“If you hear a voice within you say ‘you cannot paint’, then by all means paint and that voice will be silenced,” Vincent van Gogh.
Antiderivatives are fundamental concepts in calculus. They are the inverse operation of derivatives.
Given a function f(x), an antiderivative, also known as indefinite integral, F, is the function that can be differentiated to obtain the original function, that is, F’ = f, e.g., 3x^{2} -1 is the antiderivative of x^{3} -x +7 because $\frac{d}{dx} (x^3-x+7) = 3x^2 -1$. Symbolically, we write F(x) = $\int f(x)dx$.
The process of finding antiderivatives is called integration.
Revisiting Average Rates of Change, Using Integrals
The Fundamental Theorem of Calculus states roughly that the integral of a function f over an interval is equal to the change of any antiderivate F (F'(x) = f(x)) between the ends of the interval, i.e., $\int_{a}^{b} f(x)dx = F(b)-F(a)=F(x) \bigg|_{a}^{b}$.
Using the following notation $\Delta F = F(b) - F(a), \Delta x = b - a$, the fundamental theorem of calculus can be rewritten or expressed as $\Delta F = \int_{a}^{b} f(x)dx$, that is, the definite integral of the rate of change of a function on [a, b] is the total change of the function itself on [a, b]. and obviously $\frac{\Delta F}{\Delta x} = \frac{1}{b-a}\int_{a}^{b} f(x)dx$ where the left side of this equation is just the average of the function f over the interval [a, b] ↭ $\Delta F = Average(F’)·\Delta x$.
Suppose a car moves along a straight road, and its position at time t is given by the function s(t). The average velocity of the object over the time interval [a, b] can be represented by the average rate of change of s(t)over that interval, AvgV = $\frac{s(b)-s(a)}{b-a}$. Denoting the velocity as v(t), AvgV = $\frac{1}{b-a} \int_{a}^{b} v(t)dt$, and $\frac{\Delta F}{\Delta x} = \int_{a}^{b} v(t)dt$
Suppose the temperature at a location is given by the function T(x), where x represents time. The average temperature over a certain interval [a, b] can be expressed using integrals. The average temperature AvgTemp over the interval [a, b]is given by: AvgTemp = $\frac{1}{b - a} \int_{a}^{b} T(x)dx$, and $\frac{\Delta F}{\Delta x} = \int_{a}^{b} T(x)dx$
Exercise. Suppose that water is flowing into a vessel at a rate of (1 − e^{−t}) cm^{3}/s. What is the average flow from times t = 0 to t = 2? The average rate of change or flow is $\frac{1}{2-0}\int_{0}^{2} (1-e^{-t})dt = \frac{1}{2}(t-e^{-t})\bigg|_{0}^{2} = \frac{1}{2}(2-e^{-2}+e^{-0}) ≈ 0.56.$
If f is continuous over an integral [a, b], then there is at least one point c ∈ [a, b] such that $f(c) = \frac{1}{b-a}\int_{a}^{b} f(x)d(x) ↭ \int_{a}^{b} f(x)d(x) = f(c)(b-a)$. Futhermore, since f is continuous on [a, b], by the Extreme Value Theorem, there exist m and m such that m ≤ f(x) ≤ M, and so m(b-a) ≤ $\int_{a}^{b} f(x)d(x)$ ≤ M(b -a).
Proof.
f is continuous over an integral [a, b] ⇒[By the Extreme Value Theorem] there exist m and m such that m ≤ f(x) ≤ M ∀ x∈ [a, b] ⇒[Comparison Theorem] m(b-a) ≤ $\int_{a}^{b} f(x)d(x)$ ≤ M(b -a) ⇒ m ≤ $\frac{1}{b-a}\int_{a}^{b} f(x)d(x)$ ≤ M.
Since $\frac{1}{b-a}\int_{a}^{b} f(x)d(x)$ ∈ [m, M], f is continuous and assumes the values m and M over [a, b] ⇒[By the Intermediate Value Theorem] there is a number c over [a, b] such that $f(c) = \frac{1}{b-a}\int_{a}^{b} f(x)d(x)$.
Suppose F’(x) = $\frac{1}{1+x}, F(0) = 1$, the MVT implies A < F(4) < B, could you compute A and B?
By MVT, F(4) - F(0) = F’(c)(4 -0) = $\frac{1}{1+c}·4$ ⇒[c ∈ (0, 4)] $\frac{1}{1+4}·4 < \frac{1}{1+c}·4 < \frac{1}{1+0}·4 ⇒ \frac{4}{5} < F(4) - F(0) < 4 ⇒ \frac{4}{5} < F(4) - 1 < 4 ⇒ \frac{9}{5} < F(4) < 5$, hence A = 9/5, and B = 5.
Similarly, it could be argued by the FTC, F(4) -F(0) = $\int_{0}^{4} \frac{dx}{1+x}dx <[x = 0] \int_{0}^{4} dx = 4$ -this is the red rectangle, Figure 1.a.-. On the other hand, F(4) -F(0) = $\int_{0}^{4} \frac{dx}{1+x}dx >[x = 4] \int_{1}^{5} \frac{1}{5}dx = \frac{4}{5}$ -this is the yellow rectangle below, Figure 1.a.-, and we obtain the same result.
Find the average value of the function f(x) = 8 -2x over the interval [0, 4] and find c such that f(c) equals this value.
