Author Topic: integration by substitution...except there's a catch >.< (Read 1501 times) Tweet Share

pHysiX · « **on:** May 10, 2009, 07:54:53 pm »

hello again VNers. I have yet another question. This time though, I would like some HINTS to try and get me started.

$f(x) = \int\limits_0^{{x^2}} {\sqrt {1 + {t^2}} dt}$

Find the derivative of f(x) WITHOUT evaluating the integral.

I managed to kinda find a pattern and got the answer being $2x\sqrt {{x^4} + 1}$ . However, I need a "mathematically correct" way of arriving at this answer.

All help much appreciated =]

bigtick · « **Reply #1 on:** May 10, 2009, 08:58:19 pm »

Use the chain rule

pHysiX · « **Reply #2 on:** May 10, 2009, 09:16:28 pm »

xD

that's exactly wat was on my sheet.

problem is i don't exactly know where to start. maybe like a formula or one step of working please? =]

ed_saifa · « **Reply #3 on:** May 10, 2009, 09:32:43 pm »

The textbook gives you a really clear example. This is an assignment question right?

pHysiX · « **Reply #4 on:** May 10, 2009, 09:58:18 pm »

haha yup. are you talking about the 5th edition txtbook?

ed_saifa · « **Reply #5 on:** May 10, 2009, 10:04:27 pm »

yep. That's how I did the assignment last year =P

Mao · « **Reply #6 on:** May 10, 2009, 10:09:27 pm »

fundamental theorem I:

if $\frac{d}{du} \left( \int_{a}^{u} g(t)\; dt\right) = g(u)$

hence, letting $u=x^2$

$\implies \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \frac{d}{du} \left( \int_{0}^{u} \sqrt{1+t^2}\; dt\right) \cdot \frac{d(x^2)}{dx} = 2x\sqrt{1+x^4}$

pHysiX · « **Reply #7 on:** May 10, 2009, 10:15:34 pm »

wow thanks mao =]

here was my way:

$f(x) = \int\limits_0^{{x^2}} {\sqrt {1 + {t^2}} dt}$

Since $\int\limits_a^b {f(x)dx} = F(b) - F(a)$ ,

=> $f(x) = F({x^2}) - F(0)$
=> $f'(x) = F'({x^2})2x$

Because logically $F'(t) = \sqrt {1 + {t^2}}$

therefore, $f'(x) = 2x\sqrt {1 + {x^4}}$

Does that work? My assignment specifically said to use the chain rule so yea...

Mao · « **Reply #8 on:** May 10, 2009, 11:26:36 pm »

You used the chain rule correctly, but the line 'because logically $F'(t)=\sqrt{1+t^2}$ ' can lose you marks. I got told off for writing things like that last year. Either reference that to fundamental theorem I (i.e. $F'(t) = \frac{d}{dt} \int_{0}^{t} \sqrt{1+s^2}\; ds = \sqrt{1+s^2}$ ), or just use the fundamental theorem I from the start [which is basically the point of these few questions].

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Author Topic: integration by substitution...except there's a catch >.< (Read 1501 times) Tweet Share

pHysiX

integration by substitution...except there's a catch >.<

bigtick

Re: integration by substitution...except there's a catch >.<

pHysiX

Re: integration by substitution...except there's a catch >.<

ed_saifa

Re: integration by substitution...except there's a catch >.<

pHysiX

Re: integration by substitution...except there's a catch >.<

ed_saifa

Re: integration by substitution...except there's a catch >.<

Mao

Re: integration by substitution...except there's a catch >.<

pHysiX

Re: integration by substitution...except there's a catch >.<

Mao

Re: integration by substitution...except there's a catch >.<

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