area above the x-axis is equal to that below the x-axis.