February 2003:

Home | Problem Archive | Past Winners | Number Games | Contact Me

02/03/03) Updated 02/07/03) Find the average value of y = sin(ln x) from 1 to epi. Also, find the volume of the same region revolved about the x-axis.

Solution:

The area is the integral of y from 1 to epi. To find the integral of the function we have to use integration by parts, which says the integral of u*dv is equal to u*v minus the integral of v*du. Substituting u for sin(ln x) and v for dx, and repeating the process for a second time, we find the antiderivative to be (x/2)*(sin(ln u) - cos(ln u)). Thus from 1 to epi the area is equal to (epi + 1)/2.
The second part of the problem is integrating pi*sin2(ln x). Using trigonometric identities and the same process of integration by parts, we obtain the antiderivative as (x/5)*(5/2 - sin(ln x2) - cos(ln x2)/2). Thus the volume would equal 2pi/5 * (epi - 1)

Topic:

Applications of Integrals

______________________________________

02/10/03) Find the maximum error of integral approximation for f(x) = k x2 from x = a to x = b using the Trapezoidal rule with 10 intervals, if k is a constant.

Solution:

The formula for the maximum error is (b-a)/12 times h2 times the maximum value of the second derivative of the function, or shortly h2*M*(b-a)/12. In our case, this would be ((b-a)/10)2*(b-a)/12 times 2k, which equals k(b-a)3/60

Topic:

Trapezoidal Rule for Approximating Integrals

______________________________________

02/17/03) Solve the following differential equation:
dy/dx = (cos(x)(y+2)(y-3))/-15.

Solution:

First we start by separating the variables. In the y side of the equation, we need to use the algebraic fact that 1/(y+2)(y-3) = (1/5)/(y-3) - (1/5)/(y+2). On the x side of the equation we only integrate the cosine function. Thus we get y = 5/(1-c*e-sin(x)/3) - 2. Note that c is a constant.

Topic:

Integration

______________________________________

02/24/03) A) Knowing that the derivative of sin(x) is cos(x), derive the derivative of sin-1(x).
B) If f(sin x) = cos x, find f'(x).

Solution:

It can be derived that the derivative of a function's inverse is the reciprocal of the function's derivative. Thus, the derivative of sin-1 x is equal to 1/cos(sin-1 x). Using the Pythagorean Theorem, this equals 1/sqrt(1-x2). If f(sin x) = cos x, then f(x) = cos(sin-1 x) = sqrt(1-x2). Thus f'(x) = -x/sqrt(1-x2).

Topic:

Derivatives and Inverse Functions

______________________________________

Previous Month Archive Next Month