# Fundamental Theorems of Calculus. MVT for Integrals

“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.

# Recall

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., 3x2 -1 is the antiderivative of x3 -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.

# Alternative version of the Fundamental Theorem of Calculus.

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) cm3/s. What is the average flow from times t = 0 to t = 2? The average rate of change or flow is $\frac{1}{2-0}\int_{0}^{2} (1-e^{-t})dt = \frac{1}{2}(t-e^{-t})\bigg|_{0}^{2} = \frac{1}{2}(2-e^{-2}+e^{-0}) ≈ 0.56.$

# The Mean Value Theorem for Integrals

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 $f(c) = \frac{1}{b-a}\int_{a}^{b} f(x)d(x)$.

• Exercises.
1. 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.

2. 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

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-.

• Exercises
1. $\frac{d}{dx} \int_{1}^{x} \frac{dt}{t^2}$. 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.

2. $\frac{d}{dx} \int_{2}^{x} (cos(t) -t)dt$. 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.

3. $\frac{d}{dx} \int_{1}^{\sqrt{x}} sin(t)dt$ =[$u(x)=\sqrt{x}, F(x)=\int_{1}^{u(x)} sin(t)d(t) = G∘u, G(x)=\int_{1}^{x} sin(t)d(t)$]. By the The Second Fundamental Theorem of Calculus and the Chain Rules, $\frac{d}{dx} \int_{1}^{\sqrt{x}} sin(t)dt = sin(u(x))·\frac{du}{dx} = sin(u(x))·(\frac{1}{2}x^{\frac{-1}{2}}) = \frac{sin(\sqrt{x})}{2\sqrt{x}}$

# Proof of the Fundamental Theorem of Calculus

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) = $\int_{a}^{b} f(x)dx -\int_{a}^{a} f(t)dt = \int_{a}^{b} f(x)dx$ ∎

# Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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