# Fundamental Theorem of Calculus

What are the most beautiful theorems of calculus 1? This article reviews the Intermediate Value Theorem, the Extreme Value Theorem, the Mean Value Theorem. I then explain the first and second fundamental theorems of calculus.

We being by reviewing the Intermediate Value Theorem and the Extreme Value Theorem both of which are needed later when studying the Fundamental Theorem of Calculus. The Mean Value Theorem for Integrals and the first and second forms of the Fundamental Theorem of Calculus are then proven.

## Introduction to the Fundamental Theorem of Calculus

Before we begin, let’s review the Intermediate Value Theorem and the Extreme Value Theorem.

Theorem. (Intermediate Value Theorem) If $f$ is a continuous function on a closed interval $[a,b]$ and $M$ is a number between $f(a)$ and $f(b)$, inclusive, then there is at least one number $c$ in $[a,b]$ such that $f(c)=M.$

Theorem. (Extreme Value Theorem) If $f$ is a continuous function on a closed interval $[a,b]$, then $f$ attains an absolute maximum value $f(c)$ for some number $c$ in $[a,b]$ and an absolute minimum value $f(d)$ for some number $d$ in $[a,b].$

Definition. If $f$ is integrable on $[a,b]$, then the average value of $f$ over $[a,b]$ is the number $$f_{\text{av}}=\frac{1}{b-a} \int_a^b f(x) \, dx.$$

If we assume that $f$ is nonnegative, then we have the following geometric interpretation for the average value of a function over $[a,b].$ We see that $f_{\text{av}}$ is the height of the rectangle with base lying on the interval $[a,b]$ and having the same area as the area of the region under the graph of $f$ on $[a,b].$

Example. Find the average value of $$f(x)=\frac{x}{\sqrt{x^2+1}}$$ over the interval $[0,3].$

## Mean Value Theorem for Integrals

Cauchy discovered both the Mean Value Theorem and the Mean Value Theorem for Integrals.

Theorem. (Mean Value Theorem for Integrals) If $f$ is continuous on $[a,b]$, then there exists a number $c$ in $[a,b]$ such that $$f(c) = \frac{1}{b-a}\int_a^b f(x) \, dx.$$

Proof. Since $f$ is continuous on the interval $[a,b]$, the Extrema Value Theorem tells us that $f$ attains an absolute minimum value $m$ at some number in $[a,b]$ and an absolute maximum value $M$ at some number in $[a,b].$ So $m\leq f(x) \leq M$ for all $x$ in $[a,b].$ Hence the average value holds: $$m\leq \frac{1}{b-a}\int_a^b f(x) \, dx \leq M$$ Because the average value lies between $m$ and $M$, the Intermediate Value Theorem there exists at least one number $c$ in $[a,b]$ such that $$f(c)=\frac{1}{b-a}\int_a^b f(x) \, dx.$$ as needed.

Example. Show that the inequality $$0\leq \int_0^1 \frac{x^5}{\sqrt[3]{1+x^4}}\, dx \leq \frac{1}{6}$$ holds.

Solution. Let $f$ be the function $$f(x)=\frac{x^5}{\sqrt[3]{1+x^4}}.$$ Since $f$ is continuous on $[0,1]$, we know that $f$ must attain a maximum $M$ and a minimum value $m.$ We find the first derivative $$f'(x)= \frac{x^4(19 x^4+15)}{3(x^4+1)^{2/3}}.$$ Hence the only critical number is $0.$ We test and find $m=f(0)=0$ and $M=f(1)=1/\sqrt[3]{2}\approx 0.79.$ However, we can do better, by observing that $1+x^4\geq 1$ for $x$ in $[0,1].$ Since we know that $$0\leq \frac{x^5}{\sqrt[3]{1+x^4}} \leq x^5$$ it must also be that \begin{align} 0=\int_0^ 1 0 \, dx \leq \int_0^ 1 \frac{x^5}{\sqrt[3]{1+x^4}} \, dx \leq \int_0^ 1 x^5 \, dx=\frac{1}{6} \end{align}
as needed.

