Maths Assessment on exponentials/logs and integration (1 Viewer)

[goan_crazy]

ive uploaded this take home assessment 4 u guys

its due 2moro
one question but

i couldnt get this answer.

d) (i) Use the trapezoidal rule once to obtain an approximation for the are between the x-axis, the curve y=e^x and the lines x=1 and x=2

ii) Use the result in part (i) to show that e approximitely equals 3

im not sure i got an answer for part (i), have no idea bout the 2nd one
i jus wanna check that my answer is right

help would be appreciated

joe

 

[Trev]

Would you like answers for all the questions?

 

[FinalFantasy]

is that a take home assessment for maths? so u take it home and get da q's done and if u get dem all right u get 100%? lol

 

[Trev]

(i) Use the trapezoidal rule once to obtain an approximation for the are between the x-axis, the curve y=e^x and the lines x=1 and x=2

= (approx.) 1/2*(2-1)[e¹ + e²]
= [e + e²]/2
(ii) The exact area is given by
∫e^xdx [from 2 to 1]
[e^x] [from 2 to 1]
=e² - e
Therefore you can say;
Approximation = Exact (though it doesn't, but you know)
[e + e²]/2 = e² - e
Simplifying, you get e = 3.

 

[Trev]

i have done all the rest except that question

hey trev did u get the answer for part 1 of the trapezoidal rule question
coz we need an aproximation
i got 2.53 or somethin for that

Leaving my answer as that is still an approximation do not forget, as it was an approximation method we were using. Though, to 2 decimal places, the answer is (approx.) 5.05.

 

[Trev]

hey trev i got 1/2 for my h
i went b-a/2
so thats 2-1/2
=1/2
so i did 1/2 over 2 when calculating trapezoidal rule
=1/4

(i) Use the trapezoidal rule once to obtain an approximation for the are between the x-axis, the curve y=e^x and the lines x=1 and x=2

= (approx.) 1/4*[e1 + e2]

is that rite?

No, sorry. It is right how I did it before.
A = 1/2*h* [f(a)+f(b)]
Your h is (b-a)=(2-1)=1.
So it is 1/2*1*....(the rest).

 

[Trev]

a)
(i)
∫4/(3x-2)dx [2 to 1]
4/3*ln|3x-2| [2 to 1]
4/3[2ln2 - ln1]
8/3*ln2.

(ii)
∫xe^x²-1dx
1/2*(e^x²-1) + C

(iii)
∫2dx/(2x+1)²dx
-(2x+1)^-1 + C

 

[VLTURBO]

Does anyone know how to do question 3 b on joes sheet

 

[acmilan]

The domain is just basically all the x-values that the function can take without producing an illegal result. So the domain of y = xlnx would be the same as for y = lnx,

For stationary points you differentiate and make that equal 0.

y' = lnx + 1
y' = 0

lnx = -1
x = e^-1

So the stationary point is at x = e^-1, y = -e^-1. You can get nature by subbing into y'' and seeing its value.

Theres no inflection points because

y'' = 1/x

That has no solution of x for y'' = 0

 

[Trev]

trev i still dont get the trapazoidal question 2nd part
i got 5.05 doing it ur way

how can i explain e aprox equals 3?

I don't know how else to explain it.

I'll show working.
Approximation = Exact (though it doesn't, but you know)
[e + e²]/2 = e² - e
e + e² = 2e² - 2e
3e = e²
3 = e.

 

[HellVeN]

Joe, learn the trapeziodal rule again.

And remember, there are two trapezoidal rules.
Since this is just one application you could have used:


A = 1/2(b-a)[f(a)+f(b)]

or

A = h/2(y0,y1)
where h = (b-a)/2


It's the same thing since y0=f(a) and y1=f(b).

 

[rama_v]

a)
(i)
∫4/(3x-2)dx [2 to 1]
4/3*ln|3x-2| [2 to 1]
4/3[2ln2 - ln1]
8/3*ln2.

(ii)
∫xe^x²-1dx
1/2*ln(e^x²-1) + C

(iii)
∫2dx/(2x+1)²dx
-(2x+1)^-1 + C

hey Trev, where did the ln come from in part (ii), i just got the answer as (1/2)e^x2-1 ...

 

[Trev]

hey Trev, where did the ln come from in part (ii), i just got the answer as (1/2)e^x2-1 ...

My mistake, sorry.
I should really learn to read over what I type

 

[acmilan]

d) (i) Use the trapezoidal rule once to obtain an approximation for the are between the x-axis, the curve y=e^x and the lines x=1 and x=2

ii) Use the result in part (i) to show that e approximitely equals 3

we got the results back 2day
part i is alrite
nearly Everyone got q1d part ii wrong
explain how e aprox equals 3?

rep for the first person who gets it right

Didnt Trev explain it?

Approximation is approximately = Exact
[e + e2]/2 = e2 - e
e + e2 = 2e2 - 2e
3e = e2
3 = e.

The only other way I could think of explaining it is by printing out the result when you type e into the calculator and showing that it is close to 3

 

[acmilan]

Whatever answer given, im sure Trev's method would be just as valid as it does what it asks to do.

 

[Trev]

Whatever answer given, im sure Trev's method would be just as valid as it does what it asks to do.

I have no idea how to do it otherwise.

Putting e into your calculator and saying it equals 2.7 etc which is close to 3; I doubt this is a way which you would be awarded marks for.

 

[Riddles]

good luck mate, i hate exponentials

 

[SaHbEeWaH]

by approximation,
area = [e + e²]/2 = 5.053668964...

by exact value,
area = e² - e

equate the exact value to the approximate value

e² - e = 5.053668964...
e(e - 1) = 5.053668964...
e = [5.053668964...]/(e - 1) = 2.941117621...
approximately = 3

kind of stupid as you have to divide by e to get an approximation of it but yea

 

[VLTURBO]

exact area = e^2 - e^1

so, e^2 - e^1 = (e^2 + e^1)/2

2e^2 – 2e^1 = e^2 + e^1

e^2 = 3e^1

3e^1 - e^2 = 0

e(3 – e) = 0

so e=0 and e=3

 

[acmilan]

er thats exactly the same method? Except that Trev divided by e earlier on, therefore eliminating the whole e = 0 thingy which is impossible