5.2 The Fundamental Theorem of Calculus Part 1

Goals

• introduce the concept of an indefinite integral of a function;
• apply the methods from Section 5.1 to find antiderivatives to compute indefinite integrals;
• state and prove the Mean Value Theorem for Integrals (MVTFI);
• state, proof, and understand the role of the Fundamental Theorem of Calculus Part 1 (TFTCP1);
• combine TFTCP1 and the chain rule to compute derivatives of new functions obtained via integration.

5.2.1 Indefinite Integrals

Definition 1. The indefinite integral of a function \(f\) on \((a,b)\), denoted by
\[ \int f(x)\,dx,\]
is the most general antiderivative of \(f\) on \((a,b)\).

The expression \(f (x)\) is called the integrand, and the variable \(x\) is the variable of integration.

Remark. (1) \(\displaystyle \int f(x)\,dx\) represents the most general function \(F\) defined on \((a,b)\), such that \(F'(x)=f(x)\) for all \(x\in (a,b)\); (2) it follows that if \(F\) is an antiderivative of \(f\) on \((a,b)\), then \(\displaystyle \int f(x)\,dx = F(x)+ C\).

Example 1.
Verify the validity of each statement.

1. \(\displaystyle \int 4x e^{x^2}\arctan e^{x^2}\,dx = 2e^{x^2}\arctan e^{x^2} – \ln \left (e^{2x^2}+1\right ) + C\);
2. \(\displaystyle \int x^{x^2}\left (\frac{1}{x}+ ( x+ 2x\ln x) \ln x \right )\,dx = x^{x^2}\ln x+ C\).

Solution. (a) The statement
\[ \int 4x e^{x^2}\arctan e^{x^2}\,dx = 2e^{x^2}\arctan e^{x^2} – \ln \left (e^{x^2}+1\right ) + C\]
means that the RHS is an antiderivative of the integrand on the LHS. Thus, we need to show that
\[\frac{d}{dx}\left ( 2e^{x^2}\arctan e^{x^2} – \ln \left (e^{2x^2}+1\right ) + C\right ) = 4x e^{x^2}\arctan e^{x^2}.\]
Applying the product rule and the chain rule to compute the derivative on the LFH, we have (verify this)
\[4x e^{x^2}\arctan e^{x^2} + 2e^{x^2} \frac{2x e^{2x^2}}{1+ e^{2x^2}} – \frac{ 4xe^{x^2}}{ e^{2x^2}+1 } = 4x e^{x^2}\arctan e^{x^2}.\]

(b) Similarly, we need to show that the derivative of the RHS is equal to the integrand on the LHS. In this case, it is convenient to use logarithmic differentiation. Let \( y = x^{x^2}\ln x\), then
\[\frac{d}{dx}|y| = \frac{d}{dx}\ln \left | x^{x^2}\ln x\right |\Longrightarrow \frac{1}{y}y’ = \frac{d}{dx}\left ( x^2\ln x + \ln|\ln x|\,\right ) = 2x\ln x + x + \frac{1}{\ln x}\cdot\frac{1}{x} \Longrightarrow\]
\[\Longrightarrow y’ = x^{x^2}\ln x\left (2x\ln x + x + \frac{1}{\ln x}\cdot\frac{1}{x}\right )= x^{x^2} \left (\ln x (2x\ln x + x ) + \frac{1}{x}\right ).\]

Example 2.
Find a function \(f\) such that the equality is valid.    (a) \(\displaystyle \int f(t)\,dt= \cot^{-1}\sqrt[3]{t^2}+C\);    (b) \(\displaystyle \int f(t)\,dt= \ln \frac{x^3 e^{-x^2}}{\displaystyle \sqrt{1+x^2}}+C\).

