AQA Core 3 - June 11th 2010

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So 8d)

The curves are $e^{2x}-1$ and $4e^{-2x}+2$. The intersection, from an earlier part is $x=ln2$.

To find the shaded area is: $\displaystyle \int_0^{ln2} 4e^{-2x}+2 \, dx. - \displaystyle \int_0^{ln2} e^{2x}-1 \, dx.$

$\displaystyle \int_0^{ln2} 4e^{-2x}+2 \, dx. = \left[ -2e^{-2x} + 2x \right]_0^{ln2}$

$= (-2e^{-2lnx} + 2ln2) - (-2e^0 + 0)$

$= (-\frac{1}{2} + ln4) - (-2)$

$= \frac{3}{2} + ln4$



$\displaystyle \int_0^{ln2} e^{2x}-1 \, dx. = \left[ \frac{1}{2}e^{2x} - x \right]_0^{ln2}$

$= (\frac{1}{2}e^{2ln2} - ln2) - (\frac{1}{2}e^0 - 0)$

$= (2 - ln2) - (\frac{1}{2})$

$= \frac{3}{2} - ln2$


Therefore shaded area: $= (\frac{3}{2} + ln4) - (\frac{3}{2} - ln2)$

$= ln4 + ln2 = ln8$

So 8d)

The curves are $e^{2x}-1$ and $4e^{-2x}+2$. The intersection, from an earlier part is $x=ln2$.

To find the shaded area is: $\displaystyle \int_0^{ln2} 4e^{-2x}+2 \, dx. - \displaystyle \int_0^{ln2} e^{2x}-1 \, dx.$

$\displaystyle \int_0^{ln2} 4e^{-2x}+2 \, dx. = \left[ -2e^{-2x} + 2x \right]_0^{ln2}$

$= (-2e^{-2lnx} + 2ln2) - (-2e^0 + 0)$

$= (-\frac{1}{2} + ln4) - (-2)$

$= \frac{3}{2} + ln4$



$\displaystyle \int_0^{ln2} e^{2x}-1 \, dx. = \left[ \frac{1}{2}e^{2x} - x \right]_0^{ln2}$

$= (\frac{1}{2}e^{2ln2} - ln2) - (\frac{1}{2}e^0 - 0)$

$= (2 - ln2) - (\frac{1}{2})$

$= \frac{3}{2} - ln2$


Therefore shaded area: $= (\frac{3}{2} + ln4) - (\frac{3}{2} - ln2)$

$= ln4 + ln2 = ln8$

Oh my days!!!! I used the limits from the y-axis instead of the x! There goes the A*!

How many marks do you think I'll lose for using the wrong limits but integrating correctly?

Oh my days!!!! I used the limits from the y-axis instead of the x! There goes the A*!

same here... damn!

how many marks will i have lost for doing the VOR in degrees, i also left it in terms of pi will this affect the marks? didn't finish the last question just felt completely lost on it!!
on the simpsons rule i stupidly mixed up the odd and even ordinates so probably got no marks for that which i'm annoyed at!

what the ****... where did I go wrong

e^4x -3e^x -4 = 0

lne^4x -3lne^x -ln4 = 0

4x -3x = ln 4
x= ln 4

what have i donewrong?

what was the range on question 3? also what was the gf(x)=0 and the inverse of g?