The first four questions are worth 10 points each, and the last four are worth 1

Question

The first four questions are worth 10 points each, and the last four are worth 15 points each. Find the radius of convergence, and the interval of convergence, for sigma^infinity _n=1 2^n (x+1)^n/n^2. Find the power series expansion of 1/(2x-3) centered at 0. Find the third term of the Taylor series expansion of In x, centered at 4. Find the derivative of integral^sinx _0 x^5 dx. Find the area between the curves y = 5x and y = 3x, bounded by the vertical lines x = 2 and x = -2. Consider the region between the curves y = squareroot x and y = x^2 above the interval [0, 1]. Find the volume of the volume of the solid obtained by rotating this region about the x-axis. Consider the same region as in problem 6, and find the volume of the volume of the solid obtained by rotating this region about the y-axis. Find the arc-length of the function f(x) = 2/3 (x - 1)^3/2 on the interval [1, 4].

Explanation / Answer

8) We have given f(x)=(2/3)*(x-1)^(3/2) on the interval [1,4]

f'(x)=(2/3)*(3/2)*(x-1)^(1/2)

f'(x)=(x-1)^(1/2)

[f'(x)]^2=(x-1)

The arc-length of the function is

L =integration of (1 to 4) (sqrt(1+(f'(x))^2))dx

=integration of (1 to 4) (sqrt(1+(x-1)))dx

=integration of (1 to 4) (sqrt(x))dx

=integration of (1 to 4) (x^(1/2))dx

=[x^(3/2)/(3/2)] from (1 to 4)

=[(2/3)*x^(3/2)] from (1 to 4)

=[((2/3)*(4)^(3/2))-((2/3)*(1)^(3/2))]

=[((2/3)*(2)^3-2/3]

=[16/3-2/3]

L=14/3

The arc-length of the function is 14/3

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