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C2 soloman paper B. integration with an infinate limity

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so what about OP's question?

i integrated and got

4x^ {\dfrac -{3}{2}}

then i put values in and got

[0]-[1.414]
so is -1.414 correct as an answer?

You've messed up your

\LaTeX

. The function is positive everywhere in the domain of integration; does a negative answer really make sense? You mean

\sqrt{2}

, not 1.414.

Reply 21

thefatone

Original post by Zacken

You've messed up your

\LaTeX

. The function is positive everywhere in the domain of integration; does a negative answer really make sense? You mean

\sqrt{2}

, not 1.414.

better? i changed it

oh ... i just used my calculator but yes

\sqrt 2

so does that mean the area i got is below the x-axis?

Reply 22

Zacken

Original post by thefatone

better? i changed it

oh ... i just used my calculator but yes

\sqrt 2

so does that mean the area i got is below the x-axis?

No, it means you integrated wrong.

Reply 23

thefatone

Original post by Zacken

No, it means you integrated wrong.

aww s**t let's try again

\dfrac{6}{x^ \frac {5}{2}} = 6x^{-\frac {5}{2}}

so integrating

\displaystyle \frac{6x^{-\frac{3}{2}}}{-\frac{3}{2}}

-4x^{-\frac{3}{2}}

subbing infinity gives 0

subbing 2 in gives

-\left( \sqrt 2\right)

[0]-[-\sqrt 2]

so i get positive

\sqrt 2

????

(edited 8 years ago)

Reply 24

Zacken

Original post by thefatone

...

Where'd the negative in the denominator go?

Reply 25

thefatone

Original post by Zacken

Where'd the negative in the denominator go?

oh.... lemme change that

Reply 26

Zacken

Original post by thefatone

oh.... lemme change that

Okay, it's fine now. Now do your limits and remember the negative signs.

Reply 27

thefatone

Original post by Zacken

Okay, it's fine now. Now do your limits and remember the negative signs.

all better? (editied)

Reply 28

Zacken

Original post by thefatone

all better? (editied)

Yes.

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C2 soloman paper B. integration with an infinate limity

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