Author Topic: Can anyone show me how to do spesh VCAA 09 exam 1 Q8B? (Read 756 times) Tweet Share

98.40_for_sure · « **on:** July 17, 2010, 11:29:36 pm »

The answer is -2 + 3ln(3)

ive done the question about 40 times... i keep getting -2

WHAT THE HELL :tickedoff:

GerrySly · « **Reply #1 on:** July 17, 2010, 11:52:06 pm »

$\begin{align*} \int_{-1}^{1} \frac{2+x^2}{4-x^2}&=\int_{-1}^{1} \frac{3}{2 (x+2)}-\frac{3}{2 (x-2)}-1\\ &=\left [ \frac{3}{2}\log_e{(x+2)}-\frac{3}{2}\log_e{(x-2)}-x \right ]_{-1}^{1}\\ &=\left ( \frac{3}{2}\log_e{(1+2)}-\frac{3}{2}\log_e{(1-2)}-1\right )-\left ( \frac{3}{2}\log_e{(-1+2)}-\frac{3}{2}\log_e{(-1-2)}+1\right )\\ &=\frac{3}{2}\left ( \log_e{(3)}-\log_e{(-1)}-\log_e{(1)}+\log_e{(-3)} \right )-2\\ &=\frac{3}{2}\left ( \log_e{(-3)}+\log_e{(-3)}\right )-2\\ &=\frac{3}{2}\log_e{(9)}-2\\ &=3\log_e{(\sqrt{9})}-2\\ &=3\log_e{(3)}-2 \end{align*}$

Wow been a while since I've done any maths what so ever

98.40_for_sure · « **Reply #2 on:** July 17, 2010, 11:55:04 pm »

Quote from: GerrySly on July 17, 2010, 11:52:06 pm

$\begin{align*} \int_{-1}^{1} \frac{2+x^2}{4-x^2}&=\int_{-1}^{1} \frac{3}{2 (x+2)}-\frac{3}{2 (x-2)}-1\\ &=\left [ \frac{3}{2}\log_e{(x+2)}-\frac{3}{2}\log_e{(x-2)}-x \right ]_{-1}^{1}\\ &=\left ( \frac{3}{2}\log_e{(1+2)}-\frac{3}{2}\log_e{(1-2)}-1\right )-\left ( \frac{3}{2}\log_e{(-1+2)}-\frac{3}{2}\log_e{(-1-2)}+1\right )\\ &=\frac{3}{2}\left ( \log_e{(3)}-\log_e{(-1)}-\log_e{(1)}+\log_e{(-3)} \right )-2\\ &=\frac{3}{2}\left ( \log_e{(-3)}+\log_e{(-3)}\right )-2\\ &=\frac{3}{2}\log_e{(9)}-2\\ &=3\log_e{(\sqrt{9})}-2\\ &=3\log_e{(3)}-2 \end{align*}$

Wow been a while since I've done any maths what so ever

Yeah in your 2nd line of working, you get the 2 logs MINUSED
i keep getting them PLUSED...
i don't change my (4-x^2) to (x^2-4) but i should still be getting the same answer
farkkkkkin hell

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Author Topic: Can anyone show me how to do spesh VCAA 09 exam 1 Q8B? (Read 756 times) Tweet Share

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Can anyone show me how to do spesh VCAA 09 exam 1 Q8B?

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