Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 6

Integration

The Fundamental Theorem of Calculus

For a long time, scientists struggled with the problem of finding areas of regions bounded by irregular curves. The answer required the work of the best scientific minds of the eighteenth and nineteenth centuries, and it is known as the “Fundamental Theorem of Calculus.”

In this section, we study the problem of finding the area of regions bounded by a continuous functions on a closed interval and the $x$-axis. Such areas are referred as “areas under curves.”

Example 6.26. If ${\displaystyle f\left(x\right)=2x+5\text{,}}$ then the area under the curve on the interval ${\displaystyle \left[-4,1\right]}$ consists of the areas of the two triangles shown in Figure 6.6, below.

Figure 6.6: Area under the curve defined by ${\displaystyle f\left(x\right)=2x+5}$ on the interval ${\displaystyle \left[-4,1\right]}$—not to scale

Hence, the area under the curve is

${\displaystyle {\displaystyle \frac{|-4+5∕2|\left(3\right)}{2}}+{\displaystyle \frac{|1+5∕2|\left(7\right)}{2}}={\displaystyle \frac{29}{2}}\text{.}}$

Example 6.27. If ${\displaystyle f\left(x\right)=2x+5\text{,}}$ then area under the curve on the interval ${\displaystyle \left[2,6\right]}$ is the area of a rectangle plus the area of a triangle, as shown in Figure 6.7, below.

Figure 6.7: Area under the curve defined by ${\displaystyle f\left(x\right)=2x+5}$ on the interval ${\displaystyle \left[2,6\right]}$—not to scale

Hence, the area under the curve is

${\displaystyle \left(4\right)\left(9\right)+{\displaystyle \frac{\left(4\right)\left(8\right)}{2}}=52\text{.}}$

Finding areas under curves when the function is not a line is more challenging. We consider this problem for functions that are continuous on a closed interval and that have smooth curves; that is, for differentiable functions.

We will approach this problem by approximation, and as we look for better approximations, we will arrive at the solution; that is, we will come to the Fundamental Theorem of Calculus.

Let $f$ be a continuous function on the interval ${\displaystyle \left[a,b\right]\text{,}}$ and let $f$ be differentiable on the interval ${\displaystyle \left(a,b\right)\text{.}}$

A possible approximation to the area under this curve is the area of a rectangle with base ${\displaystyle \left(b-a\right)}$ and height ${\displaystyle f\left(x*\right)\text{,}}$ where ${\displaystyle x*}$ is any number ${\displaystyle a<x*<b\text{,}}$ as shown in Figure 6.8, below.

Figure 6.8: Estimating the area under the curve of a continuous function $f$—first approximation

If $A$ is the area under the curve, then we know that

${\displaystyle A\approx f\left(x*\right)\left(b-a\right)\text{.}}$

If we want a better approximation, we should consider two rectangles instead of one. So we must divide the interval ${\displaystyle \left(a,b\right)}$ into two parts, say

${\displaystyle a=x_1<x_2<x_3=b\text{.}}$

The bases of the rectangles are ${\displaystyle \left(x_2-x_1\right)}$ and ${\displaystyle \left(x_3-x_2\right)}$; their heights are ${\displaystyle f\left(x_1*\right)}$ and ${\displaystyle f\left(x_2*\right)}$ for any ${\displaystyle x_1<x_1*<x_2<x_2*<x_3\text{,}}$ as shown in Figure 6.9, below.

Figure 6.9: Estimating the area under the curve of a continuous function $f$—second approximation

For these two rectangles.

${\displaystyle A\approx f\left(x_1*\right)\left(x_2-x_1\right)+f\left(x_2*\right)\left(x_3-x_2\right)\text{.}}$

Three rectangles give an even better approximation. We divide the interval as follows:

${\displaystyle a=x_1<x_2<x_3<x_4=b\text{.}}$

The bases of the rectangles are ${\displaystyle \left(x_2-x_1\right)\text{,}}$ ${\displaystyle \left(x_3-x_2\right)}$ and ${\displaystyle \left(x_4-x_3\right)}$; their heights are ${\displaystyle f\left(x_1*\right)\text{,}}$ ${\displaystyle f\left(x_2*\right)}$ and ${\displaystyle f\left(x_3*\right)}$ for any ${\displaystyle x_1<x_1*<x_2<x_2*<x_3<x_3*<x_4\text{,}}$ as shown in Figure 6.10, below.

