1. Calc Final Review - Problems (Please Explain)

3. Evaluate the integral sin^5x cos^3x dx , interval [pi/2, 3pi/4]

14. Evaluate the integral (x^3)/sqrt(16-x^2) dx, interval [0, 2sqrt3]

19. Evaluate the integral 1/(x^2-1) dx, interval [2, 3]

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6. Use midpoint rule and simpsons rule to approximate the given integral with the specified value of n.
e^(-srt(x)) dx, interval [0,1] n=6

8. Use the trapezoidal rule and the midpoint rule and the simpsons rule to approximate the given integral with the specified vaule of n.
sin(x^2)dx, interval [0, 1/2] n=4

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Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

6. 1/(2x-5) dx, interval [0, -infinity]

20. lnx/x^3 dx, interval [1, infinity]

30. dx/sqrt(1-x^2), interval [0,1]

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4. Functions given x=y^2 - 4y and x=2y - y^2 points of cross over (-3,3) and (0,0). Find the area of which curves overlap.

14. Sketch the region enclosed by the given curves. Decide whether to intergrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the are of the region.
y=singx, y=2x/pi , x is greater of equal to 0

38. Find the area of the region bounded by the parabola y=x^2, the tangent line to this parabola at (1,1) and the x-axis.

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8. Find the volume of the solid obtained by rotating the region bounded by the given curvbes about the specified line. Sketch the region, the solid, and a typical disk or washer.
y=x^(2/3), x=1, y=0, about the y-axis

22. Each integral represents the volume of a solid. Describe the solid.
a. pi y dy, interval [2,5] b. pi [(1+cosx)^2 - 1^2]dx, interval [0,pi/2]

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1. Use the arc length formula (2) to find the length of the curve y=2-3x, interval [-2,1]

formula L = sqrt [ 1+ (dy/dx)^2 ] dx interval [a,b]

8. Graph teh curve and find its exact length.
y=(x^2)/2 - (lnx)/4, interval from 2 to 4

20. Use either a CAS or a table of integrals to find the exact length of the curve.
y = lnx interval from 1 to sqrt(3)

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Please explain how to do these problems, no need for solutions but if you have some spare time it would be helpful.
Thank you

2. 3.) $\int sin^4xcos^3xdx = \int sin^4x(1 - sin^2x)cosxdx = \int (sin^4xcosx - sin^6xcosx)dx$

14.) $\int \frac{x^3}{\sqrt{16 - x^2}}dx = \int \frac{(4sin\theta)^3}{4cos\theta}4cos\theta d\theta = 64\int sin^3\theta d\theta$

$x = 4sin\theta$

$dx = 4cos\theta d\theta$

19.) $\int \frac{1}{x^2 - 1}dx = \int \frac{1}{(x - 1)(x + 1)}dx$

$\frac{A}{x - 1} + \frac{B}{x + 1} = \frac{1}{x^2 - 1}$

$A(x + 1) + B(x - 1) = 1$

$A = 1/2 ... B = -1/2$

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4.) $\int_{0}^{3} [(2x - x^2 + 4) - (x^2 - 4x + 4)]dx$

38.) $f(x) = x^2$

$1 = f'(1)(1) + b$

$g(x) = 2x - 1$

$f(x) = g(x)$ @ $x = 1$

$\int_{0}^{1} [x^2 - (2x - 1)]dx$

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8.) $f(x)^{2/3}$

$\int_{0}^{1} (1 - y^{3/2})dy$

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1.) $\int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2}dx$

$y = 2 - 3x$

$\frac{dy}{dx} = -3$

$\int_{-2}^{1} \sqrt{1 + (-3)^2}dx$

8.) $y = \frac{x^2}{2} - \frac{lnx}{4}$

$\frac{dy}{dx} = x - \frac{1}{4x}$

$\int_{2}^{4} \sqrt{1 + (x - \frac{1}{4x})^2}dx$

3. This thread is going to become an absolute shambles if replies and counter replies are posted to all these questions. Furthermore, the use of identical numbering (despite the ---------------------) is going to create a lot of problems.

@OP: Work out what questions haven't been replied to and re-post them in new threads if you still need help. No more than three questions per thread. Note also rule #1 here: http://www.mathhelpforum.com/math-he...php?do=cfrules, so please don't re-post any questions that have already been given a reply.