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HSCAREA · Mar 10, 2011

Find the area bounded by the curve y=x/ (x^2 + 1), the x-axis and the lines x = 2 and x = 4. 2 dec place.

what I ended up with is 1/2(ln(17) - ln(5))

which is wrong

Aquawhite · Mar 10, 2011

Have you drawn the line of the equation, that can help finding the area.

You are correct in identifying that you must integrate using a log, however.

HSCAREA · Mar 10, 2011

Aquawhite said:
$A = \int_{4.2}^{2} \frac{x}{x^2 + 1} dx$

$A = 2ln[2x]_{2}^{4.2}$

$A = 2 [ln(8.4) - ln(4)]$

$A = 2 ln(\frac{8.4}{4})$

$A = 2 ln(2.1)$

nonono not 4.2 its just 4

Aquawhite · Mar 10, 2011

I removed that post because I used 2x in the calculation - whereas all I had to use was x^2 + 1. I confused myself - it has been a long time since I did maths.

It should read: $A = \frac{1}{2} ln[x^2 + 1]_2^4$

My apologies - then just substitute stuff in. I was watching Judge Judy at time >_>. Rookie mistake which I should not have made.

Edit: I cannot believe I actually made that mistake in my initial post, haha.

HSCAREA · Mar 10, 2011

ah cheers, I had it right. 1/2 * (ln(17) - ln(5)) = 0.61 units^2

I went one step further and made it 1/2 *(ln(12)) ' which is indeed very very wrong. thanks.

HSCAREA · Mar 12, 2011

Solve tan^2 x - 1 = 0 for 0 <= x <= 2pi

AAEldar · Mar 12, 2011

HSCAREA said:
Solve tan^2 x - 1 = 0 for 0 <= x <= 2pi

$tan^2{x}-1=0 \\ tan^2{x}=1 \\ tan{x}=\pm1 \\ x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$

HSCAREA · Mar 12, 2011

AAEldar said:
$tan^2{x}-1=0 \\ tan^2{x}=1 \\ tan{x}=\pm1 \\ x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$

nice thanks I like your thinking.

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