Integral Calculus (Antiderivatives, areas, applications) - 10/99
  • Antiderivatives
  • Integration Rules
  • Calculus of Areas
  • Fundamental Thm of Calculus
  • Other Applications of Integrals
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    Antiderivatives

    The derivative f'(x) is used to calculate the slope, or rate of change, of a function f(x). But what is the function f(x) the derivative OF? Or, if you prefer better grammar, describe a function F(x) having F'(x) = f(x). Such a function F(x) is called an antiderivative of f(x).

    Example: What function has a derivative of x^3 ? We know the deriv goes dowm one
    power, and we might think: (x^4) = 4 x^3. But we want x^3 not 4 x^3.
    How do we avoid the 4? Put a 1/4 at the beginning to wait and pounce on the 4 when we do the derivative. Thus if F(x) = (1/4) x^4 , we get F'(x) = (1/4) 4 x^3 = x^3 = f(x).
    Actually F(x) could be (1/4) x^4 + c for any constant c, and F'(x) would still be x^3.
     
    The antiderivative F(x) is also called the integral of f(x).
    The usual notation for the antiderivative of f(x) is F(x) =f(x) dx .
    We would writex^3 dx = (1/4) x^4 + c.(Check out my new integral.gif!)
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    Integration rules! (10/99)

    I mean it can be fun, challenging, and useful! I also mean there are rules for what do do to look for an antideriv. We just discovered the

    Power rule:x^n dx = (1/(n+1)) x^(n+1) + c

    Example: x dx = x^(1/2) dx = (2/3) x^(3/2) + c

    Constant multiple rule: c f(x) dx = cf(x) dx (Pulling a constant outside)

    Sum rule: ( f(x) + g(x)) dx = f(x) dx +g(x) dx

    Chain rule: (aka "u-substitution")

    F '(g(x)) g'(x) dx = F '(u) (du/dx) dx =F '(u) du = F(u) + c.

    Example: (3x + 4)^20 dx = (1/3)(3x + 4)^20 (3 dx)
    = (1/3)u^20 du = (1/3) (1/21) u^21 = (1/63) (3x + 4)^21 + c
    Here I used u = g(x) = 3x + 4 , so that du = g'(x) dx = 3 dx.
     
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    Calculus of Areas (10/99)

    Historically, an important problem was figuring out areas. A long time before Newton, Archimedes (c.287 BC, from Greece) and many Chinese mathematicians, were chopping regions up into little slices to estimate their areas.

    We can chop up the area A under the graph of y = f(x) by using a bunch of rectangles. Each rectangle has height f(x) and base Dx, so the area of each "strip" is DA = f(x) Dx.

    The area A is about the same as the sum of the areas of all those rectangles;

    A :=:DA = f(x) Dx . . . (This is called a Riemann Sum.)

    Here's a Mathematica->QuickTime->GIF Builder animation showing:

    The more rectangles we use, the better the approximation to the true area!

     

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    The Fundamental Theorem of Calculus (10/99)

    Let's assume that F(x) is an antiderivative of f(x), that is, F'(x) = f(x).

    The surprising part is that F(x) represents the area under the graph of y = f(x). Can this be true? let's define A(x) to be the area under y = f(x) from 0 to x (let's pretend we're in the first quadrant). Then A'(x) would be dA/dx. But the extra area under y = f(x) from x to x + dx is about the same as the area of the skinny rectangle of height f(x) and base dx, so dA = f(x) dx, and so dA/dx = f(x).

    Here we have that the area function is an antiderivative of f(x); A'(x) = f(x).

    Wow. So to compute an area, we can just compute an antiderivative of f(x)! (not factorial)

    If we want the area A from x = a to x = b, we can subtract area functions;

    if F(x) is the area from 0 to x : F(x) =f(x) dx , then A = F(b) - F(a) .

