Chapter 1
Integration

The antiderivative or indefinite integral

• We know that
  • xⁿ ® n x^n-1,
  • sin(x) ® cos(x),
  • cos(x) ® -sin(x).

  We want to know given
  

  dy dx = f(x)

  what is y? We want the inverse of differentiation or the antiderivative.

• This is not uniquely defined since we can always add a constant to y since [dy/dx] would remain the same which is why it is called the indefinite integral.
• If we are given what [dy/dx] is and what y is at some point then we have a unique answer. This is known as an initial value problem.
• We can flip the table above and say
  • xⁿ ® [(xⁿ⁺¹)/(n+1)] + C,
  • sin(x) ® -cos(x) + C,
  • cos(x) ® sin(x) + C.

  These are our elementary functions and are the only things we know how to work with.

  For example If [dy/dx] = 4x² then we have 4xⁿ where n = 2 and so get y = 4x³/3+C.

• If we have a sum of a collection of elementary functions we can deal with them term by term. So
  

  dy dx = 3x-7+6x4

  means that
  

  y = 3x2 2 -7x+ 6x5 5 + C.

  This is known as the general solution. If we also know that y(0) = 0 then we can fix the constant since using the general solution we know that
  

  y(0) = 3(0)2 2 -7(0)+ 6(0)5 5 + C = C,

  but we know that y(0) = 0 so we must have C = 0.
• Substitution:Here we use the fact that
  

  dy dx = dy du du dx

  to simplify our problem.

  If we have [dy/dx] = (7x-3)^-2 then what is y? We cannot expand out the bracket and do it term by term. Instead we say u = 7x-3. Then we have
  

  du dx = 7

  which means that du = 7dx or dx = du/7. So
  

  dy dx = (u)-2

  or
  

  dy = u-2dx = u-2du/7

  . We can integrate both sides to get
  

  y = u-1 -1 1 7 + C.

  Substituting u = (7x-3) back into this we have
  

  y = -1 7 (7x-3)-1 + C.

  So there are 4 steps to this. First we pick u, then we rewrite the integral in terms of u and du. We can then integrate to get y in terms of u before rewriting in terms of x to get our final answer.

  A trickier example:
  

  dy dx = xdx (x2-3)3 dx.

  1. Pick u = x²-3.
  2. So du = 2xdx or xdx = [du/2] . Then
     

     dy = 1 2 du u3 .

  3. Integrate this to get
     

     y = 1 2 u-4-4 + C.

  4. Plug in u = x²-3 to get
     

     y = - 1 8 (x2-3)-4 + C.

  Areas or the definite integral

• Integration can also be used to find the area under a curve. If we want to the area under the curve y(x) from x = a to x = b we just evaluate the antiderivative of y at b and subtract from this the value at a.

  If we have y(x) = 3 from 0 to 3 then this corresponds to a 3×3 square whose area is of course 9. The antiderivative of y(x) = 3 is 3x+C. Evaluating this at x = 0 and subtracting it from this at x = 3 we get (9+C)-(0+C) = 9. So we get the same answer. We can also check with a triangle. Here would have y(x) = x say. Then if we want to find the area of a right-angled triangle with width and height one we would integrate y = x from 0 to 1. The antiderivative is x²/2 so we get 1/2, the expected answer.

  The definite integrals involve computing indefinite ones so we have the same problems and can again use substitution to help us out. If we have y = f(x), we need to compute the antiderivative F(x). The definite integral from a to b is simply F(b)-F(a).

  Tricks

• Integration by parts:We know from the product rule that
  

  d(uv) dx = u dv dx +v du dx .

  This means that
  

  u dv dx = d(uv) dx -v du dx .

  If we integrate both sides we get
  

  ó õ b a  u dv dx = ó õ b a  d(uv) dx - ó õ b a  v du dx .

  The antiderivative of [d(uv)/dx] = uv by definition so we have
  

  ó õ b a  u dv dx = [uv]ab- ó õ b a  v du dx .

  This can be very useful since if we pick u and v carefully we can replace a tricky integral by something easier. Since the uv term doesn't require any integration we have replaced the integral of u[dv/dx] by v[du/dx]. If we want to integrate xsin(x) we can pick u = x and v = -cos(x). Then u[dv/dx] = xsin(x) while v[du/dx] = -cos(x) which is an elementary function.
• Trapezoid rule:We can integrate elementary functions, use substitutions or integration by parts but what if we are still stuck. For example if we have ò[x/((x³-7x)²)] dx then we can't substitute u = x³-7x since this leads to du = (3x²-7)dx so we don't end up with a simple integral.

  Since we know that Integration is finding the area under the curve we can determine this computationally. We evaluate the function at n+1 evenly spaced points x₀, ..., x_n then the definite integral from a to b is
  

  (b-a) 2n [y0+2y1+¼+2yn1+yn],

  where y_i = y(x_i).

  The more points we take, the more accurate an approximation we have. On a computer it is easy to just say n = 1000000 but of course by hand you only need to take n = 4 or something like that to get a reasonable answer. You would like (b-a)/n to be less than 1 since the error is proportional to this squared.

  If want to integrate x² from 0 to 1:

  1. Decide how many intervals we want to break 0 to 1 into. For this example say we chose 4 so [(1-0)/4] = 1/4. Then we would take the points 0,0.25,0.5,0.75 and 1. These are out x_i's.
  2. Next we need so work out what the y_i's are where y_i = f(x_i). In this case f(x) = x² so we have that y₀ = 0² = 0, y₁ = [1/4]² = [1/16], y₂ = [2/4]² = [4/16], y₃ = [3/4]² = [9/16] and y₄ = 1² = 1.
  3. Now we have everything we need to plug into the formula
     

     (b-a) 2n [y0+2y1+2y2+2y3+y4]

     to get
     

     1 8 [0+2/16+8/16+18/16+16/16] = 1 8 [10/16+34/16] = 1 8 [44/16] = 11 32 .

     Exact answer is [(1³)/3] = 1/3 = 11/33. So pretty close!