1) d/dx loge 2x/x+1

2) if l=w=x and h=y of a rectangular prism I worked out the volume to be V=(125x-x^3)/2. Find the dimensions for the maximum volume.

3) for the curve y=x^2-x-2.
a) Explain why the area bounded by the curve, the x axis and the lines x=1 and x=3 is not given by the integral with bounds 3 and 1 (x^2-x-2)dx.
b) Find the area bounded by the curve the x axis and the lines x=1 and x=3.

4) Find the point on the curve y= 3 +4x-x^2 where the tangent is perpendicular to y=x/3+1

Originally Posted by HeroWise

For question two the shape is a rectangular prism what will the height be if h= 125/2x-x/2?

I still don’t get Q3. The question says you can not use 1 and 3 as the bounds to calculate the area and yet you did

Here are the questions can you do the one with the stars please (questions listed below). Maybe seeing the questions might make it clearer.
Q11 b ii
Q11 c ii, iii
Q12 a and b

https://imgur.com/gallery/G4TwR

OOHHH
Ok see in q3, if u see its graph, a section is below the x axis, thus getting subtracted from the portion which is above the x axis:

A1 and A2 are the two areas we are tryna find but A1 is belo the axis, it WILL get subtracted when we integrate from 3 to 1
If u notice just after the integration step I have

(-3/2)-(1/3-1/2-2)

The (-3/2) is later absolute valued to give (3/2), i just removed the negative sign, if it were to stay it would subtract the original area and anyway u cant have neg area
Yeah sure ill do em rn

Doing the last one now

Thank you so much for this. You are awesome

And here is the last questions answer:

Thank you so much. Okay last two part b and d

https://imgur.com/gallery/9h7mn

b)

i) use the arc length formula $l = r \theta$ to find $AC$ and $DE$.

ii) use the area formula $A = \frac{1}{2} r^2 \theta$ to find the areas of the large sector $ABC$ and the small sector $DBE$.

For d), be careful when notating angles using one letter ($\angle A$) - it's technically ambiguous since that could be referring to any of two angles.

iii) $AD$ and $CA$ are corresponding sides of the two similar triangles. So to find their ratio you only need to find the ratio of any other two corresponding sides of the triangles, e.g. $AB$ and $DB$.

for first step in the formula, i tend to derive it and not memorise it cos idk i cant get to stick it in my head, like l=rtheta, htats simple but my assurance increases if I derive it haha

Originally Posted by HeroWise
OOHHH
Ok see in q3, if u see its graph, a section is below the x axis, thus getting subtracted from the portion which is above the x axis:

A1 and A2 are the two areas we are tryna find but A1 is belo the axis, it WILL get subtracted when we integrate from 3 to 1
If u notice just after the integration step I have

(-3/2)-(1/3-1/2-2)

The (-3/2) is later absolute valued to give (3/2), i just removed the negative sign, if it were to stay it would subtract the original area and anyway u cant have neg area
Yeah sure ill do em rn
So why is this wrong??? Why can’t you work out the area seperately and add them???

https://imgur.com/gallery/P2Kou

You can, but writing

$\int_1^2 x^2-x-2 \, dx = -\frac{7}{6} = \frac{7}{6} \mathrm{u^2}$

is a little questionable, and some markers may penalise you for this.

Alternatively you could take the absolute value

$\left| \int_1^2 x^2-x-2 \, dx \right| = \frac{7}{6}$

or even swap the bounds of integration

$\int_2^1 x^2-x-2 \, dx = \frac{7}{6}$

But then our areas are different? Cause wouldn’t you have to take both areas A1 and A2 get the absolute values and add them. For A1 I get 7/6 and A2= 11/6 so the total is 3 u^2. Your area was a fraction

My area was a fraction because we still need to add the other area, the one under the curve from 3 to 2.

Technically you only need to take the absolute value of
$\int_1^2 x^2-x-2 \, dx$
since the other part is known to be positive.

\begin{aligned} A &= \left| \int_1^2 x^2-x-2\, dx\right| + \int_2^3 x^2-x-2\, dx \\ &= \left|-\frac{7}{6}\right| + \frac{11}{6} \\ &= 3\, \mathrm{u^2}\end{aligned}

ofc u can integrate them separately, thats what you are meant to do, BUt if u do it in a step you need to abs the value which gives -ve

So 3u^2 is correct?

Ja senpai

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