Solution. (a) The equation \(\displaystyle \int f(t)\,dt= \cot^{-1}\sqrt[3]{t^2}+C\) means that \(A(t)=\cot^{-1}\sqrt[3]{t^2}+C\) is an antiderivative of \(f\), that is, \(A'(t)=f(t)\). Thus,
\[f(t)=\frac{d}{dt}\cot^{-1}\sqrt[3]{t^2} \underset{?}{=}-\frac{2}{3}\frac{1}{\sqrt[3]{t}}\frac{1}{1+\left (\sqrt[3]{t^2}\right )^2} =-\frac{2}{3}\frac{1}{\sqrt[3]{t}\left (1+t\sqrt[3]{t}\right )}.\]

(b) Since \(\displaystyle F(x) =\ln \frac{x^3 e^{-x^2}}{\displaystyle \sqrt{1+x^2}}+C = 3\ln x -x^2 – \frac{1}{2}\ln (1+x^2) +C\),
\[\Longrightarrow f(x) = \frac{d}{dx}\left (3\ln x -x^2 – \frac{1}{2}\ln (1+x^2) +C \right ) =\frac{3}{x}-2x -\frac{x}{1+x^2}.\]

Example 3.
Evaluate the indefinite integral.   (a) \(\displaystyle \int x\sin(x^2) \,dx\);     (b) \(\displaystyle \int \frac{\arcsin^2 3x}{ \sqrt{ 1-9x^2}}\,dx\);   (c) \(\displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx\);     (d) \(\displaystyle \int \left(\cos^4x – \sin^4x\right ) \,dx\).

Solution. The common procedure to evaluate these indefinite integrals require (i) to rewrite and simplify the integrands, and (ii) the use of the chain rule: \((f\circ g)'(x) = f'(g(x))g'(x)\).

(a) Let \(g(x)=x^2\), then \(g'(x)=2x\), and \(\displaystyle f'(x)=\frac{1}{2}\sin x\). Thus \(\displaystyle f(x)=-\frac{1}{2}\cos x+C\), and
\[\int x\sin(x^2) \,dx =f(g(x))+C= -\frac{1}{2}\cos(x^2)+C.\]

(b) Let \(g(x)=\arcsin 3x\), then \(\displaystyle g'(x) = \frac{3}{\sqrt{1-9x^2}}\), and le t\(\displaystyle f'(x)=\frac{1}{3} x^2\). Then, \(\displaystyle f(x)=\frac{1}{9} x^3+C\), and
\[\int \arcsin^2 3x\frac{1}{ \sqrt{ 1- 9x^2}}\,dx =f(g(x))+C= \frac{1}{9}\arcsin^3 3x+C.\]

(c) Let \(g(x)=\sqrt{x}\), then \(\displaystyle g'(x) = \frac{1}{2\sqrt{x}}\), and let \(\displaystyle f'(x)=2 e^x\). Then, \(\displaystyle f(x) = 2e^x+C\), and
\[\int \frac{e^{\sqrt{x}}}{2\sqrt{x}}\,dx =f(g(x))+C= 2 e^{\sqrt{x}}+C.\]

(d) \( \cos^4x – \sin^4x\ = \left(\cos^2x – \sin^2x\right )\left(\cos^2x + \sin^2x\right ) = \cos^2x – \sin^2x = \cos 2x\). Thus
\[\int \left(\cos^4x – \sin^4x\right )\,dx = \int \cos 2x \, dx = \frac{1}{2}\sin 2x + C.\]

Example 4.
Evaluate the indefinite integral \(\displaystyle \int \frac{ \cos \pi x}{\displaystyle \sin^3\pi x} \,dx\) in three different ways and verify that your answers are compatible.

Solution.
(1) Let \(g(x)=\sin \pi x, g'(x)= \pi \cos \pi x\), and \(\displaystyle f'(g(x))= \frac{1}{g(x)^3} = g(x)^{-3}\). Thus \(\displaystyle f(x)= -\frac{1}{2}x^{-2}+C\), and
\begin{align*}
\int \frac{ \cos \pi x}{\sin^3\pi x} \,dx
& =\frac{1}{\pi} \int \pi \cos \pi x \frac{1}{\sin^3 \pi x} \,dx = \frac{1}{\pi} \cdot\left (-\frac{1}{2}(\sin \pi x)^{-2}\right ) +C = -\frac{1}{2\,\pi} \csc^2\pi x+C
\end{align*}

