I have to work this problem 3 ways, and I've gotten two, but am not sure about the third way.
the first way I worked by setting u = x^2 - 9
the second way I worked by setting x = 3sec(theta)
but the third way I have no clue, the book gave us a hint: Let x^2 - 9 =...
I've almost gotten through this one integral problem, but i've seem to have gotten stuck:
integral ((sqrt(4-x^2))/x) dx
i let x = 2sin(theta), and dx = 2cos(theta)d(theta)
sqrt(4-x^2) = 2cos(theta)
integral ((2cos(theta))/(2sin(theta)) * 2cosd(theta)
This one has me stumped, I dont even know where to start.
-Find the area of the triangular region in the first quadrant that is bounded above by the curve y=e^(2x), below by y=e^x, and on the right by the line x=ln(3).
Thanks for any help
A = L(h)
A = L(16 - L²)
A = 16L - L³
A` = 16 – 3L²
16 – 3L² = 0
√(L²) = √(16/3)
L = 2.3094011
h = (16 – 2.3094011²)
h = 10.6666
A = 24.63361146
Then I times it by 2 to account for the other side of the parabola (right?)
A = 49.26722297
Well, I know that the f(x) is going to be an upside down parabola. The area of a rectangle is obviously A = l(w). You would have to find the points on the graph to have your top line...but I'm not sure how you would do this. I'm thinking somthing with the derivative but that's where I'm stuck.