Integration and Logarithmic Help! (1 Viewer)

Lemiixem · Feb 14, 2012

Hey Guys,
I am just having trouble making X the subject in these questions.

Q10. Find the Area between the curve y=lnx, the y-axis and the lines y=2 and y=4, correct to 3 significant figures.

Q11. Find the exact volume of the solid formed when the curve y=logx is rotated about the y-axis from y=1 to y=3

Q12. Show that (3x+3)/(x^2-9) = 1/(x+3) + 2/(x-3)
Hence find the integral of (3x+3)/(x^2-9) dx.

nightweaver066 · Feb 14, 2012

$$Remember that: $ log_ba = x$

$b^x = a$

$$Q10. $ y = lnx$

$x = e^y$

$I'm sure you can do the rest yourself.$

$Q11. use the same property above$

$You can do this method by splitting the LHS in to two fractions or combining the RHS$

$If you want to split the LHS, do it this way:$

$\frac{3x + 3}{x^2 - 9}$

$=\frac{3x + 3}{(x - 3)(x + 3)} \equiv \frac{A}{x -3} + \frac{B}{x + 3}$

$= 3x + 3 = A(x + 3) + B(x - 3)$

$From there, solve for A and B and you should obtain the fractions$

$\int \frac{3x + 3}{x^2 - 9}dx$

$= \int \frac{1}{x + 3} + \frac{2}{x - 3} dx$

$= ln|x + 3| + 2ln|x - 3| + c$

D94 · Feb 14, 2012

With the third question, since I assume you only do Maths 2U, combine the RHS to form the LHS.

Integration and Logarithmic Help! (1 Viewer)

Lemiixem

Member

nightweaver066

Well-Known Member

D94

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)