Exponential and Logarithmic Functions (1 Viewer)

[foram]

Another question;

Find ∫x²e^{3 - 1} dx.

Don't forget the question in my last post though, I still need help with that.

EDIT: ^ I'm really not sure. I don't suppose you could go through it...?

e^2 is a constant. so it's just e^2 . 1/3 . x^3 + C

 

[FDownes]

Sorry, I made a mistake when I first typed out the question, it should be Find ∫x²e^{x3 - 1} dx. It might mean ∫x²e^{x3 - 1} dx, I'm not sure.

The textbook answer is 1/3 * e^{x3 - 1} + C, however it might also mean 1/3 * e^{x3 - 1}...

 

[tacogym27101990]

Sorry, I made a mistake when I first typed out the question, it should be Find ∫x²e^{x3 - 1} dx. It might mean ∫x²e^{x3 - 1} dx, I'm not sure.

The textbook answer is 1/3 * e^{x3 - 1} + C, however it might also mean 1/3 * e^{x3 - 1}...

id say let u= x^3 -1

 

[midifile]

id say let u= x^3 -1

See thats what I was thinking, and I just tried it out and it works, but in 3 unit you should never be expected to come up with a substitution, it should always be given. But other than that I cant think of another way to do it

 

[cwag]

Ah... Right. Silly mistake again.

EDIT: Hmm... How would I go about integrating (ln y)²? Would it be 1/2(ln y) or am I completely off the track?

EDIT x 2: Oh crap, I'm completely forgetting my log laws again. (ln y)² = 2ln y, right? So how do I go about integrating that?

(ln y)² is not equal to 2 ln y....ln y² = 2 ln y

(ln y)² is not the same as ln y²

you cannot integrate a log....and the quesiton wants you to use simpsons rule anyway

 

[webby234]

I assume it's x²e^x3-1

Observe that when you differentiate x³ - 1 you get 3x²
So the substitution is u = x³- 1

 

[webby234]

See thats what I was thinking, and I just tried it out and it works, but in 3 unit you should never be expected to come up with a substitution, it should always be given. But other than that I cant think of another way to do it

True, but you don't explicitly need to use substitution (despite my solution ). It's like 2u questions that ask you to integrate things like x/(x² + 1) So just change the original question to 1/3(3x²e^x3-1) and then just observe that you have an exponential times the derivative of the power of the exponential.

 

[cwag]

Sorry, I made a mistake when I first typed out the question, it should be Find ∫x²e^{x3 - 1} dx. It might mean ∫x²e^{x3 - 1} dx, I'm not sure.

The textbook answer is 1/3 * e^{x3 - 1} + C, however it might also mean 1/3 * e^{x3 - 1}...

you do not need to subsitute....i think the question is ∫x²e^{x3 - 1} dx

in this case...by observation we can see that the x² is generated from the derivative of e^{x3 - 1}

if we differentiate e^{x3 - 1} we get.....3x²e^{x3 - 1}

to cancel the 3 we divide..

so the integral becomes. 1/3 e^{x3 - 1}

 

[cwag]

oh..sorry webby234...you already answered

 

[tacogym27101990]

yeah its not neccesary.
ive just found when i explainit to ppl they understand that easier

 

[FDownes]

Thanks guys. Ugh... I don't know how I'm gonna make it though my extension 1 exam without you.

 

[FDownes]

One more time;

Graph y = log_e(x - x²).

I'm not expecting anyone to be able to show me a completed diagram of the graph (there's one in the back of my book anyways), but could someone explain how I figure out things like the vertex and asymptotes? I can see that x =/= 0 and x =/= 1, but I'm not sure how to express it mathematically.

 

[tommykins]

One more time;

Graph y = log_e(x - x²).

I'm not expecting anyone to be able to show me a completed diagram of the graph (there's one in the back of my book anyways), but could someone explain how I figure out things like the vertex and asymptotes? I can see that x =/= 0 and x =/= 1, but I'm not sure how to express it mathematically.

Let x-x^2 = b

For ln b, b can be any number > 0

Thus, x-x^2 > 0
x>x^2

Therefore, the only possible way of doing this is by thinking logically. x =/= 0 is established, but what about the numbers more than 0?

we cannot have b<0 so therefore, any value of x must be greater than x squared.

Squaring a fraction = smaller number, thus the domain for this graph is 0<x<1.
You then think of it logically once again
if b is a really small fraction, then the value of y is infinitely small.

Sub in extremeties, such as x = 0.0000001, .9999999, .5

 

[Mark576]

i don't think you can intergrate log, unless you use the simpsons rule or somthing like that, but thats not really intergrating.

&int;ln(x)dx = xln(x) - x + c

 

[Slidey]

One more time;

Graph y = log_e(x - x²).

I'm not expecting anyone to be able to show me a completed diagram of the graph (there's one in the back of my book anyways), but could someone explain how I figure out things like the vertex and asymptotes? I can see that x =/= 0 and x =/= 1, but I'm not sure how to express it mathematically.

y=ln(x-x^2)
y=lnx+ln(1-x)

Graph y=lnx and y=ln(1-x)=ln(-(x-1)) so draw ln(x), flip it around the y-axis, then translate it one unit to the right.
Now graph the addition of those two functions by adding the co-ordinates of each one.

It helps to have some reference points first, so find the asymptotes and maxima/minima: y'=1/x-1/(1-x)=0, 1-x=x, x=1/2. Stat point at (1/2, -2ln(2)). Quickly test either side on a calculator to see if it is max, min or inflexion. Since there's only one stat point, and it is a maxima, the rest of the graph is below this point and there is no curvy funny business.

As for asymptotes: now we have the form y=ln(x)+ln(1-x), we can see that the asymptotes are going to be the asymptotes of y=ln(x) and y=ln(1-x). What are they? They are x=0 and x=1 (because x>=1 makes ln(1-x) undefined).

So in summary: one maxima, 2 asymptotes equally spaced either side of the maxima, and a completely negative, continuous (within 0 < x < 1) function (because the y value of the maxima is negative). This should be enough information to tell you that the function basically looks like a parabola (more like y=-(5)*(x-1/2)^2-1.4 actually).