(2) First, we rewrite the integrand:
\[\frac{ \cos \pi x}{\sin^3\pi x} = \frac{ \cos \pi x}{\sin \pi x}\frac{1}{\sin^2 \pi x} = \cot \pi x \csc^2 \pi x.\]
Then, the chain rule immediately implies that
\[ \int \cot \pi x \csc^2 \pi x \,dx
=\color{red}{\pmb{-\frac{1}{2\,\pi}}} \int \color{red}{\pmb{2}}\cot \pi x \cdot \left ( \color{red}{\pmb{-\pi}}\csc^2 \pi x \right )\,dx =\color{red}{\pmb{-\frac{1}{2\,\pi}}} \cot^2\pi x + C.\]

(3) Note that
\[ \cot \pi x \csc^2 \pi x =\csc \pi x \cdot \cot \pi x \csc \pi x.\]
Then, by the chain rule
\[ \int \csc \pi x \cdot \cot \pi x \csc \pi x \,dx
=\color{red}{\pmb{-\frac{1}{2\,\pi}}} \int \color{red}{\pmb{2}}\csc \pi x \cdot \left ( \color{red}{\pmb{-\pi}}\cot \pi x \csc \pi x \right )\,dx =\color{red}{\pmb{-\frac{1}{2\,\pi}}} \csc^2\pi x + C.\]

To verify that the answers are compatible, note that since \(\cot^2x=\csc^2x-1\), we have
\[-\frac{1}{2\,\pi} \cot^2\pi x + C= -\frac{1}{2\,\pi} (\csc^2x-1)+ C = -\frac{1}{2\,\pi} \csc^2x+ D\]
for the new constant \(\displaystyle D= C-\frac{1}{2\, \pi}\). Thus, the answers in (1), (2), and (3) are compatible.

5.2.3 The Mean Value Theorem for Integrals

Goals. In this subsection, we introduce the average value of a continuous function over a closed interval, and then state and prove the Mean Value Theorem of Integrals (MVTFI). The proof of the MVTFI makes use of the Extreme Value Thoerem (EVT) for continuous functions on closed intervals; which we state without proof. The MVTFI will be used in the proof of The Fundamental Theorem of Calculus Part 1.

Let \(f\) be a continuous function defined on in the interval \(I=[a,b]\). Let \(a=x_0, x_1, \dots , x_n=b\) be the endpoints of a partition of \(I\) into \(n\) subintervals of equal length \(\displaystyle \Delta x=\frac{b-a}{n}\), and let \(x_i^*\) be a random point in the \(i\)-th intervals \([x_{i-1},x_i]\).
The sum
\[\frac{f(x_1^*) + f(x_2^*) + \cdots + f(x_n^*)}{n} = \sum_{i=1}^{n} f(x_i^*)\frac{1}{n}\]
provides an approximation to the average of the function \(f\) over \(I\). Note that this sum can be rewritten as follows:
\[\sum_{i=1}^{n} f(x_i^*)\frac{1}{n} = \frac{1}{b-a}\sum_{i=1}^{n} f(x_i^*)\Delta x \]
Thus, we have
\[\lim_{n\to \infty} \sum_{i=1}^{n} f(x_i^*)\frac{1}{n} = \lim_{n\to \infty}\frac{1}{b-a}\sum_{i=1}^{n} f(x_i^*)\Delta x = \frac{1}{b-a}\int_a^bf(x)\,dx.\]
This calculation motivates the following definition.
Definition 2.
Let \(f (x)\) be continuous over the interval \(⎣a, b⎤\). Then, the average value of \(f \) on \([a, b]\) is
\[\frac{1}{b-a}\int_a^bf(x)\,dx.\]

Theorem 1. (The Mean Value Theorem for Integrals.) If \(f (x)\) is continuous over the interval \([a,b]\), then there exists at least one point \(c \in [a,b]\) such that
\begin{equation}
f(c) = \frac{1}{b-a}\int_a^bf(x)\,dx.\label{ch05-02-eq01}\tag{5.1}
\end{equation}

For the proof of the MVTFI we will use the following theorem about continuous functions, which we state without proof.