Figure 6.10: Estimating the area under the curve of a continuous function $f$—third approximation

For these three rectangles

${\displaystyle A\approx f\left(x_1*\right)\left(x_2-x_1\right)+f\left(x_2*\right)\left(x_3-x_2\right)+f\left(x_3*\right)\left(x_4-x_3\right)\text{.}}$

We can continue with four, five and more rectangles, obtaining better and better approximations—the greater the number of rectangles, the better the approximation. In general, if we take $n$ rectangles, then we divide the interval ${\displaystyle \left(a,b\right)}$ into $n$ parts

${\displaystyle a=x_1<x_2<x_3<.....<x_n<x_{n+1}=b\text{.}}$

The bases of the $n$ rectangles are

${\displaystyle \left(x_2-x_1\right),\left(x_3-x_2\right),\left(x_4-x_3\right),.....,\left(x_{n+1}-x_n\right)\text{.}}$

Their heights are

${\displaystyle f\left(x_1*\right),f\left(x_2*\right),f\left(x_3*\right),....,f\left(x_n*\right),}$

for any

${\displaystyle x_1<x_1*<x_2<x_2*<x_3<x_3*<x_4<....<x_n*<x_{n+1}\text{.}}$

See Figure 6.11, below.

Figure 6.11: Estimating the area under the curve of a continuous function $f$—$n$th approximation

For these $n$ rectangles

${\displaystyle \begin{array}{ll}A\approx f\left(x_1*\right)\left(x_2-x_1\right)\hfill & +f\left(x_2*\right)\left(x_3-x_2\right)\hfill \\ \hfill & +f\left(x_3*\right)\left(x_4-x_3\right)+\text{. }\text{. }\text{. }+f\left(x_n*\right)\left(x_{n+1}-x_n\right)\text{.}\hfill \end{array}}$

In sigma notation,

(6.7) ${\displaystyle A\approx {\displaystyle \sum _{i=1}^{n}f\left(x_i*\right)\left(x_{i+1}-x_i\right)\text{.}}}$

Let us now use the Mean Value Theorem to find the sum in Equation 6.7.

Let $F$ be the antiderivative of the function $f$; that is, ${\displaystyle F^{\prime }=f\text{.}}$ Hence, $F$ is continuous in the interval ${\displaystyle \left[a,b\right]}$ and differentiable in ${\displaystyle \left(a,b\right)\text{,}}$ and for any ${\displaystyle 1\le i\le n\text{,}}$ the function $F$ is continuous in the interval ${\displaystyle \left[x_i,x_{i+1}\right]\text{,}}$ and differentiable in ${\displaystyle \left(x_i,x_{i+1}\right)}$ for ${\displaystyle i=1,2,3...n\text{.}}$

By the Mean Value Theorem

${\displaystyle F\left(x_{i+1}\right)-F\left(x_i\right)=F^{\prime }\left(x_i*\right)\left(x_{i+1}-x_i\right)=f\left(x_i*\right)\left(x_{i+1}-x_i\right)}$

for some

${\displaystyle x_i<x_i*<x_{i+1}\text{.}}$

Hence,

${\displaystyle {\displaystyle \sum _{i=1}^{n}f\left(x_i*\right)\left(x_{i+1}-x_i\right)=}{\displaystyle \sum _{i=1}^{n}F\left(x_{i+1}\right)}-F\left(x_i\right)\text{.}}$

The sum ${\displaystyle {\displaystyle \sum _{i=1}^{n}F\left(x_{i+1}\right)}-F\left(x_i\right)}$ is telescopic, because

\begin{align*} \sum_{i = 1}^n F(x_{i + 1}) - F(x_i) & = [F(x_2)- F(x_1)] + [F(x_3) - F(x_2)] + \; . \ . \ . \\ & \ . \ . \ . \; + \; [F(x_n) - F(x_{n - 1})] + [F(x_{n + 1}) - F(x_n)] \\ & = F(x_{n + 1}) - F(x_1) \\ & = F(b) - F(a) \end{align*}

Therefore,

${\displaystyle A\approx {\displaystyle \sum _{i=1}^{n}f\left(x_i*\right)}\left(x_{i+1}-x_i\right)=F\left(b\right)-F\left(a\right)\text{.}}$

As we said, the more rectangles (the larger the $n$), the better the approximation, and the more rectangles, the smaller ${\displaystyle \left(x_{i+1}-x_i\right)\text{.}}$ Since ${\displaystyle \left(x_{i+1}-x_i\right)}$ is a small quantity of the variable ${\displaystyle x\text{,}}$ it is a differential ${\displaystyle \text{Δ}{x}_i=\left(x_{i+1}-x_i\right)\text{.}}$

