Topic 6: Fundamental Theorem of Calculus Part 2

Fundamental Theorem of Calculus - Part 2

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.

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).

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.

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.

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.

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

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}
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)

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

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

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.

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}

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}$$