Author Topic: 4U Maths Question Thread (Read 665242 times) Tweet Share

RuiAce · « **Reply #120 on:** April 03, 2016, 05:17:29 pm »

Quote from: mq123 on April 03, 2016, 05:01:34 pm

`hey, can someone please help me with this...i have little/to no idea how to do it...
find d/dx of:
$\int_{1}^{1+{x}^{2}}{\frac{1}{\sqrt{1+{e}^{t}}}\,dt}$

thankyou so much in advance

$\text{Mentally processing the integral I could not find a method to use the fundamental theorem of calculus}\\\text{So the rigorous way to compute this integral is as follows}$
${u}^{2}=1+{e}^{t}\Rightarrow 2u\,du=e^{t}\,dt$
$\begin{align*}\int_{1}^{1+{x}^{2}}{\frac{{e}^{t}dt}{{e}^{t}\sqrt{1+{e}^{t}}}}&=\int_{\sqrt{1+e}}^{\sqrt{1+{e}^{1+{x}^{2}}}}{\frac{2u\,du}{{\left({u}^{2}-1\right)}u}}\\&=\int_{\sqrt{1+e}}^{\sqrt{1+{e}^{1+{x}^{2}}}}{\frac{2}{{u}^{2}-1}du}\\&=\int_{\sqrt{1+e}}^{\sqrt{1+{e}^{1+{x}^{2}}}}{\left(\frac{1}{u-1}-\frac{1}{u+1}\right)du}\\&={\left[\ln{\left|u-1\right|}-\ln{\left|u+1\right|}\right]}_{\sqrt{1+e}}^{\sqrt{1+{e}^{1+{x}^{2}}}}\\&=\ln{\left(\sqrt{1+{e}^{1+{x}^{2}}}-1\right)}-\ln{\left(\sqrt{1+{e}^{1+{x}^{2}}}+1\right)}-C\end{align*}$
$\text{where }C=\ln{\left(\sqrt{1+{e}}-1\right)}-\ln{\left(\sqrt{1+{e}}+1\right)}$

$\text{The derivatives are MUCH harder to compute, as it is essentially the chain rule within the chain rule.}\\\text{This is why I imagine that there'd be a way to use the first fundamental theorem of calculus.}\\\text{But right now I can't figure out what...nor is it in the HSC syllabus.}$

Maz · « **Reply #121 on:** April 03, 2016, 05:21:25 pm »

okay...thankyou
do you know how to do it without use using logs or In?

RuiAce · « **Reply #122 on:** April 03, 2016, 05:22:08 pm »

Quote from: mq123 on April 03, 2016, 05:21:25 pm

okay...thankyou
do you know how to do it without use using logs or In?

Do they teach you hyperbolic functions in VCE?

$\int{\frac{dx}{{x}^{2}-1}}=\tanh^{-1}{x}$

Maz · « **Reply #123 on:** April 03, 2016, 05:26:47 pm »

no they don't...is there a way to integrate the function without using any of those things?
there's another question you might be able to use to explain it (i attached it) ...because we haven't done logs at school yet...so i'm thinking they may not ask those type of questions

RuiAce · « **Reply #124 on:** April 03, 2016, 05:30:36 pm »

$\int_{1}^{{e}^{x}}{{\left(1+\sqrt{t}\right)}^{5}\,dt}$

I would've expanded that one out tbh. What was your method to attempting this one?

Maz · « **Reply #125 on:** April 03, 2016, 05:39:46 pm »

i just subbed e^x into (1+√t)^5 into the equation...then subbed 1 in and then just F(e^x) - F(1)...but it's not working

RuiAce · « **Reply #126 on:** April 03, 2016, 05:47:31 pm »

Quote from: mq123 on April 03, 2016, 05:39:46 pm

i just subbed e^x into (1+√t)^5 into the equation...then subbed 1 in and then just F(e^x) - F(1)...but it's not working

Facepalm
Of course! Just use the second fundamental theorem in the more clever way...

Ok let's see, in that case the answer would crumble down to
$\frac{d}{dx}\int_{1}^{1+{x}^{2}}{\frac{dt}{\sqrt{1+{e}^{t}}}}=\frac{1}{\sqrt{1+e^{1+{x}^{2}}}}-\frac{1}{\sqrt{1+e}}$

Then what does your given answer say?