Hence, as ${\displaystyle {\displaystyle \sum _{i=1}^{n}f\left(x_i*\right)\text{Δ}{x}_i}}$ approaches the area ${\displaystyle A\text{.}}$ This is precisely a limit, and we have found that

${\displaystyle A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i*\right)\text{Δ}{x}_i=}\underset{n\to \infty }{\lim }F\left(b\right)-F\left(a\right)\text{=}F\left(b\right)-F\left(a\right)\text{.}}$

This is the Fundamental Theorem of Calculus. It is fundamental, because a simple geometrical concept—area—is given by the antiderivative of a function that is not a geometrical concept. That these two concepts are so intimately related is what makes this theorem so outstanding. Moreover, to arrive to this theorem was not easy, we needed to learn about limits—a concept that took humankind a long time to understand fully—together with the Mean Value Theorem.

We write ${\displaystyle F{\left(x\right)|}_a^b}$ to indicate the difference ${\displaystyle F\left(b\right)-F\left(a\right)\text{.}}$

That is, ${\displaystyle F{\left(x\right)|}_a^b=F\left(b\right)-F\left(a\right)\text{.}}$

What we have shown so far is that, if $f$ is a continuous positive function on the interval ${\displaystyle \left[a,b\right]}$ and has antiderivative ${\displaystyle F\text{,}}$ then the area under the curve on the interval is the limit.

${\displaystyle \underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f}{\left(x_i*\right)\text{Δ}{x}_i=F\left(x\right)|}_a^b=F\left(b\right)-F\left(a\right)\text{.}}$

The notation for this limit must indicate the function ${\displaystyle f\text{,}}$ the interval ${\displaystyle \left[a,b\right]}$ and the fact that the limit is the limit of a sum; therefore, instead of ${\displaystyle \Sigma \text{,}}$ we use the integral symbol ${\displaystyle \int \text{,}}$ and we indicate the interval ${\displaystyle \left[a,b\right]}$ by writing $b$ as an upper limit of integration and $a$ as a lower limit of integration, as is shown below.

${\displaystyle {\int }_a^b}$

We next indicate the integrand function as $f$ and the differential as ${\displaystyle dx\text{.}}$ Once we have done so, we have the definite integral of the function $f$ on the interval ${\displaystyle \left[a,b\right]\text{,}}$ as shown in Definition 6.10, below.

Area Under Curve Graph

Example 6.28. We recognize the limit

$\displaystyle{\lim_{n \to \infty} \sum_{i = 1}^n x_i^* \sin(x_i^*) \Delta x_i}$

on the interval ${\displaystyle \left[0,\pi \right]}$ as the definite integral

${\displaystyle {\int }_0^{\pi }x\sin xdx\text{.}}$

Observe that the function ${\displaystyle x\sin x}$ is positive on the interval ${\displaystyle \left[0,\pi \right]}$; thus, the definite integral is the area under this function on the interval ${\displaystyle \left[0,\pi \right]\text{.}}$

Example 6.29. The function ${\displaystyle f\left(x\right)=2x+5}$ is positive on the interval ${\displaystyle \left[2,6\right]\text{.}}$ So, by the Fundamental Theorem of Calculus, the definite integral

${\displaystyle {\int }_2^{6}2x+5dx}$

is the area under the curve on the interval ${\displaystyle \left[2,6\right]\text{.}}$ Hence, the area is

${\displaystyle {\displaystyle {\int }_2^{6}2x+5dx=x^2+{5x|}_2^6}={\left(6\right)}^2+5\left(6\right)-\left({\left(2\right)}^2+5\left(2\right)\right)=66-14=52\text{,}}$

as we found in Example 6.27.

