calculus 2 Evaluate the given integral Determine the area bounded by the curves

Question

calculus 2

Evaluate the given integral Determine the area bounded by the curves f(x) = x + 1 and g(x) = (x - 1)2 Determine the volume of the solid formed by rotating the region bounded by f(x) = x + 1 and g(x) = (x - 1)2 about the z-axis. Determine the volume of the solid formed by rotating the region bounded by f(x) = x + 1 and g(x) = (x - 1)2 about the y-axis. A parabolic tank with upper radius 2 ft and height 4 ft is full of water. Determine the work required to pump the water out of the tank. (Use the fact that water weighs approximately 62.5 lb/ft3.) Determine the arclength of the curve y = Ax3/2 on 0 x 2 Evaluate the integral if it converges. Show divergence otherwise.

Explanation / Answer

1a The integral of 3xe^-2x, integrating by parts is -3x/2e^-2x + the integral of 3/2e^-2x =

-3x/2e^-2x -3/4e^-2x + c

b) Solve (x+3)/(x^2+3x+2) = A/x+1 + B/x+2

Ax+2A + Bx + B = 1x + 3

A+B = 1

2A + B = 3

This has solution B=-1, A = 2

The integral of 2/x+1 -1/x+2 = 2 ln(x+1) - ln(x+2) + c

You can also write this ln((x+1)^2/(x+2)) + c

c) The integral of 3x/sqrt(1+x^2) = as the derivative of sqrt(1+x^2) = x/sqrt(1+x^2),

3 sqrt(1+x^2) + c

2 We find where f(x) = g(x) and which function is above which function where

f(x) = x+1

g(x) = (x-1)^2

Note that g(x) is a parabola opening up with vertex at (1, 0) and f(x) = 2 there. As f(x) is a line, we see that f(x) is above g(x) on a certain interval and below it elsewhere, so we integrate f(x) - g(x) between points of intersection.

f(x) = x+1 g(x) =(x-1)^2

(x+1) - (x-1)^2 = 0

x+1 - (x^2 - 2x + 1) = 0

3x - x^2 = 0

-x(x-3) = 0

x = 0, 3

Using I[a,b] for the integral from a to b and E[a,b] for the evaluation,

I[0, 3] 3x - x^2 dx = 3/2x^2 - x^3/3 E[0, 3] = 3/2*3^2 - 3^3/3 - (0) = 9/2

3. From problem 2, we are integrating from 0 to 3, f(x) is above g(x) and both are non-negative (g(x) is quadratic and f(x) is above)

Then, I[0, 3] pi f^2(x) - pi g^2(x) dx = I[0,3] pi(x+1)^2 - pi(x-1)^4 dx =

pi((x+1)^3/3 - (x-1)^5/5)E[0, 3] =

pi(4^3/3 - 2^5/5 - ((1)^3/3 - (-1)^5/5) =

pi (64/3 - 32/5 - 1/3 + 1/5) =

74/5 pi

4. Here, we use the shell method and integrate 2 pi x(f(x) - g(x)) dx

2 pi I[0, 3] x(3x-x^2)dx = 2 pi I[0, 3] 3x^2 - x^3 dx =

2pi (x^3 - x^4/4) E[0, 3] =

2pi (27 - 81/4 - (0)) =

27/2 pi

5. Let the parabola be z=ax^2

As 4= 2^2a, a = 1

As sqrt(x^2+y^2)^2 = z is the description of the paraboloid, note that the cross-section of the paraboloid = pi(x^2+y^2) =z pi

Then, the water is pumped a height of 4 - z

Thus, we have I[0, 4] 62.5 pi z(4-z) dz = 62.5 pi I[0,4] 4z - z^2 dz =

62.5 pi (2z^2 - z^3/3) E[0, 4] = 62.5pi(32 - 64/3 - (0)) =

2000/3 pi ft-lbs

6. y = 4x^(3/2)

dy/dx = 6 sqrt(x)

I[0,2] sqrt(1+(dy/dx)^2) dx =

I[0, 2] sqrt(1 + 36x) dx =

1/54(1+36x)^(3/2) E[0, 2] =

1/54(73 sqrt(73) -1)

= 11.5317087668181

7.

a) I[1, 3] 8/(x-1)^3

As the power in the denominator is >= 1 and this point is the endpoint of the interval, this diverges

I[0, inf] 3xe^-2x = referring to problem 1 a),

-3x/2e^-2x -3/4e^-2x E[0, inf] = 3/4

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