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FAO Matt4478 and Nick R

needs nick and matts help on some working out for my new super b*****d exhaust can hence its in general clio disscusion

sorry about keep picking yours brain, but can some one explain this?

4 4
Area=|4x-x²dx=[2x²- x³/3]
0 0
={2(4)²- 4³/3}-(0)=10 2/3.

what i dont get is why x²dx becomes "x³/3", actually if u can can u explain the whole working out

and nick how do u get the dy/dx for b)?

dont know why it keeps putting the 4 and 0 to the begining
-----4................. 4
Area=|4x-x²dx=[2x²- x³/3]
just ignore the lines and dots
  BMW 320d Sport

OK 4X definitely integrates to 2X squared. If you differentiate back it confirms: (2x2)X to the power (2-1) other words 4X.

Integrating X squared gives you (X cubed)/3, because when you divide anything by 3, its equivalent to multiplying it by 1/3. (Whenever youre dealing with int. or diff. its a lot easier to rid yourself of divisions, and substitute them for multiples of if its X/4, call it 1/4X instead.)

Anyway at first glance X squared will integrate to X cubed. But if you diff that back to check it gives you 3(X squared). In other words three times bigger than the number you want. So to get round this problem, divide it by 3...hence (X cubed)/3

Cant remember what the b equation was. I know it was tricky though.


Missed this post earlier - sorry m8! Bit confused about the question (HTML is bollox for maths notation!) but here goes...

You asked why dy/dx=x^2 becomes y=(x^3)/3 after integration: Firstly, the most important thing to grasp is that integration is effectively the reverse of differentiation, so you just do the following:

Add one to the power (aka the exponent) and then divide the multiplier in front of the x by the new number. For example:

If: dy/dx = x^2 = 1x^2
Then add one to the power to make that part x^3 and divide it by the new power so you get: y=(x^3)/3 or y=(1/3)x^3

You might be able to see whats happening better if you now differentiate y=(1/3)x^3. Obviously, to differentiate we just multiply by the power then remove one from the power which gives us dy/dx=((1/3)*3)x^(3-1) or dy/dx=1x^2 or dy/dx=x^2. Which is where we started!

Integration is just differentiation in reverse! The only tricky bits are when you get to differentiating/integrating trigonometric functions and logarithmic functions (which arent actually that bad )

Also, on your question b) y = x^(1/3) - x^(-2/5) you just differentiate this by exactly the same rules you used before: Multiply the bit infront of the x by its power then remove one from the power. So:

y = x^(1/3) - x^(-2/5)
dy/dx = (1/3)x^((1/3)-1) - (-2/5)x^((-2/5)-1)
dy/dx = (1/3)x^(-2/3) - (-2/5)x^(-7/5)
dy/dx = (1/3)x^(-2/3) + (2/5)x^(-7/5)

Which is hopefully the same answer Nick got! Hope that made some sense...