Sunday, January 31, 2010

Posting...#24

Area between curves: formula: bSa top equation-bottom equation
example: Find the area of the region enclosed by y=-4x^2+41x+94 and y=x-2 between x=1 and x=7.

Steps:
1. draw a picture
2. subtract the two equations: (-4x^2+41x+94-(x-2)-don't square, it's not volumee! (-4x^2+41x+94-(-x+2)
3.combine like terms and integrate: S(-4x^2+41x+94-(-x+2))(-4/3)x^3+20x^2+96x
4. plug in 1 and 6 and subtract: (3584/3)-(344/3)= 1080


The formula for the volume of disks is S (top)^2 - (bottom)^2 dx

The formula for the area of washers is S (top) - (bottom)

STEPS:1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate

I don't know why but lately i been having trouble with your integration so if yall got tips are a better way of explaining it thanks.
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3 comments:

  1. dylan bFebruary 3, 2010 at 8:20 PM

    when taking the calculator portion you can plug a definite integral into your calculator but hitting the math button first.
    then hit the 9 button. Next plug in your equation followed by a comma then x then another comma your top bound then another comma your bottom bound then close parenthesis and hit enter

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  2. jessieFebruary 3, 2010 at 8:38 PM

    e integration:

    whatever is raised to the e power will be your u and du will be the derivative of u. For example:

    e^2x-1dx
    u=2x-1 du=2
    rewrite the function as:
    1/2{ e^u du, therefore
    1/2e^2x-1+C will be the final answer.

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  3. merfFebruary 3, 2010 at 11:10 PM

    integration is easy, you just take the equation, and do this to it, lets say the equation is x^3+2x^2

    to take the integral of something, first you have to raise the exponent by one, so lets take the first part of the problem, x^3, and do that to it. it becomes x^4. Now, you just multiple the whole thing by the inverse of the new exponent. so it would become (1/4)x^4.

    then you do the same thing to the second part of the problem. 2x^2 becomes 2x^3, then you have to multiply, and it becomes (2/3)x^3

    then you put the two parts back together like normal, and it becomes (1/4)x^4+(2/3)x^3.

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