The average value of the function over [0, 4] = $\frac{1}{4-0}\int_{0}^{4} (8-2x)dx = \frac{1}{4}·(8x-x^2)\bigg|_{0}^{4} = \frac{1}{4}·(8·4-16) = 4$. Next, we are going to set this average value to f(c) and solve for c, 8 -2c = 4 ⇒ c = 2.
The Second Fundamental Theorem of Calculus. If f is a continuous function and c is a constant, then f has a unique antiderivative A that satisfies A(c) = 0, and that antiderivative is given by the rule A(x) = $\int_{c}^{x} f(t)dt$.
Proof.
A’(x) = $\lim_{h \to 0} \frac{A(x+h)-A(x)}{h} = \lim_{h \to 0} \frac{\int_{c}^{x+h} f(t)dt-\int_{c}^{x} f(t)dt}{h}$ =[Integrating Definite Integral Backwards] $\lim_{h \to 0} \frac{\int_{c}^{x+h} f(t)dt+\int_{x}^{c} f(t)dt}{h}$ =[Definite Integrals on Adjacent Intervals] $\lim_{h \to 0} \frac{\int_{x}^{x+h} f(t)dt}{h}$ =🚀
Now, observe that for very small values of h, $\int_{x}^{x+h} f(t)dt ≈ f(x)·h$ by a simple left-hand approximation of the integral.
$\lim_{h \to 0} \frac{\int_{x}^{x+h} f(t)dt}{h}$ =🚀 = $\lim_{h \to 0} \frac{f(x)·h}{h} = f(x)$∎
The reader should notice that A(x) solves the differential equation y’ = f, with the initial condition y(c) = 0 ($\int_{c}^{c} f(t)dt = 0$). If we differentiate A’(x) = $\lim_{\Delta x \to 0}\frac{\Delta A}{\Delta x} = f(x)$ -By assumption, f is continuous-.
$\frac{d}{dx} \int_{1}^{x} \frac{dt}{t^2}$. Notice that A(x) = $\int_{c}^{x} \frac{dt}{t^2} = \int_{a}^{x} f(t)dt$ where f(t) = $\frac{1}{t^2}$, c = 1, and by the previous result, A’(x) = f(x), that is, $\frac{d}{dx} \int_{1}^{x} \frac{dt}{t^2} = \frac{1}{x^2}$.
Let’s check our previous result. $\int_{1}^{x} \frac{dt}{t^2} = \int_{1}^{x} t^{-2} = -t^{-1}\bigg|_{1}^{x} = -\frac{1}{x}-(-1) = 1 -\frac{1}{x} = A(x)$
⇒ $\frac{d}{dx}A(x) = \frac{d}{dx}(1 -\frac{1}{x}) = \frac{1}{x^2}$∎ ⇒ A is an antiderivative of f, and since A(1) = $\int_{1}^{1} f(t)dt = 0$, A is the only antiderivative of f for which A(1) = 0.
$\frac{d}{dx} \int_{2}^{x} (cos(t) -t)dt$. Notice that A(x) = $\frac{d}{dx} \int_{2}^{x} (cos(t) -t)dt$ where f(x) = cos(t) -t and c = 2, and by the previous result A’(x) = f(x), that is, $\frac{d}{dx} \int_{2}^{x} (cos(t) -t)dt = cos(x)-x.$
Let’s check our previous result. $\int_{2}^{x} (cos(t) -t)dt = sin(t) -\frac{1}{2}t^2\bigg|_{2}^{x} = sin(x)-\frac{1}{2}x^2-(sin(2)-2)= A(x)$
⇒ $\frac{d}{dx}A(x) = cos(x)-x$ ∎ ⇒ A is an antiderivative of f, and since A(2) = $\int_{2}^{2} f(t)dt = 0$, A is the only antiderivative of f for which A(2) = 0.
$\frac{d}{dx} \int_{1}^{\sqrt{x}} sin(t)dt$ =[$u(x)=\sqrt{x}, F(x)=\int_{1}^{u(x)} sin(t)d(t) = G∘u, G(x)=\int_{1}^{x} sin(t)d(t)$]. By the The Second Fundamental Theorem of Calculus and the Chain Rules, $\frac{d}{dx} \int_{1}^{\sqrt{x}} sin(t)dt = sin(u(x))·\frac{du}{dx} = sin(u(x))·(\frac{1}{2}x^{\frac{-1}{2}}) = \frac{sin(\sqrt{x})}{2\sqrt{x}}$
The Fundamental Theorem of Calculus states roughly that the integral of a function f over an interval is equal to the change of any antiderivate F (F'(x) = f(x)) between the ends of the interval, i.e., $\int_{a}^{b} f(x)dx = F(b)-F(a)=F(x) \bigg|_{a}^{b}$.
Proof. (assuming f is continuous)
Let’s F be an antiderivate, F’ = f, and let’s define G(x) = $\int_{a}^{x} f(t)dt$. By the Fundamental Theorem of Calculus, G’(x) = f(x) ⇒[F’ = f] F’(x) = G’(x) ⇒[Uniqueness of antiderivatives] F(x) = G(x) + c where c is a constant.
Therefore, F(b) - F(a) =[F(x) = G(x) + c] (G(b) + c) - (G(a) + c) = G(b) -G(a) = $\int_{a}^{b} f(x)dx -\int_{a}^{a} f(t)dt = \int_{a}^{b} f(x)dx$ ∎