## The First Fundamental Theorem of Calculus

Theorem. (First Fundamental Theorem of Calculus) If $f$ is continuous on $[a,b]$, then the function $F$ defined by $$F(x)=\int_a^x f(t) \, dt, \quad a\leq x \leq b$$ is differentiable on $(a,b)$ and $$F'(x)=\frac{d}{dx} \int_a^x f(t) \, dt = f(x).$$

Proof. Fix $x$ in $(a,b)$ and suppose that $x+h$ is in $(a,b)$, where $h\neq 0.$ Then \begin{align} F(x+h)-F(x) & =\int_a^{x+h} f(t) \, dt -\int_a^x f(t)\, dt \\ & = \int_a^x f(t)\, dt +\int_x^{x+h} f(t)\, dt -\int_a^x f(t)\, dt \\ & =\int_x^{x+h} f(t)\, dt. \end{align} By the Mean Value Theorem for Integrals there exists a number $c$ between $x$ and $x+h$ such that $$f(c)=\int_x^{x+h} f(t)\, dt$$ Therefore, $$\frac{F(x+h)-F(x)}{h}=\frac{1}{h} \int_{x}^{x+h} f(t)\, dt =f(c).$$ Next, observe that as $h$ approaches $0$, the number $c$, which is squeezed between $x$ and $x+h$ approaches $x$, and by continuity, $f(c)$ approaches $f(x).$ Therefore, \begin{align} F'(x) & =\lim_{h\to 0} \frac{F(x+h)-F(x)}{h} \\ & =\lim_{h\to 0} \frac{1}{h} \int_x^{x+h} f(t)\, dt =\lim_{h\to 0} f(c)=f(x). \end{align} as needed.

Example. Find the derivative of the following functions.

$(1) \quad \displaystyle f(x)=\int_{-1}^x t \sqrt{t^2+1} \, dt$

$(2) \quad \displaystyle g(x)=\int_{0}^{x^2} t \sin t \, dt$

$(3) \quad \displaystyle F(x)=\int_{2}^{\sqrt{x}} \frac{\sin t}{t} \, dt$

$(4) \quad \displaystyle G(x)=\int_{x^2}^{x^3} \ln t\, dt$, $x>0$

Solution. We find that $f'(x)=x\sqrt{x^2+1}$ and $$g'(x)=(x^2 \sin x^2) (2x) = 2x^3\sin x^2.$$ For the third function, we find that $$F'(x) =\frac{\sin \sqrt{x}}{\sqrt{x}} \frac{d}{dx}\left(\sqrt{x}\right) =\frac{\sin \sqrt{x}}{\sqrt{x}} \left(\frac{1}{2\sqrt{x}}\right).$$ For the last function we first write, $$G(x)=\int_{x^2}^c \ln t\, dt +\int_c^{x^3} \ln t\, dt = -\int_{c}^{x^2} \ln t\, dt +\int_c^{x^3} \ln t\, dt$$ and so we find $$G'(x) = – 2x \ln x^2 +3x^2\ln x^3 = -4x \ln x+9 x^2\ln x =x(9x-4) \ln x$$ as needed.

Example. Evaluate $\displaystyle \lim_{h\to 0} \frac{1}{h} \int_2^{2+h} \sqrt{5+t^2} \, dt.$

Solution. Let $F$ be the function $$F(x)=\int_2^x \sqrt{5+t^2}\, dt.$$ Then it follows that \begin{align} F'(2) & =\lim_{h\to 0} \frac{F(2+h)-F(2)}{h} \\ & =\lim_{h\to 0} \frac{\int_0^{2+h} \sqrt{5+t^2}\, dt -\int_0^2 \sqrt{5+t^2}\, dt}{h} \\ & =\lim_{h\to 0} \frac{\int_2^{2+h} \sqrt{5+t^2}\, dt }{h}. \end{align} Next using the Fundamental Theorem, we have $$F'(x)=\frac{d}{dx} \int_2^x \sqrt{5+t^2}\, dt =\sqrt{5+x^2}$$ and so $F'(2)=3$, hence $$\lim_{h\to 0} \frac{1}{h} \int_2^{2+h} \sqrt{5+t^2} \, dt=3.$$ as desired.

