math doubts [NEW]

[SGVaibhav]

i wanted to know what all they teach in AS level mathematics  EDEXCEL
im talking about the normal maths which everyone takes

do they have integration, differentiation and limits

if they have, what all do they have?

[astarmathsandphysics]

Yes all of thos but it is all so simple. Sleep easy

[cooldude]

if anyone uses Pure mathematics 1 by Hugh Neill & Douglas Quadling can u plz turn over to page 253 Ex.16D Q4.
i got rong answer prolly cause one of my limit is rong and thts whr im confused, can summon plz do n lemme kno?

if u still want the answer, here it is (i dont know how to use the integral sign, sorry)

ive attached a drawing, (please refer to that while reading the explanation)
first of all we'll find the area K+L by integration, we then subtract the area L from this area to find the area K.


first of all we find the area K+L by integration----->>>


the integral of the function (2x-5)^4---> [(2x-5)^5)/10]
the limits will be b/w Q and R, we can find the co-ordinate of Q by equating the equation of the curve to 0.
then the area K+L comes out to be---->


=> 24.3-0=24.3
we then find the area L. (we first have to find the equation of PR though).
since f(x)=(2x-5)^4
f'(x)=[4(2x-5)^3]*2

therefore the gradient of the tangent=216 at (4,81)
then the equation of PQ becomes---> y=216x-783, we find out the x-coordinate of R by equating the equation of PQ to 0.
then we find the area  L-->
0.5*0.375*81=15.1875
therefore K=24.3-15.1875=9.1125

hope i helped

p.s. sorry about the bad handwriting in the drawing, so just for reference Q(2.5,0), P(4,81), R(4,0), S(3.625,0)

[SGVaibhav]

i wanted to know what all they teach in AS level mathematics  EDEXCEL
im talking about the normal maths which everyone takes

do they have integration, differentiation and limits

if they have, what all do they have?

ok but then i wanted to know what all they teach in limits, differentitaion and integration

only the topic names of content...

[zara]

if u still want the answer, here it is (i dont know how to use the integral sign, sorry)

ive attached a drawing, (please refer to that while reading the explanation)
first of all we'll find the area K+L by integration, we then subtract the area L from this area to find the area K.


first of all we find the area K+L by integration----->>>


the integral of the function (2x-5)^4---> [(2x-5)^5)/10]
the limits will be b/w Q and R, we can find the co-ordinate of Q by equating the equation of the curve to 0.
then the area K+L comes out to be---->


=> 24.3-0=24.3
we then find the area L. (we first have to find the equation of PR though).
since f(x)=(2x-5)^4
f'(x)=[4(2x-5)^3]*2

therefore the gradient of the tangent=216 at (4,81)
then the equation of PQ becomes---> y=216x-783, we find out the x-coordinate of R by equating the equation of PQ to 0.
then we find the area  L-->
0.5*0.375*81=15.1875
therefore K=24.3-15.1875=9.1125

hope i helped

p.s. sorry about the bad handwriting in the drawing, so just for reference Q(2.5,0), P(4,81), R(4,0), S(3.625,0)

thanks a lot!!

[cooldude]

thanks a lot!!

np