| 2005/07/07 12:47:01 PDT by gandalf [manager] |
Since the theme of the day seems to be the upcoming midterm, let's start off with a question that may aid you in your studies.
Problem #1: Evaluate the following integral.
![[image]](http://tol2mac8.soe.berkeley.edu:8081/balooch/16227966/Calc%20Prob%20of%20Day%201%20Image.JPG)
Calculus Problem of the Day, 7/7
| 2005/07/07 16:06:31 PDT by stuopidget [0/6] Awards: 5 from gandalf |
8/3 x root 2?
| 2005/07/07 17:37:18 PDT by stuopidget [0/6] Awards: 1 from gandalf |
Quote from stuopidget:
8/3 x root 2?
uh nvm, should be not possible with our present knowledge?
| 2005/07/07 20:41:13 PDT by gandalf [manager] |
Quote from stuopidget:
Quote from stuopidget:
8/3 x root 2?
uh nvm, should be not possible with our present knowledge?
No, that answer is not correct, but it is certainly solvable with the knowledge you have thus far. You just have to think outside the box - it is a test of how well you know the concepts behind integration.
Hint: Think about it geometrically. :)
| 2005/07/08 17:42:59 PDT by gandalf [manager] |
Anyone get a chance to try the problem yet? Even if you have a partial solution or some ideas, post them. I don't want to post the answer just yet - I'm interested in seeing what people have come up with. Keep my previous hint in mind. Remember geometrically, the definite integral is the area under the curve from a to b. Can you visualize this curve? That's a couple more hints right there. :D
| 2005/07/09 23:25:30 PDT by gandalf [manager] |
Quote from afantasys:
is it π? maybe?
Close. Very Close. You are half-way there. ::cough:: You seem to be on the right track. Look back to how you arrived at π - if you post your reasoning, I'll have a better idea of what you were thinking and I'll be able to point you in the right direction.
By the way, did you copy and paste the π symbol in from MSWord? Or were you able to find another way to enter it in? I'm asking because I'm looking for ways to post mathematical symbols on the forums. These forums unfortunately don't have LaTeX (a mathematical type-setter) enabled so I've been resorting to making images of the problems.
| 2005/07/10 12:11:41 PDT by afantasys [0/19] Awards: 3 from gandalf |
I got π because I figured that the graph of ∫(√(1-x2)dx) was a half circle above the x-axis from -1 to 1. but come to think of it and with help from your ::hint:: I think the graph should be a circle and the answer 2π.
PS: I copied the pie symbol from the some website, and i copied ∫(√(1-x2)dx) from MSWord symbols.
| 2005/07/10 12:32:00 PDT by gandalf [manager] |
Quote from afantasys:
I got π because I figured that the graph of ∫(√(1-x2)dx) was a half circle above the x-axis from -1 to 1. but come to think of it and with help from your ::hint:: I think the graph should be a circle and the answer 2π.
PS: I copied the pie symbol from the some website, and i copied ∫(√(1-x2)dx) from MSWord symbols.
Your first thought was correct when you said that the curve is a half-circle above the x-axis. The bottom half of the circle is given by - √(1-x^2) . Though your reasoning after this fact is faulty. What is the area of the unit circle with radius 1? ( Not 2π. )Then what is the area of half the unit circle (the top half as you had figured correctly before).
P.S: Sorry if my hint led you the other way. I suppose I should have said you were double of the way there - if that makes any sense. :P
| 2005/07/11 22:16:50 PDT by gandalf [manager] |
Quote from afantasys:
woopsie, for some reason i thought the area of a circle was 2π. i think i was confusing circumference with area. so the answer's π/2??
Indeed. Good Job. Now what has this problem taught us? (Incidentally, in case you were curious, this integral is also solvable using Trigonometric Substituion, a method which you guys will learn in Ch. 7. But nonetheless, the integral is solvable using much easier, more basic concepts.)