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    Other applications of integrals (9/00)

    "What are we gonna use this stuff for?"
    That's what students always ask (or at least wonder), so here are some examples:

    1. Work: The amount of "work" you do pushing an object is equal to (how hard you push) times (how far you push it). Then Work = Force * Distance . If the force is not constant then we have to add up lots of little bits of (force) * (distance) for short distances along our path.

    (Work done) = (force at x) (Dx) --> W = F(x) dx

    2. Savings account buildup: If you put money into an account and leave it there, it grows by compound interest, exponentially. But if you deposit a little bit, say each month, then each little bit earns interest for a diffferent amount of time (like my students!). This cries out for an integral, where r is annual int rate and n = # of years...

    (Amount in account)

    = (from t = 0 to 12n) [ (depos at month t) * (int for n - t mos) ]

    --> (from t = 0 to 12n) e^(r(n - t)) dt

    Vector Calculus (updated 3/99) . . . (top of page)

    The xyz-coordinate system; points, lines, planes (back to vec calc)

    Parametric curves in 2D and 3D; arclength, curvature (back to vec calc)

    A curve can be parametrized by a variable t (such as time), where its coords (two or three) are functions of t. Let's show one 2D curve and one 3D curve:
    Here's a spiral curve in the xy-plane:
    x = t cos(t) , y = t sin(t) ; 0 < t < 6 Pi . 
    (click here for picture of spiral)
    x = 3 cos(t) , y = 3 sin(t) , z = t ; 0 < t < 6 Pi .
    (click here for picture of helix)

    The curvature is how fast the ant has to turn, in, say, degrees per centimeter.

    Functions of two variables; surfaces (back to vec calc)

    Think of z as a function of x and y ; z = f(x,y).
    For example we could have z = x^2 - y^2 + 10.
    Then for each point (x,y) in the plane, we put a point (x, y, x^2 - y^2 + 10) in 3-space. One is (3, 2, 15). The collection of all these points forms a surface, which is the graph of z = f(x,y). Here's part of this surface:

    (click here for picture of surface)

    Level curves and gradient vectors (back to vec calc)

    A level curve for level c of a function z = f(x,y) is the set of points (x,y) in the plane for which f(x,y) = c. These are also called contours.

    Here are some level curves for z = x^2 - y^2:

    (click here for level curve diagram)

    Double, triple integrals; area, volume, surface area (back to vec calc)

    In first-year calculus you learned that an area under a curve y = f(x) over an interval [a,b] can be calculated by doing the integral f(x) dx , where the integral goes from a to b.

    Now in 3D calculus we can calculate a volume by doing an integral of z over a certain region in the xy-plane: V = f(x,y) dA , where the (x,y) is over a region R and dA = dx dy. This double integral is evaluated from the "inside out."

    The surface area is more complicated, involving the slope and direction of the surface. There are also triple (and higher!) integrals.

    Cylindrical and spherical coordinates (9/98) (back to vec calc)

    These are extensions of the idea of polar coordinates; see Trigonometry.

    In 3-dimensional (x, y, z) space, we can use polar coords (r, t) on the xy-plane, together with the z-coordinate z, to get a point (r, th, z), called "cylindrical coordinates," where th is a "theta" and we have the relations

    x = r cos th , y = r sin th , z = z (cylindrical to rectangular) . . . and . . .

    r^2 = x^2 + y^2 , tan th = y/x , z = z (rectangular to cylindrical)

    They're called cylindrical because the graph of r = c is a cylinder of radius c , because th and z aren't mentioned, so they can be anything; around and around or up and down, but always c units away from the z-axis. (picture when I get to it . . . if ever.)

    An alternative to polar coords is to use the distance p (rho, not "p") in space from the origin O to the point P, the angle th (theta) from the x-axis to the projection of the ray OP onto the xy-plane, and the angle ø (phi) from the z-axis down to the ray OP. The point (p, th, ø) is said to be in "spherical coordinates," because the graph of p = c is a sphere of radius c, the longitude th and the latitude ø being free to range. (picture next in line.)

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    Well, that's it for now. Check back often for new stuff!
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