Past paper 1, 2003, Question 8 I'm seeking help on integration
xkarenx
Posted 29 January 2006 - 07:58 PM
The Wedge Effect
Posted 29 January 2006 - 08:12 PM
First of all, you move the dx to the side, so it becomes:
Then you change it around so that:
You then integrate the expression and substitue the upper and lower limit for the values of x, with the lower limit's result being subtracted from the upper limit's result and you have your answer.
Edited by Wedge37, 29 January 2006 - 08:42 PM.
The Wedge Effect
Posted 29 January 2006 - 08:19 PM
That was a coding mistake. Sorry! The above post has now been edited. Thanks for pointing that out.
xkarenx
Posted 29 January 2006 - 08:30 PM
The Wedge Effect
Posted 29 January 2006 - 08:44 PM
George
Posted 30 January 2006 - 09:25 AM
You can see the rule, and a couple of examples, on page 4 of our Further Calculus notes.
xkarenx
Posted 30 January 2006 - 07:54 PM
Solve the equation 3cos(2 )+10cos( )-1=0 for 0 , correct to 2 decimal places.
If anyone can be bothered to explain in steps how to tackle this question I'd be very grateful.
The Wedge Effect
Posted 30 January 2006 - 08:27 PM
You then simplify the expression to get:
As you can see, this is a quadratic of the form
Substitue into the expression:
You can then factorise this expression and then replace the y values with the cos x values. It can then be simplified to get two values of x, in radians, of course. But be cafeful of the limits of the range of values of x you solve for. In this case, it's from 0 to
Edit: If you want me to clarify any further, feel free to ask me and I'll edit above post to final solutions and add a few more pointers.
Edited by Wedge37, 30 January 2006 - 08:31 PM.
xkarenx
Posted 30 January 2006 - 08:36 PM
The Wedge Effect
Posted 30 January 2006 - 09:07 PM
Just remember the limits, from 0 to That's probs why you end up with a different answer. Also, maybe you forgot to change your calculator to radians mode when calculating the answers to this questions. Values between 0 to 3.141 are correct.
Steve
Posted 30 January 2006 - 09:23 PM
I've attached my solution to this question, let me know if it helps (or doesn't!).
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The Wedge Effect
Posted 31 January 2006 - 02:37 PM