Topic 6: Definite Integrals and the Fundamental Theorem of Calculus

Definite Integrals

• Definite integrals involve integrating a function between specified limits for x.
• Graphically, the definite integral represents the area under the curve between those x limits.
• The fundamental theorem of calculus allows us to calculate these areas using calculus rather than geometric methods.

The Fundamental Theorem of Calculus

• The fundamental theorem of calculus states that integration and differentiation are inverse operations.
• Theorem Statement: If f(t) is a continuous function and F(t) is its antiderivative, then \int_{a}^{b} f(t) dt = F(b) - F(a).
• This implies that integrating the derivative of F returns F itself.
• The "F(b) - F(a)" part involves evaluating the antiderivative at the upper limit b, evaluating it at the lower limit a, and finding the difference.
• Example (Physics): Integrating velocity (the rate of change of displacement) between two time points t = a and t = b gives the net change in displacement during that time.

Properties of the Definite Integral

• Sum of Terms: \int{a}^{b} [f(x) + g(x)] dx = \int{a}^{b} f(x) dx + \int_{a}^{b} g(x) dx (Integrate each term separately).
• Constant Multiple: \int{a}^{b} cf(x) dx = c \int{a}^{b} f(x) dx (Bring constants outside the integral).
• Reversing Limits: \int{a}^{b} f(x) dx = -\int{b}^{a} f(x) dx (Flipping the limits introduces a minus sign).
• If the lower limit is larger than the upper limit, flip the limits and negate the integral.
• Splitting the Integral: \int{a}^{b} f(x) dx = \int{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx, where a < c < b (Split the area under the curve into pieces using an intermediate value c).
• Zero Width: \int_{a}^{a} f(x) dx = 0 (The integral is zero when the upper and lower limits are the same).

Examples

• Exercise 8a: Evaluate \int_{1}^{3} t^2 dt
  • Find the antiderivative: \int t^2 dt = \frac{t^3}{3} (This is capital F(t)).
  • Evaluate using the fundamental theorem:
    \int_{1}^{3} t^2 dt = F(3) - F(1) = \frac{3^3}{3} - \frac{1^3}{3} = \frac{26}{3}
  • Shorthand Notation:
    \int{1}^{3} t^2 dt = \left[ \frac{t^3}{3} \right]1^3 = \frac{3^3}{3} - \frac{1^3}{3} = \frac{26}{3}
  • Steps:
    1. Integrate the function (find the antiderivative).
    2. Substitute the limits and subtract (upper limit minus lower limit).
  • The final answer to a definite integral is always a number (an area).
  • Arbitrary constants are not needed in definite integrals, as they will always cancel out during the subtraction step.
• Exercise 8b: Evaluate \int_{1}^{4} (x - 2) dx
  • Integrate term by term: \int (x - 2) dx = \frac{x^2}{2} - 2x
  • Apply limits:
    \int{1}^{4} (x - 2) dx = \left[ \frac{x^2}{2} - 2x \right]1^4 = \left( \frac{4^2}{2} - 2(4) \right) - \left( \frac{1^2}{2} - 2(1) \right) = \frac{3}{2}

Integration Techniques

• When using substitution, convert the limits of integration to be in terms of the new variable.
• When using integration by parts, apply the limits to the entire expression:
  \int{a}^{b} u dv = [uv]a^b - \int_{a}^{b} v du
• Exercise 9a: Evaluate \int_{0}^{1} \frac{x}{x^2 + 5} dx
  • Substitution: Let u = x^2 + 5, then \frac{du}{dx} = 2x and dx = \frac{1}{2x} du
  • Change limits:
    • When x = 0, u = 0^2 + 5 = 5
    • When x = 1, u = 1^2 + 5 = 6
  • The integral becomes:
    \int{5}^{6} \frac{x}{u} \cdot \frac{1}{2x} du = \frac{1}{2} \int{5}^{6} \frac{1}{u} du = \frac{1}{2} [\ln|u|]_5^6 = \frac{1}{2} (\ln 6 - \ln 5) = \frac{1}{2} \ln \frac{6}{5}
  • Alternative method: Find the indefinite integral in terms of x and then apply the original limits on x.
• Exercise 9b: Evaluate \int_{0}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} dx
  • Substitution: Let u = \sqrt{x}, then \frac{du}{dx} = \frac{1}{2\sqrt{x}} and dx = 2\sqrt{x} du
  • Change limits:
    • When x = 0, u = \sqrt{0} = 0
    • When x = 4, u = \sqrt{4} = 2
  • The integral becomes:
    \int{0}^{2} \frac{e^u}{u} \cdot 2u du = 2 \int{0}^{2} e^u du = 2 [e^u]_0^2 = 2 (e^2 - e^0) = 2(e^2 - 1)
• Exercise (Integration by parts): Evaluate \int_{0}^{1} \arctan(x) dx
  • Integration by parts: Let u = \arctan(x) and dv = 1 dx. Then du = \frac{1}{x^2 + 1} dx and v = x.
  • Apply integration by parts formula:
    \int{0}^{1} \arctan(x) dx = [x \arctan(x)]0^1 - \int_{0}^{1} \frac{x}{x^2 + 1} dx
  • Evaluate the first term:
    [x \arctan(x)]_0^1 = (1 \cdot \arctan(1)) - (0 \cdot \arctan(0)) = \frac{\pi}{4} - 0 = \frac{\pi}{4}
  • Evaluate the integral (using the formula that you have for the integral of a fraction, where the top of the fraction is the derivative of the bottom):
    \int{0}^{1} \frac{x}{x^2 + 1} dx = \frac{1}{2} \int{0}^{1} \frac{2x}{x^2 + 1} dx = \frac{1}{2} [ln(x^2+1)]_0^1 = \frac{1}{2}ln(2)
  • Combine the results:
    \int_{0}^{1} \arctan(x) dx = \frac{\pi}{4} - \frac{1}{2} \ln 2