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Arisa_90 · « **Reply #1410 on:** March 11, 2017, 11:37:29 am »

Could i have some help with this question please?

RuiAce · « **Reply #1411 on:** March 11, 2017, 11:46:38 am »

Quote from: Arisa_90 on March 11, 2017, 11:37:29 am

Could i have some help with this question please?

What are we doing with this expression?

J.B · « **Reply #1412 on:** March 11, 2017, 04:23:29 pm »

I was just wondering how do you integrate this expression in respect to x?
(iii)

jakesilove · « **Reply #1413 on:** March 11, 2017, 04:30:05 pm »

Quote from: J.B on March 11, 2017, 04:23:29 pm

I was just wondering how do you integrate this expression in respect to x?
(iii)

(Image removed from quote.)

We want to integrate

$xe^{3x^2+5}$

Now, we need to remember that when we differentiate an exponential function, the power stays the same, and we multiply be the derivative of the power. In formal terms;

$\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$

This is the only way we know how to integrate exponentials; if the derivative of the power is in front of the exponential! Turning to the question at hand, let's differentiate the power.

$\frac{d}{dx}3x^2+5=6x$

Now, we don't quite have 6x in front of the exponential function. However, we can rewrite the question like this

$xe^{3x^2+5}=\frac{1}{6}6xe^{3x^2+5}$

Finally, we can integrate!

$\int \frac{1}{6}6xe^{3x^2+5} dx =\frac{1}{6} \int 6xe^{3x^2+5}=\frac{1}{6} e^{3x^2+5}+C$

For questions like this, practice makes perfect

J.B · « **Reply #1414 on:** March 11, 2017, 04:35:18 pm »

Quote from: jakesilove on March 11, 2017, 04:30:05 pm

We want to integrate

$xe^{3x^2+5}$

Now, we need to remember that when we differentiate an exponential function, the power stays the same, and we multiply be the derivative of the power. In formal terms;

$\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$

This is the only way we know how to integrate exponentials; if the derivative of the power is in front of the exponential! Turning to the question at hand, let's differentiate the power.

$\frac{d}{dx}3x^2+5=6x$

Now, we don't quite have 6x in front of the exponential function. However, we can rewrite the question like this

$xe^{3x^2+5}=\frac{1}{6}6xe^{3x^2+5}$

Finally, we can integrate!

$\int \frac{1}{6}6xe^{3x^2+5} dx =\frac{1}{6} \int 6xe^{3x^2+5}=\frac{1}{6} e^{3x^2+5}+C$

For questions like this, practice makes perfect

Awesome thank you

RuiAce · « **Reply #1415 on:** March 11, 2017, 04:39:08 pm »

Note that at the 2U level, you are NOT expected to do that sort of question without guidance.

jakesilove · « **Reply #1416 on:** March 11, 2017, 05:23:32 pm »

Quote from: RuiAce on March 11, 2017, 04:39:08 pm

Note that at the 2U level, you are NOT expected to do that sort of question without guidance.

I honestly didn't know that. Thanks Rui! In that case, they'd always get you to do something like differentiate the function, and then integrate a (slightly) different function.

Arisa_90 · « **Reply #1417 on:** March 11, 2017, 05:49:11 pm »

Oops. the question asks to anti differentiate it

Quote from: RuiAce on March 11, 2017, 11:46:38 am

What are we doing with this expression?

VydekiE · « **Reply #1418 on:** March 11, 2017, 07:28:45 pm »

Hi, it would be great if I could get some help on this question
I'm not sure how to evaluate e^log3
Thank you!!

jakesilove · « **Reply #1419 on:** March 11, 2017, 07:33:02 pm »

Quote from: VydekiE on March 11, 2017, 07:28:45 pm

Hi, it would be great if I could get some help on this question
I'm not sure how to evaluate e^log3
Thank you!!

Let's let the value equal some value, x. Also, can I assume that you mean the natural log of 3?

$e^{ln(3)}=x$

Now, we can ln both sides

$ln(e^{ln(3)})=ln(x)$

Using log laws,

$ln(3)*ln(e)=ln(x)$

Now, ln(e) is just 1, so

$ln(3)=ln(x)$
$x=3$

As a general rule,

$e^{ln(x)}=x$

jakesilove · « **Reply #1420 on:** March 11, 2017, 07:40:05 pm »

Quote from: VydekiE on March 11, 2017, 07:28:45 pm

Hi, it would be great if I could get some help on this question
I'm not sure how to evaluate e^log3
Thank you!!

If you meant log base 10, then I don't think there's a straight forward answer.

