Integration Help (1 Viewer)

[qwerty44]

Can I please get help with Q5.C and F. Could you just explain what to do before actually doing it.
I'm learning ahead so i haven't actually done any of this before.
Thanks

 

[SpiralFlex]

Can I please get help with Q5.C and F. Could you just explain what to do before actually doing it.
I'm learning ahead so i haven't actually done any of this before.
Thanks

View attachment 24221

*GROOOWLL*

I will first point of So be careful.

This means we will treat k as a constant. Eg a number. It could be anything really...1, a half, 60.







Let's substitute our limits!







Second question!






Substitute our limits!








A QUADRATIC! This is where YEAR 8 MATHS KICKS IN!

 

[bleakarcher]

c) integral[k-3] dx from x=2 to x=3
=[(k-3)x] from x=2 to x=3
=(k-3)3-(k-3)2=k-3
But integral[k-3] dx from x=2 to x=3 is equal to 5 it follows that:
=>k-3=5. Hence k=8.
f) integral[k+3x] dx from x=1 to x=k
=[kx+(3/2)x^2] from x=1 to x=k
=k^2+(3/2)k^2-k-(3/2)=(5/2)k^2-k-(3/2)
But integral[k+3x] dx from x=1 to x=k is equal to 13/2 it follows that:
=>5k^2-2k-3=13 i.e. 5k^2-2k-16=0
i.e. (k-2)(5k+8)=0. Hence, k=-8/5, 2

 

[qwerty44]

*GROOOWLL*

I will first point of WE ARE INTEGRATING WITH RESPECT TO dx not dk! So be careful.

This means we will treat k as a constant. Eg a number. It could be anything really...1, a half, 60.






Let's substitute our limits!






Second question!





Substitute our limits!







A QUADRATIC!

Ahhhhh. haha you dont even want to know what i had as the primitive for the first one. I did it right be i messed up the primitive.

And i understand the second one too. THANX

 

[Carrotsticks]

Spiral, may I suggest that for the limits after integration (involving brackets), you use:



As opposed to:



I just think it is a lot more aesthetically pleasing =)

Code:

\left [\frac{x^{2}}{3}^ \right ]^{1}_{0} <--------- Pretty one

versus

[\frac{x^{2}}{3}]^{1}_{0}]

 

[SpiralFlex]

Spiral, may I suggest that for the limits after integration (involving brackets), you use:



As opposed to:



I just think it is a lot more aesthetically pleasing =)

Code:

\left [\frac{x^{2}}{3}^ \right ]^{1}_{0} <--------- Pretty one

versus

[\frac{x^{2}}{3}]^{1}_{0}]

YOU KNOW HOW LONG I'VE BEEN WAITING FOR SOMEONE TO TELL ME HOW TO DO THAT?! THANKS!

 

[qwerty44]

btw, with the first question, is it just a coincidence that 8-3=5? if you know what i mean????

 

[SpiralFlex]

btw, with the first question, is it just a coincidence that 8-3=5? if you know what i mean????

For this case yes.

HOWEVER,

If k was indeed 8.













Turns out factor by grouping can give us this.

This is just a method of CHECKING your answers.

 

[RealiseNothing]

btw, with the first question, is it just a coincidence that 8-3=5? if you know what i mean????

Coincidence.

If you put the limits as 4 and 2 you wouldn't get k=8.

 

[Carrotsticks]

btw, with the first question, is it just a coincidence that 8-3=5? if you know what i mean????

Perhaps you can try finding out yourself. Playing around with things is the best way to learn.

To find out if your little 'theory' is true, try to evaluate:



If your theory is true, you should arrive with the result:

 

[RealiseNothing]

Also a quick way to do the first one:

Notice that since we are integrating with respect to "x" that "k" is a constant? Therefore, the expression we are integrating is a constant, ie y=m (m is just any number).

So we notice that the area under the line will form a rectangle with the x and y axis. Since the limits are 3 and 2, the width of the rectangle is 1.

But we know that the area is 5, so the length must also be 5 (length times width = area)

So y=5 is the line we are integrating, hence k-3=5 and k is 8.

 

[qwerty44]

Yep i get it now..

 

[RealiseNothing]

But if u made the limit 3 and 4 you would get k=8. Or if you made it 5 and 6. or 10 and 11. Or 24 and 25. As long as there is one difference between the limits, then k=8. I tried it with a different question, and as long as the difference of the roots is one, then it holds true and not coincidence.

That's because you are integrating a constant. It will always form a rectangle and when the limits differ by only 1, the width is 1. So the length will always equal the resulting area.

It's a coincidence in general, not if you applied this rule.

 

[SpiralFlex]

As he will see in the next chapter of the Cambridge book.