integration and logarithmic function (1 Viewer)

ozidolroks · Mar 27, 2009

hi can someone pleasse help me with this question ?

Find the indefinite integral of 3 ^2x-1

and 2Q) Find correct to 2 decimal places, the area enclosed by the curve y= log base 2 x, the x-axis and the lines x=1 and x=3 by using Simpson's rule with 3 function values.

and 3 Q)find the exact area between the curve y= 1/x, the x-axis and the lines y= X and x= 2 in the first quadrant.

thankx

Drongoski · Mar 27, 2009

$Q1.\ \ Now:\ \ \ \int a^{u}\ du\ \ =\ \ \frac{a^{u}}{ln\ a}\ \ +\ \ C\\ \\ \therefore \ \ \ \int 3^{2x-1}\ dx\ \ =\ \ \frac{3^{2x=1}}{2\ ln\ 3}\ +\ C$

Edit: as pointed out by jetblack2007 the denominator should have a '2' as well.. Above now corrected.

Drongoski · Mar 27, 2009

jet · Mar 27, 2009

Drongoski said:
$Q1.\ \ Now:\ \ \ \int a^{u}\ du\ \ =\ \ \frac{a^{u}}{ln\ a}\ \ +\ \ C\\ \\ \therefore \ \ \ \int 3^{2x-1}\ dx\ \ =\ \ \frac{3^{2x=1}}{ln\ 3}\ +\ C$

Here, you have forgotten about the function in the place of u. So:
$\int 3^{2x-1} \, dx = \frac{3^{2x - 1}}{2 \, ln \, 3} + C = \frac{3^{2x - 1}}{ln \, 9} + C$

Drongoski · Mar 27, 2009

You are right .. I stand corrected. Thank you for pointing out jetblack2007

integration and logarithmic function (1 Viewer)

ozidolroks

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