Theorem 2.
(The Extreme Value Theorem.) If \(f\) is a continuous function over a closed interval \([a, b]\), then there exist points \(x_1\) and \(x_2\) in \([a, b]\) such that
\[f(x_1) \leq f(x) \leq f(x_2), \;\; \forall \; x\in [a,b].\]
\(f(x_1)\) and \(f(x_2)\) are called an absolute minimum and an absolute maximum, respectively, of \(f\) over \([a, b]\).

Proof.
(Of the MVTFI.)

Step 1. Application of the Extreme Value Theorem.
Since \(f\) is continuous on \([a,b]\), there exist points \(x_1\) and \(x_2\) in \([a, b]\) such that
\[f(x_1) \leq f(x) \leq f(x_2), \;\; \forall \; x\in [a,b].\]

Step 2. Application of the comparison properties for definite integrals.
\[f(x_1)(b-a) \leq \int_a^bf(x)\,dx \leq f(x_2)(b-a).\]
Hence
\[\hspace{1in}\hphantom{(*)}f(x_1) \leq \frac{1}{(b-a)}\int_a^bf(x)\,dx \leq f(x_2). \hspace{1in}(*)\]

Step 3. Application of the Intermediate Value Theorem. Since \([x_1, x_2] \subset [a,b]\), \(f\) is continuous on \([x_1,x_2]\). Thus, \(f\) satisfies the assumptions of the IVT on the interval \([x_1,x_2]\) for the value as in \((*)\). It follows that there exists a number \(c \in [x_1,x_2]\) such that (\ref{ch05-02-eq01})) is satisfied.

5.2.4 The Fundamental Theorem of Calculus Part 1

Theorem 3. (The Fundamental Theorem of Calculus Part 1.) Let \(f (x)\) be a continuous function over the interval \([a,b]\), \(c\in [a,b]\), and define the function \(F\) by
\begin{equation}
F(x)=\int_c^xf(t)\,dt,\;\; x\in(a,b).\label{ch05-02-eq02}\tag{5.2}
\end{equation}
Then, \(F\) is an antiderivative of \(f\) on \((a,b)\).

Proof. We need to prove that \(F'(x) = f(x)\) for all \(x\in (a,b)\).

Step 1.
Let \(x\in (a,b)\), then, by definition, the derivative of \(F\) at \(x\) is given by
\[F'(x) = \lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=
\lim_{h\to 0}\frac{1}{h}\left (\int_c^{x+h}f(x)\,dx -\int_c^{x}f(x)\,dx\right ).\]

Step 2.
Application of the additive property of definite integrals (see equation (3.4) in 3.4.2).
\[\int_c^{x+h}f(x)\,dx -\int_c^{x}f(x)\,dx =\int_x^{c}f(x)\,dx + \int_c^{x+h}f(x)\,dx
= \int_x^{x+h}f(x)\,dx\]
Thus
\[\hspace{1in}\hphantom{(*)}
F'(x) =
\lim_{h\to 0}\frac{1}{h}\int_x^{x+h}f(x)\,dx.\hspace{1in} (*)\]

Step 3.
Application of the Mean Value Theorem for Integrals (MVTFI).

Assume that \(h > 0\), the proof for \(h < 0 \) is similar). Note that the right-hand-side of \((*)\) is precisely the mean value of \(f\) over the interval \([x, x+h]\). Thus, by the MVTFI, there exists a number \(d\in [x, x+h]\) such that \[\frac{1}{h}\int_x^{x+h}f(x)\,dx= f(d).\] Step 4.
Application of the continuity of \(f\).

Given that \(d \in [x, x+h]\), then, as \(h\to 0\), \(d\to x\). And, since \(f\) is continuous at \(x\), we have
\[\lim_{h\to 0} f(d) = \lim_{d\to x}f(d) = f(x).\]
Thus
\[F'(x) =
\lim_{h\to 0}\frac{1}{h}\int_x^{x+h}f(x)\,dx = \lim_{h\to 0} f(d) = f(x),\]
and the proof is complete.

