here it is: http://img266.imageshack.us/img266/435/untitledvm3.jpg
any help please.
here it is: http://img266.imageshack.us/img266/435/untitledvm3.jpg
any help please.
I think I can read it correctly.
In latex (when it works again) $\displaystyle \int_{0}{1} \sqrt{9+4t^2}dx$.
In easier to read type: int sqrt(9+4t^2)dt, evaluated from 0 to 1.
Before I do this, have you studied trig substitution, because if you haven't this might be slightly difficult to get at first?
Alright. Well draw a right triangle with the horizontal leg being 3 and the vertical leg being 2t. So it follows that the hypotenuse is sqrt(9+4^2).
Now let's make some substitutions. Call the angle between the horizontal leg and the hypotenuse theta.
tan(theta) = (2t)/3, and taking the derivative...
sec^2(theta)d(theta) = (2/3)dt.
So t = [3tan(theta)]/2 and dt = [3sec^2(theta)]/2
Now plug this back into the original integral.
int sqrt(9+4t^2)dt = int sqrt[(9/4)tan^2(theta)+9] *(3/2)sec^2(theta)d(theta).
Is this making sense?