Topic 6: Fundamental Theorem of Calculus Part 2

Fundamental Theorem of Calculus - Part 2

Differentiating an Integral

• The fundamental theorem of calculus part 1 states that integrating a derivative returns the original function:
  \int_{a}^{b} \frac{df}{dx} dx = F(b) - F(a)
• Part 2 explores the result of differentiating an integral.
• Differentiation and integration are inverse operations.

Area as a Function

• A definite integral corresponds to an area under a curve.
• For a general shape where a geometric formula doesn't exist, early mathematicians struggled to calculate the area.
• The breakthrough came with the connection between area calculation and differentiation.
• The area under a curve y = f(t) can be considered as a function of x, where x is the upper limit of the area.
• Varying x changes the area under the curve.
• If A(x) represents the area function, then its derivative A'(x) = f(x).

Approximating Area Change

• Consider increasing x by a small amount \Delta x.
• The area increases by a small additional piece, which can be approximated by a rectangle.
• The rectangle's height is f(x), and its width is \Delta x.
• The area of the rectangle is approximately f(x) \cdot \Delta x.
• A small triangle is ignored, that shrinks to zero as \Delta x \rightarrow 0.

Derivative of the Area Function

• The additional area can be expressed as:
  \Delta A \approx f(x) \Delta x
• The derivative of A(x) is found by:
  \frac{dA}{dx} = \lim{\Delta x \to 0} \frac{\Delta A}{\Delta x} = \lim{\Delta x \to 0} \frac{f(x) \Delta x}{\Delta x} = f(x)
• Thus, the rate of change of the area with respect to x is equal to the height of the area at that point.

Fundamental Theorem of Calculus - Part 2 Expression

• The area A(x) can be expressed as a definite integral:
  A(x) = \int_{a}^{x} f(t) dt
• t is a dummy variable (can be any variable that doesn't conflict with the limits).
• The derivative of this area function is:
  \frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)
• This equation demonstrates what happens when you differentiate an integral.
• Differentiating and integrating are opposite operations.

Example

• Given g(x) = \int_{0}^{x} \sqrt{1 + t^2} dt, find g'(x).
• Applying the fundamental theorem of calculus, g'(x) = \sqrt{1 + x^2}.
• Functions defined as integrals occur in various areas, such as the error function in statistics:
  erf(x) = \int_{0}^{x} e^{-t^2} dt
• The sine integral function in electrical engineering:
  Si(x) = \int_{0}^{x} \frac{sin(t)}{t} dt

Exercise 12

• Given F(x) = \int_{\frac{\pi}{4}}^{x} cos(2t) dt, find F'(x).
• Since cos(2t) is continuous, the fundamental theorem of calculus applies.
• F'(x) = cos(2x).

Verification by Integration and Differentiation

1. Evaluate the integral:
   F(x) = \int{\frac{\pi}{4}}^{x} cos(2t) dt = \left[ \frac{1}{2} sin(2t) \right]{\frac{\pi}{4}}^{x} = \frac{1}{2} sin(2x) - \frac{1}{2} sin(\frac{\pi}{2}) = \frac{1}{2} sin(2x) - \frac{1}{2}
2. Differentiate the result:
   F'(x) = \frac{d}{dx} \left( \frac{1}{2} sin(2x) - \frac{1}{2} \right) = \frac{1}{2} \cdot 2 cos(2x) = cos(2x)

• This confirms the fundamental theorem of calculus.

Exercise 14

• Given f(x) = \int_{1}^{x^2} cos(t) dt, find f'(x).
• The fundamental theorem of calculus cannot be directly applied because the upper limit is x^2 not x.
• Let g(u) = \int_{1}^{u} cos(t) dt, where u = x^2. Thus, f(x) = g(x^2).
• Use the chain rule:
  f'(x) = g'(x^2) \cdot \frac{d}{dx}(x^2) = g'(x^2) \cdot 2x
• By the fundamental theorem of calculus, g'(u) = cos(u).
• Therefore,
  f'(x) = cos(x^2) \cdot 2x

Application of Integration: Work

• Work is the transfer of energy, often associated with applying a force to a moving object.
• Constant Force: If a constant force F is applied to an object that moves a distance D, the work done is:
  W = F \cdot D
• Variable Force: If the force F(x) varies with position x, the work done is calculated as an integral:
  W = \int_{a}^{b} F(x) dx

Exercise 15: Work Done in Stretching a Spring

• Hooke's Law: The force required to stretch a spring a distance x beyond its natural length is:
  F = kx
  where k is the spring constant.

Part A

• If a force of 2.5 N stretches a spring 0.5 m, find the spring constant.
• k = \frac{F}{x} = \frac{2.5}{0.5} = 5 \text{ N/m}

Part B

• Calculate the work required to stretch the spring 1 m beyond its equilibrium position.
• W = \int{0}^{1} 5x dx = \left[ \frac{5x^2}{2} \right]{0}^{1} = \frac{5}{2} - 0 = 2.5 \text{ Joules}

Exercise 16: Work Done Against Gravity

• Calculate the energy required to lift an astronaut from the Earth's surface to the International Space Station (400 km above the surface).

• The gravitational force on the astronaut at a distance x from the Earth's center is:
  F = \frac{2.8 \times 10^{16}}{x^2} \text{ Newtons}

• Radius of Earth: 6.4 \times 10^6 \text{ meters}

• Distance to ISS: 6.4 \times 10^6 + 4 \times 10^5 = 6.8 \times 10^6 \text{ meters}

• \text{Work} = \int_{6.4 \times 10^6}^{6.8 \times 10^6} \frac{2.8 \times 10^{16}}{x^2} dx

• = -2.8 \times 10^{16} \left[ \frac{1}{x} \right]_{6.4 \times 10^6}^{6.8 \times 10^6}

• = -2.8 \times 10^{16} \left( \frac{1}{6.8 \times 10^6} - \frac{1}{6.4 \times 10^6} \right)

• = 2.57 \times 10^8 \text{ Joules}$$