Topic 6: Reduction Formulas and Definite Integrals

Reduction Formulas

• Reduction formulas are an application of integration by parts.
• They simplify integrals with a constant (n) by producing another integral with a smaller value of that constant.
• The goal is to simplify the integral into a form that can be evaluated.

Example: Integral of x^n e^x

• Integral: \int x^n e^x dx, where n is a positive integer.

• We anticipate that the solution will depend on n

• Use integration by parts: choose u and \frac{dv}{dx}.

  • Let u = x^n.
  • Let \frac{dv}{dx} = e^x.

• Differentiate u: \frac{du}{dx} = nx^{n-1}.

• Integrate \frac{dv}{dx}: v = e^x.

• Apply integration by parts formula: \int u dv = uv - \int v du
  \int x^n e^x dx = x^n e^x - \int e^x nx^{n-1} dx
  \int x^n e^x dx = x^n e^x - n \int x^{n-1} e^x dx

• This is a reduction formula because it reduces the power of x in the integral from n to n-1.

• Second part of the example: evaluate the integral \int x^2 e^x dx

  • Set n = 2 in the reduction formula.
  • \int x^2 e^x dx = x^2 e^x - 2 \int x e^x dx
  • From a previous lecture, \int x e^x dx = xe^x - e^x.
  • \int x^2 e^x dx = x^2 e^x - 2(xe^x - e^x) + C
  • \int x^2 e^x dx = x^2 e^x - 2xe^x + 2e^x + C

Example: Integral of \cos^n(x)

• Integral: \int \cos^n(x) dx which is equivalent to \int (\cos(x))^n dx
• Apply integration by parts: split into u and \frac{dv}{dx}.
  • Let u = \cos^{n-1}(x).
  • Let \frac{dv}{dx} = \cos(x).
• Differentiate u using the chain rule:
  • \frac{du}{dx} = (n-1) \cos^{n-2}(x) * (-\sin(x))
• Integrate \frac{dv}{dx}:
  • v = \sin(x)
• Apply integration by parts formula:
  \int \cos^n(x) dx = \sin(x) \cos^{n-1}(x) - \int \sin(x) (n-1) \cos^{n-2}(x) (-\sin(x)) dx
  \int \cos^n(x) dx = \sin(x) \cos^{n-1}(x) + (n-1) \int \cos^{n-2}(x) \sin^2(x) dx
• Use the trig identity: \sin^2(x) = 1 - \cos^2(x)
  • \int \cos^n(x) dx = \sin(x) \cos^{n-1}(x) + (n-1) \int \cos^{n-2}(x) (1 - \cos^2(x)) dx
  • \int \cos^n(x) dx = \sin(x) \cos^{n-1}(x) + (n-1) \int (\cos^{n-2}(x) - \cos^n(x)) dx
  • \int \cos^n(x) dx = \sin(x) \cos^{n-1}(x) + (n-1) \int \cos^{n-2}(x) dx - (n-1) \int \cos^n(x) dx
• Notice that the integral on the left-hand side also appears on the right-hand side. Grouping these terms together and adding (n-1) \int \cos^n(x) dx to both sides:
  • n \int \cos^n(x) dx = \sin(x) \cos^{n-1}(x) + (n-1) \int \cos^{n-2}(x) dx
• Divide both sides by n:
  • \int \cos^n(x) dx = \frac{1}{n} \sin(x) \cos^{n-1}(x) + \frac{n-1}{n} \int \cos^{n-2}(x) dx
• This is a reduction formula.
• Specifically, imagine we want to integrate \cos^9(x).
  • One could use the reduction formula repeatedly to reduce the power until we have an integral we know how to evaluate.
• Evaluate \int \cos^4(x) dx
  • Set n = 4:
    • \int \cos^4(x) dx = \frac{1}{4} \sin(x) \cos^3(x) + \frac{3}{4} \int \cos^2(x) dx
  • Use the trig identity: \cos^2(x) = \frac{1}{2} (1 + \cos(2x))
    • \int \cos^4(x) dx = \frac{1}{4} \sin(x) \cos^3(x) + \frac{3}{4} \int \frac{1}{2} (1 + \cos(2x)) dx
    • \int \cos^4(x) dx = \frac{1}{4} \sin(x) \cos^3(x) + \frac{3}{8} \int (1 + \cos(2x)) dx
    • \int \cos^4(x) dx = \frac{1}{4} \sin(x) \cos^3(x) + \frac{3}{8} (x + \frac{1}{2} \sin(2x)) + C
    • \int \cos^4(x) dx = \frac{1}{4} \sin(x) \cos^3(x) + \frac{3}{8} x + \frac{3}{16} \sin(2x) + C

Definite Integrals

• A definite integral has limits associated with it: a lower limit (a) and an upper limit (b).
• It can be interpreted as the area under a curve between those limits.
• A definite integral is a real number.
• This is different from indefinite integrals, where the answer is a function.

Geometric Interpretation

• The definite integral of a function f(x) from x = a to x = b is the area between the function and the x-axis from a to b
• Area above the x-axis is positive, and area below the x-axis is negative.
• The definite integral is equal to the signed area.

Examples: Finding Signed Areas

Example 1

• Evaluate the definite integral \int_1^4 (x - 2) dx
• Graph the function y = x - 2 between x = 1 and x = 4.
  * The function is a straight line with x-intercept at x = 2.
  * Two triangles are formed, one below the x-axis (from x = 1 to x = 2) and one above (from x = 2 to x = 4).
• Calculate the areas of the triangles.
  • Triangle 1 (below the x-axis): base = 1, height = 1, area = A_1 = \frac{1}{2} * 1 * 1 = \frac{1}{2}.
  • Triangle 2 (above the x-axis): base = 2, height = 2, area = A_2 = \frac{1}{2} * 2 * 2 = 2.
• The integral is the signed area:
  \int1^4 (x - 2) dx = -A1 + A_2 = -\frac{1}{2} + 2 = \frac{3}{2}

Example 2

• Evaluate the definite integral \int_0^2 \sqrt{4 - x^2} dx
• Graph the function y = \sqrt{4 - x^2}.
  * y = \sqrt{4 - x^2}. Square both sides and you get x^2 + y^2 = 4, which is a circle with radius 2.
  * Because the limits on the integral are 0 to 2 for x, and the square root is positive, one should only consider the first quadrant of the circle.
• Calculate the area of the quarter circle.
  \text{Area} = \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (2^2) = \pi
• So, \int_0^2 \sqrt{4 - x^2} dx = \pi

Limitations:

• This geometric approach works well for simple shapes.
• For more complex functions, calculus methods are needed to evaluate these integrals. This will be covered in the lecture.