Example 1.
Compute the derivative of the function.

1.   \(\displaystyle A(x)=\int_1^x\sin t^2\,dt\); \(\vphantom{A(x)=\int_x^{\sqrt{\arccos 2x}}\cos t^2\,dt}\)
2.   \(\displaystyle A(x)=\int_x^{\pi^2} \cos \sqrt{t}\,dt. \); \(\vphantom{A(x)=\int_x^{\sqrt{\arccos 2x}}\cos t^2\,dt}\)
3.   \(\displaystyle A(x)=\int_x^{\sqrt{\arccos x}}\cos t^2\,dt\);
4.   \(\displaystyle B(x)=\int_{\sqrt{\ln x}}^{x^2}e^{t^2}\,dt\).

Solution.
(a) \(\displaystyle A'(x)=\frac{d}{dx}\int_1^x\sin t^2\,dt = \sin x^2\).

(b) \(\displaystyle A'(x)=\frac{d}{dx}\int_x^{\pi^2} \cos \sqrt{t}\,dt =
-\frac{d}{dx}\int_{\pi^2}^x \cos \sqrt{t}\,dt = -\cos \sqrt{x}\).

(c) In this problem we apply the additive property with respect to the interval of integration for definite integrals and the Chain Rule.
\[\displaystyle A(x)=\int_x^{\sqrt{\arccos x}}\cos t^2\,dt
= \int_x^{0}\cos t^2\,dt + \int_0^{\sqrt{\arccos x}}\cos t^2\,dt \Longrightarrow\]
\[\Longrightarrow A'(x)
= -\frac{d}{dx} \int_0^{x}\cos t^2\,dt + \frac{d}{dx}\int_0^{\sqrt{\arccos x}}\cos t^2\,dt.\]
While it is immediate from TFTCP1 that
\[-\frac{d}{dx} \int_0^{x}\cos t^2\,dt= -\cos x^2,\]
To evaluate the second derivative, we need to apply the chain rule.
Let
\[u=g(x)=\sqrt{\arccos x} \;\;\;\; \text{and} \;\;\;\; y=f(u)=\int_0^{u}\cos t^2\,dt\Longrightarrow \]
\[\Longrightarrow g'(x) = -\frac{1}{2}\frac{1}{\displaystyle \sqrt{\arccos x}}\frac{1}{\displaystyle \sqrt{1-x^2}} \;\;\;\; \text{and} \;\;\;\; f'(u) = \cos u^2.\]
Thus
\[ (f\circ g)'(x) = g'(x)f'(u)
= -\frac{1}{2}\frac{1}{\displaystyle \sqrt{\arccos x}}\frac{1}{\displaystyle \sqrt{1-x^2}} \cdot \cos u^2
\underset{?}{=} -\frac{1}{2}\frac{1}{\displaystyle \sqrt{\arccos x}}\frac{1}{\displaystyle \sqrt{1-x^2}} \cdot x .\]
\[ \Longrightarrow A'(x) =-\cos x^2 -\frac{x}{\displaystyle2\,\sqrt{1-x^2}\, \sqrt{\arccos x}}.\]

