Integration

No, the fraction is $\frac{3}{2x^2}$

not on the q33 im looking at it isnt :L

edit: but if the question had a fraction like that:

$\frac{3}{2x^2} = \frac{3}{2}x^{-2}$

My bad, i have a mac so isn't compatible with this document, thus hard to read...

So, how would i go about it?


${3}x^{-3} - {6}x^{-2} + {2x} - 1$

same way you would if the powers were positive times it by the power, then take 1 off the power.

e.g. $\frac{d}{dx}3x^{-3} = -9x^{-4}$

So I got

$-9x^{-2} + 12x^{-1} + 2$

No you need to take one away from the power so -3 goes to -4, -2 goes to -3 etc..

No you need to take one away from the power so -3 goes to -4, -2 goes to -3 etc..

when you differentiate you take one away from the power

so if the power is 2 (e.g. x^2) it differentiates to 2x (because 2 - 1 = 1)

so if the power is -3 (e.g. x^-3) it differentiates to -3x^-4 (because -3 - 1 = -4)

easy to get confused by it but it actually makes perfect sense

I got you, my mind was elsewhere...

I got

$-9x^{-4} + 12x^{-3} + 2$ which = 0 @ turning point

I subbed in -2,-1,0,1,2 and none equaled 0

when you differentiate you take one away from the power

so if the power is 2 (e.g. x^2) it differentiates to 2x (because 2 - 1 = 1)

so if the power is -3 (e.g. x^-3) it differentiates to -3x^-4 (because -3 - 1 = -4)

easy to get confused by it but it actually makes perfect sense

are we still on question 33?

because that question is $\frac{1}{3}x^3-\frac{3}{2}x^2+2x-1$ :L

Which question is it?

Which question is it?

Your thoughts on 36

Area $\displaystyle = \int^3_1 (x^2+1)dx$

As we need to find area hence we need to integrate the expression, with limits.

= 4

how about Q)38.... normal is tangent right?

No, the area is not 4.

y=$x^{2} + 1$

dy/dx= $2x$ right?

Yes, thats the one, i think I may have confused you,

y = $\frac{1}{3}x^3-\frac{3}{2}x^2+2x-1$ :L