Edexcel C3,C4 June 2013 Thread

Just one more question

What if it wasn't symmetrical? Or would we never such a graph in C4? (There's a question like this in the mixed exercise for integration which is why I'm asking)
Thank you again

How do you integrate this?


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help me with differentiating brackets -

i know you can differentiate a term like (x-1)^3 without expanding

but what if it was x^2 or x^3 term in the bracket, or if it were a quadratic/cubic? Does the differentiating method still work?

Yh i get dat bit buh wah if cot3x came up in the exam? would it be 1/3ln[sin(3x)]?

Just one more question

What if it wasn't symmetrical? Or would we never such a graph in C4? (There's a question like this in the mixed exercise for integration which is why I'm asking)
Thank you again

Someone PLEASE post their solution to question 7c from paper E!!! Emailed my teacher and they havent replied -_-

Well, it differentiates to -2 x 2^(2-2x)ln(2), so you would have to divide by -2ln(2).

Well, it differentiates to -2 x 2^(2-2x)ln(2), so you would have to divide by -2ln(2).

can someone explain to me why the x goes in the bolded line?

Where did the -2 come from?

if you differentiate (2-2x) you get -2

if you differentiate (2-2x) you get -2

In an experiment a scientist considered the loss of mass of a collection of picked leaves. The
mass M grams of a single leaf was measured at times t days after the leaf was picked.
The scientist attempted to find a relationship between M and t. In a preliminary model she
assumed that the rate of loss of mass was proportional to the mass M grams of the leaf.
(a) Write down a differential equation for the rate of change of mass of the leaf, using this
model.
(2)
(b) Show, by differentiation, that M = 10(0.98)^t
satisfies this differential equation.

It's differentiated with respect to x, so x is removed.

If you swap limits by substitution and you end up with limits that have a different size order, do you still go from high to low/top to bottom? For example $\int^5_3 \ dx$ turns into $\int^2_4 \ du$ Is that right? Or should it be $\int^4_2 \ du$