(d) Note that $D_B= [1,\infty)$. Thus, we can rewrite $B$ as follows (applying the additive property with respect to the interval of integration for definite integrals):
\[B(x)=\int_{\sqrt{\ln x}}^{x^2}e^{t^2}\,dt
= \int_{\sqrt{\ln x}}^{1}e^{t^2}\,dt + \int_{1}^{x^2}e^{t^2}\,dt
= -\int_{1}^{\sqrt{\ln x}}e^{t^2}\,dt + \int_{1}^{x^2}e^{t^2}\,dt.
\]
Applying the Chain Rule as in (c), we obtain
\[-\frac{d}{dx}\int_{1}^{\sqrt{\ln x}}e^{t^2}\,dt =- \frac{d}{dx}\sqrt{\ln x}\cdot\frac{d}{du}\int_{1}^{u}e^{t^2}\,dt = -\frac{1}{\displaystyle 2x \,\sqrt{\ln x}} \cdot e^{u^2} \underset{?}{=} -\frac{1}{\displaystyle 2 \,\sqrt{\ln x}}\]
\[\frac{d}{dx}\int_{1}^{x^2}e^{t^2}\,dt = \frac{d}{dx}x^2\cdot \frac{d}{du}\int_{1}^{u}e^{t^2}\,dt = 2xe^{u^2} = 2xe^{x^4}.\]
\[\Longrightarrow B'(x) = -\frac{1}{\displaystyle 2 \,\sqrt{\ln x}} + 2xe^{x^4}. \]

5.2.5 Net Accumulated Area

In Definition 3 in 3.4.4, we introduced the concept of net accumulate starting at \(c\) for a function \(f\) continuous on \([a,b]\), with \(c\in [a,b]\), as, precisely,
\[F(x)= \int_c^xf(t)\,dt, \;\;\; x\in [a,b].\]
In particular, if \(f(x) \geq 0 \) for all \(x\in [a,b]\), and \(c=a\), then, \(F(x)\) is the accumulated area as defined in 1.3.3. Furthermore, in Example 3 (in 1.3.3), we obtained the following expression for the net accumulated area for \(\displaystyle f(x)=\sqrt{r^2-x^2}\), \(x\in[-r,r]\):
\[A(x) =
\frac{1}{2}r^2\left(\pi – \arccos \frac{x}{r}\right)+\frac{1}{2}x\sqrt{r^2-x^2}.\]
That is,
\begin{equation}\label{ch05-02-eq03}\tag{5.3}
\int_{-r}^x\sqrt{r^2-t^2}\,dt =\frac{1}{2}r^2\left(\pi – \arccos \frac{x}{r}\right)+\frac{1}{2}x\sqrt{r^2-x^2} , \;\;\; x\in [a,b].
\end{equation}
In view of TFTCP1, this means that the right-hand-side is an antiderivative of \(\displaystyle f(x)=\sqrt{r^2-x^2}\).

Example 1.
Verify that (\ref{ch05-02-eq03}) is valid.

Solution. We need to show that the RHS in (\ref{ch05-02-eq03}) is indeed an antiderivative of \(\displaystyle f(x)=\sqrt{r^2-x^2}\), and that it coincides with the LHS at one point (and hence everywhere on \([-r,r]\)).
\[\frac{d}{dx}\left (\frac{1}{2}r^2\left(\pi – \arccos \frac{x}{r}\right)+\frac{1}{2}x\sqrt{r^2-x^2}\right ) =
\frac{1}{2} r^2\frac{1}{\sqrt{r^2-x^2}} + \frac{1}{2}\sqrt{r^2-x^2} – \frac{1}{2}x\frac{x}{\sqrt{r^2-x^2}}= \]
\[=\frac{1}{2} \frac{r^2}{\sqrt{r^2-x^2}} + \frac{1}{2} \frac{r^2-2x^2}{\sqrt{r^2-x^2}}= \frac{1}{2}\frac{2r^2-2x^2}{\sqrt{r^2-x^2}} = \sqrt{r^2-x^2}. \]
Since the LHS and the RHS in (\ref{ch05-02-eq03}) are antiderivatives of \(\displaystyle f(x)=\sqrt{r^2-x^2}\) it follows that they differ by a constant (see Theorem 1 in 5.1.1), that is LHS \(=\) RHS\(\,+C\). But, since
\[\int_{-r}^{-r}\sqrt{r^2-t^2}\,dt = 0, \;\;\; \frac{1}{2}r^2\left(\pi – \arccos \frac{-r}{r}\right)+\frac{1}{2}(-r)\sqrt{r^2-(-r)^2} = 0, \Longrightarrow C=0.\]
From this, it follows that (\ref{ch05-02-eq03}) is valid.