## The Second Fundamental Theorem of Calculus

Theorem. (The Second Fundamental Theorem of Calculus) If $f$ is continuous on $[a,b]$, then $$\int_a^b f(x) \, dx=F(b)-F(a)$$ where $F$ is any antiderivative of $f$, that is $F’=f.$

Proof. Let $G$ be the function defined by $$G(x)=\int_a^x f(t)\, dt.$$ We know that $G$ is an antiderivative of $f.$ If $F$ is any other antiderivative of $f$, then $F$ and $G$ must differ by a constant. In other words, $F(x)=G(x)+C.$ To determine $C$, we put $x=a$ to obtain $$F(a) = G(a)+C = \int_a^a f(t)\, dt +C=C.$$ Therefore, evaluating $F$ at $b$, we have $$F(b) = G(b)+C=\int_a^b f(t)\, dt +F(a)$$ from which we conclude that $$F(b)-F(a)=\int_a^b f(x) \, dx.$$ as needed.

## Exercises on the Fundamental Theorem of Calculus

Exercise. Evaluate the following integrals.

$(1) \quad \displaystyle \int_{-2}^0 (2x-3)\, dx$

$(2) \quad \displaystyle \int_{0}^2 (2-4u+u^2)\, du$

$(3) \quad \displaystyle \int_1^2 \frac{3x^4-2x^2+1}{2x^2} \, dx$

$(4) \quad \displaystyle \int_2^0 \sqrt{x}(x+1)(x-2) \, dx$

$(5) \quad \displaystyle \int_0^{\pi} \sin 2x \cos x \, dx$

$(6) \quad \displaystyle \int_0^{\pi} |\cos x| \, dx$

$(7) \quad \displaystyle \int _0^{4\sqrt{3}} \frac{1}{x^2+16}\, dx$

$(8) \quad \displaystyle \int _1^e \frac{\ln x}{x} e^{(\ln x)^2}\, dx$

Exercise. Show that each of the following inequalities hold.

$(1) \quad \displaystyle 0\leq \int_0^1 \frac{x^5}{\sqrt[3]{1+x^4}}\, dx \leq \frac{1}{6}$

$(2) \quad \displaystyle 0\leq \int_0^1 \frac{1}{\sqrt{4-3x+x^2}}\, dx \leq \frac{2}{3}$

Exercise. For each of the following, find the area of the region under the graph of $f$ on $[a,b].$

$(1) \quad \displaystyle f(x)=x^2-2x+2; [-1,2]$

$(2) \quad \displaystyle f(x)=\frac{1}{x^2}; [1,2]$

$(3) \quad \displaystyle f(x)=2+\sqrt{x+1}; [0,3]$

$(4) \quad \displaystyle f(x)=\frac{1}{4+x^2}; [0,1]$

Exercise. Find $\displaystyle \frac{dx}{dy}$ if $$\int_0^x \sqrt{3+2\cos t} \, dt+\int_0^y \sin t \, dt =0.$$

Exercise. Find all functions $f$ on $[0,1]$ such that $f$ is continuous on $[0,1]$ and $$\int_0^x f(t) \, dt = \int_x^1 f(t)\, dt$$ for every $x$ in $(0,1).$

Exercise. Evaluate $$\lim_{h\to 0} \frac{1}{h} \int_2^{2+h} \sqrt{5+t^2} \, dt.$$

Exercise. Evaluate $$\int_{-1}^1 \frac{2x^5+x^4-3x^3+2x^2+8x+1}{x^2+1} \, dx.$$

Exercise. Use the identity $$\frac{x \sin\left(n+\frac{1}{2}\right)}{2\sin \frac{x}{2}} = \frac{1}{2} +\cos x +\cos 2x +\cdots + \cos n x$$ to show that $$\int_0^\pi \frac{x \sin\left(n+\frac{1}{2}\right)}{2\sin \frac{x}{2}} \, dx =\pi.$$

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.