$e^{log(3)}=x$
$ln(e^{log(3)})=ln(x)$
$log(3)*ln(e)=ln(x)$
$log(3)=ln(x)$
$log_{10}(3)=ln_e(x)=\frac{log_{10}(x)}{log_{10}(e)}$
$log_{10}(3)*log_{10}(e)=log_{10}(x)$

From there, I don't think there are any basic operations you can use to move forward. Hmm...

$10^{log(3)*log(e)}=10^{log(x)}$
$(10^{log(3)})^{log(e)}=x$
$3^{log(e)}=x$

Well... I guess that's something

Edit: That's sort of interesting, I've just shown that

$a^{log(b)}=b^{log(a)}$

Interesting, here, is obvious a relative term.

VydekiE · « **Reply #1421 on:** March 11, 2017, 08:02:54 pm »

Quote from: jakesilove on March 11, 2017, 07:33:02 pm

Let's let the value equal some value, x. Also, can I assume that you mean the natural log of 3?

$e^{ln(3)}=x$

Now, we can ln both sides

$ln(e^{ln(3)})=ln(x)$

Using log laws,

$ln(3)*ln(e)=ln(x)$

Now, ln(e) is just 1, so

$ln(3)=ln(x)$
$x=3$

As a general rule,

$e^{ln(x)}=x$

Thank you so much!!

kinky_khan · « **Reply #1422 on:** March 11, 2017, 08:26:42 pm »

Hi, I'm having some trouble with this question, I have answered it but I am unsure as to whether I am doing it correctly. Thank you for your time and help.

jakesilove · « **Reply #1423 on:** March 11, 2017, 08:34:09 pm »

Quote from: kinky_khan on March 11, 2017, 08:26:42 pm

Hi, I'm having some trouble with this question, I have answered it but I am unsure as to whether I am doing it correctly. Thank you for your time and help.

I like to think about this year-by-year.

In the first year, $1000 is deposited, and then 6% interest is added.

$A_1=1000(1.06)$

The second year, another $1000, another 6%

$A_2=(1000(1.06)+1000)(1.06)=1000(1.06)^2+1000(1.06)$

And again

$A_3=1000(1.06)^3+1000(1.06)^2+1000(1.06)$

See a pattern? For the nth year, we'll get

$A_n=1000(1.06)^n+1000(1.06)^{n-1}+....1000(1.06)=1000(1.06^n+1.06^{n-1}+...1.06)$

For n=18, we get

$A_18=1000(1.06^{18}+1.06^{17}+...1.06)($

Using sums of geometric series, we get

$A_{18}=1000(\frac{1.06(1-1.06^{18})}{1-1.06}=$32,760$

For the second part, we invest a different amount each time.

$A_1=1000(1.06)$
$A_2=(1000(1.06)+1000(1.06))(1.06)=1000(1.06)^2+1000(1.06)^2$
$A_3=(1000(1.06)^2+1000(1.06)^2+1000(1.06)^2)(1.06)=1000(1.06^3)+1000(1.06^3)+1000(1.06^3)=$3574.05$

This answers the second part. Can you see a pattern? Clearly,

$A_n=n*1000(1.06)^n$

So, for n=18,

$A_{18}=18*1000(1.06)^{18}=$51,478.10$

That's one lucky son.

kinky_khan · « **Reply #1424 on:** March 11, 2017, 08:48:37 pm »

Quote from: jakesilove on March 11, 2017, 08:34:09 pm

I like to think about this year-by-year.

In the first year, $1000 is deposited, and then 6% interest is added.

$A_1=1000(1.06)$

The second year, another $1000, another 6%

$A_2=(1000(1.06)+1000)(1.06)=1000(1.06)^2+1000(1.06)$

And again

$A_3=1000(1.06)^3+1000(1.06)^2+1000(1.06)$

See a pattern? For the nth year, we'll get

$A_n=1000(1.06)^n+1000(1.06)^{n-1}+....1000(1.06)=1000(1.06^n+1.06^{n-1}+...1.06)$

For n=18, we get

$A_18=1000(1.06^{18}+1.06^{17}+...1.06)($

Using sums of geometric series, we get

$A_{18}=1000(\frac{1.06(1-1.06^{18})}{1-1.06}=$32,760$

For the second part, we invest a different amount each time.

$A_1=1000(1.06)$
$A_2=(1000(1.06)+1000(1.06))(1.06)=1000(1.06)^2+1000(1.06)^2$
$A_3=(1000(1.06)^2+1000(1.06)^2+1000(1.06)^2)(1.06)=1000(1.06^3)+1000(1.06^3)+1000(1.06^3)=$3574.05$

This answers the second part. Can you see a pattern? Clearly,

$A_n=n*1000(1.06)^n$

So, for n=18,

$A_{18}=18*1000(1.06)^{18}=$51,478.10$

That's one lucky son.

Thanks heaps and I agree that is one lucky son

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Arisa_90

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