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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] ↭ $\Delta F = Average(F’)·\Delta x$.

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

  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-.

  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

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
  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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