Maz · « **Reply #127 on:** April 03, 2016, 05:51:57 pm »

the given answer says:

RuiAce · « **Reply #128 on:** April 03, 2016, 06:04:41 pm »

Since the method isn't wrong at this point I have to presume the answers are because...
http://www.wolframalpha.com/input/?t=crmtb01&i=1%2Fsqrt(1%2Be%5E(1%2Bx%5E2))-1%2Fsqrt(1%2Be)%3D2x%2Fsqrt(1%2Be%5E(1%2Bx%5E2))

Not true for ALL x

Maz · « **Reply #129 on:** April 03, 2016, 06:14:38 pm »

okay thankyou- so essentially u always just sub the limits in? because a lot of my answers are coming up as wrong that way...maybe they are all wrong...but then it is a published textbook...hmmm

RuiAce · « **Reply #130 on:** April 03, 2016, 06:30:36 pm »

Quote from: mq123 on April 03, 2016, 05:26:47 pm

no they don't...is there a way to integrate the function without using any of those things?
there's another question you might be able to use to explain it (i attached it) ...because we haven't done logs at school yet...so i'm thinking they may not ask those type of questions

Wait I'm just thinking, shouldn't the answer for this one have been

${\left(1+\sqrt{{e}^{x}}\right)}^{5}\cdot{e}^{x}$

Because I'm inclined to say we missed something important - the chain rule

Maz · « **Reply #131 on:** April 03, 2016, 06:40:43 pm »

that is the answer!

so do u do the chain rule on that

Maz · « **Reply #132 on:** April 03, 2016, 06:42:40 pm »

Quote from: RuiAce on April 03, 2016, 06:30:36 pm

Wait I'm just thinking, shouldn't the answer for this one have been

${\left(1+\sqrt{{e}^{x}}\right)}^{5}\cdot{e}^{x}$

Because I'm inclined to say we missed something important - the chain rule

but how does the chain rule fit...we don't even need to do any differentiation or integration cos h equestion is saying to differentiate then integrate...?

RuiAce · « **Reply #133 on:** April 03, 2016, 06:43:23 pm »

$\text{Ok, I think this is how to do it. According to the fundamental theorem of calculus, where }a\text{ is a constant:}\\\frac{d}{dx}\int_{a}^{x}{f(t)\,dt}=f(x)$

$\text{For question 1, proceed as follows:}\\\text{Let }y=\int_{1}^{{e}^{x}}{{\left(1+\sqrt{t}\right)}^{5}\,dt}\\\text{Then let }u={e}^{x}\Rightarrow \frac{du}{dx}={e}^{x}\\y=\int_{1}^{u}{{\left(1+\sqrt{t}\right)}^{5}\,dt}\\\Rightarrow \frac{dy}{du}={\left(1+\sqrt{u}\right)}^{5}\\\text{using the fundamental theorem}$
$\text{By the chain rule:} \\ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}={e}^{x}{\left(1+\sqrt{u}\right)}^{5}={e}^{x}{\left(1+\sqrt{{e}^{x}}\right)}^{5}$
$\frac{d}{dx}\int_{1}^{{e}^{x}}{{\left(1+\sqrt{t}\right)}^{5}\,dt}={e}^{x}{\left(1+\sqrt{{e}^{x}}\right)}^{5}$

$\text{For question 2:}\\\text{Let }y=\int_{1}^{1+{x}^{2}}{\frac{dt}{\sqrt{1+{e}^{t}}}}$
$\text{Then let }u=1+{x}^{2}\Rightarrow \frac{du}{dx}=2x\\ y=\int_{1}^{u}{\frac{dt}{\sqrt{1+{e}^{t}}}}\Rightarrow \frac{dy}{du}=\frac{1}{\sqrt{1+{e}^{u}}}$
$\text{If you use the chain rule again you get the correct answer}$

Basically, I must apologise - my lack of experience with the fundamental theorem meant I applied it wrongly! I learnt something new today.

RuiAce · « **Reply #134 on:** April 03, 2016, 06:45:10 pm »

Quote from: mq123 on April 03, 2016, 06:42:40 pm

but how does the chain rule fit...we don't even need to do any differentiation or integration cos h equestion is saying to differentiate then integrate...?

Actually it said to integrate then differentiate. So the fundamental theorem of calculus allows us to work our way around that.

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