Hedgecutter wrote: » =3/2 LN +c =3/2 LN [1+x^2]+c Brackets are incorrect but all I can find.
TheBody wrote: » I'm out on the sauce for New Year's Eve but I will reply to your posts tomorrow if nobody gets there before me. Happy new year!
Hedgecutter wrote: » Lovely. Thanks again for your help. I have another turning point question think I might be close. calculate the turning points and max and min value on the curve Y=x^3-3x^3-9x+10.
Hedgecutter wrote: » using integration find the area between curves y=x2+1 and y=7-x x2 +1=7-x x2+1-7+x=0 x2-6+x=0 X2+x-6=0 (x-3)(x+2) x=-3 x=2 wondering if im on the right track
TheBody wrote: » This is fine so far. You have found the x coordinates where the two curves meet. When finding the area betwen two curves, it's really useful to draw a rough sketch of the curves first. You need to know which curve is "on top" and which curve is on "the bottom". Keep going and see how you get on.
TheBody wrote: » I have just changed post #38. I missed a small mistake. X=0 is NOT a root of the quadratic equation. What you had done the first time (apart from the small error pointed out in post #38 was correct. With regards finding the y values, the ony use they serve is in helping you draw a more accurate sketch. They won't play any part in the actual integration. Also in the post above, you ran x=1 through the function. I don't know why you did this. Run the x values found earlier through one of the functions. It doesn't matter which one you use.
Hedgecutter wrote: » On the formula I need to plug in the coordinates, What i'm not sure of is where to get the other two coordinates. Ill use two parts of the area formula, on that formula I have a,b,b,c where I plug in the coordinates are the other two coordinates 0?
Hedgecutter wrote: » Sketched the graph and nothing below the line Still confused to the importance of knowing this.
TheBody wrote: » Sketch looks ok. You need to do this to find out which curve was "on top". Now you know the [latex]7-x[/latex] is on top and the [latex]x^2+1[/latex] is on the bottom (at least over the interval from -3 to 2). At any rate now you find the area by working out: [latex]\int_{-3}^2(7-x)-(x^2+1)dx[/latex]
Hedgecutter wrote: » so because 7-x is on top it's plugged in to the formula first?
TheBody wrote: » Yes, that's exactly it. The best way to see that is with a sketch.
Hedgecutter wrote: » how do you know where the coordinates go a=-3 b=2 is it because -3 is first on the table working out y+7-x so -3=a?
TheBody wrote: » Very close!! You made a small mistake near the beginning when you were removing the brackets. You should have, [latex]\int_{-3}^2(7-x)-(x^2+1) dx=\int_{-3}^2 7-x-x^2-1 dx=\int_{-3}^2-x^2-x+6 dx[/latex]. Other than that your method was correct. Also, you should have equal signs as you progress from step to step.
Hedgecutter wrote: » Why did the sign change? [latex]\int_{-3}^2(7-x)-(x^2+1) dx=\int_{-3}^2 7-x-x^2-1 dx=\int_{-3}^2-x^2-x+6 dx[/latex]. is it because -+=-
TheBody wrote: » Yes, the minus sign before the second bracket causes a sign change of everything inside the bracket.
Hedgecutter wrote: » The equivalent of second year. If I pass this year I'll have a level 6 higher certificate.
Hedgecutter wrote: » What method do I use for part (1)
TheBody wrote: » When it comes to deciding if an integral is to be done using regular substitution, you ask yourself, is there anything there that if DIFFERENTIATED, would give you the other bit, and your not too bothered about constants? You have to deal with any constants but they don't cause any trouble. If there is, then you let that bit be "u". So, With that in mind and looking at your problem, do you notice anything between the top and bottom?
Hedgecutter wrote: » do we use partial fractions method. factorise the top line? but you can only factorise the denominator