Example 6.30. If ${\displaystyle f\left(x\right)=m}$ is any constant function, then it is continuous on any interval ${\displaystyle \left[a,b\right]\text{,}}$ and the area under the curve is equal to the area of a rectangle with base ${\displaystyle \left(b-a\right)}$ and height ${\displaystyle \left|m\right|\text{.}}$ Hence, ${\displaystyle \left|m\right|\left(b-a\right)\text{.}}$ On the other hand, ${\displaystyle \left|m\right|}$ is a positive constant, and its antiderivative is ${\displaystyle F\left(x\right)=\left|m\right|x\text{.}}$ So, by the Fundamental Theorem of Calculus:

${\displaystyle {\displaystyle {\int }_a^b\left|m\right|}dx=\left|m\right|x{|}_a^b=\left|m\right|b-\left|m\right|a=\left|m\right|\left(b-a\right)\text{.}}$

For any constant ${\displaystyle m\text{,}}$ the definite integral

${\displaystyle {\int }_a^{b}mdx=m\left(b-a\right)}$

is the area above the $x$-axis if ${\displaystyle m>0}$; it is minus the area below the $x$-axis if ${\displaystyle m<0\text{.}}$

We know that for a negative function ${\displaystyle g\left(x\right)}$ on an interval ${\displaystyle \left[a,b\right]\text{,}}$ with antiderivative ${\displaystyle G\left(x\right)\text{,}}$ the function ${\displaystyle -g\left(x\right)}$ is positive on ${\displaystyle \left[a,b\right]\text{,}}$ its antiderivative is ${\displaystyle -G\left(x\right)\text{,}}$ and the areas under the curves of $g$ and ${\displaystyle -g}$ are equal. [See Definition 2.12.] Thus, the definite integral

${\displaystyle {\int }_a^b-g\left(x\right)dx=-G\left(b\right)+G\left(a\right)=-\left(G\left(b\right)-G\left(a\right)\right)}$

is positive.

Since the areas under the curves of $g$ and ${\displaystyle -g}$ are equal, we know that the area under the curve of $g$ on the interval ${\displaystyle \left[a,b\right]}$ is the positive number

${\displaystyle {\int }_a^b-g\left(x\right)dx=-\left(G\left(b\right)-G\left(a\right)\right)\text{.}}$

Therefore, if the function $g$ is negative on ${\displaystyle \left[a,b\right]\text{,}}$ the definite integral of $g$ is the negative number

${\displaystyle {\int }_a^{b}g\left(x\right)dx=G\left(b\right)-G\left(a\right),}$

and the area under the curve of $g$ in the interval $[a,b]$ is

$-{\displaystyle {\int }_a^{b}g(x)dx}$.

Therefore,

${\displaystyle {\int }_a^b-g\left(x\right)dx=-{\int }_a^{b}g\left(x\right)dx\text{.}}$

Theorem 6.11. The Fundamental Theorem of Calculus, Part 2.

[Note that we discuss Part 1 later in this unit.]

If $f(x)$ is continuous on the interval ${\displaystyle \left[a,b\right]\text{,}}$ with antiderivative ${\displaystyle F\left(x\right)\text{,}}$ then

\[\lim_{n \to \infty} \sum_{i = 1}^n f(x_i*)\Delta x_i = \int_a^b f(x)\;dx = F(b) - F(a)\]

where

$\Delta x_i = x_{i + 1} - x_i$ and $a \leq x_i \lt x_i* \lt x_{i +1} \leq b$ for $i = 1 \ . \ . \ . \ n$.

Proof. Observe that the Mean Value Theorem holds for the antiderivative function ${\displaystyle F\text{,}}$ for any continuous function $f$ not necessarily positive. Hence, the argument we presented earlier is the proof of this theorem. We used a positive function, because we wanted to give a geometric meaning to the limit of the definite integral.

Q.E.D

Example 6.31. The function ${\displaystyle f\left(x\right)=2x+5}$ on the interval ${\displaystyle \left[-4,-5∕2\right]}$ is negative, and its graph is below the $x$-axis (see Figure 6.6).

Hence, the definite integral

${\displaystyle \begin{array}{ll}{\displaystyle {\int }_{-4}^{-5/2}2x+5dx}\hfill & =x^2+5x{|}_{-4}^{-5/2}\hfill \\ \hfill & ={\displaystyle \frac{25}{4}}-{\displaystyle \frac{25}{2}}-\left({\left(-4\right)}^2+5\left(-4\right)\right)=-{\displaystyle \frac{9}{4}}\hfill \end{array}}$

is a negative number, and the area under the curve is

${\displaystyle \left|-{\displaystyle \frac{9}{4}}\right|={\displaystyle \frac{9}{4}}\text{.}}$

The area under the curve on the interval ${\displaystyle \left[-5∕2,1\right]}$ is the definite integral

${\displaystyle {\displaystyle {\int }_{-5/2}^{1}2x+5dx=x^2+5x{|}_{-5/2}^1}=1+5-\left({\displaystyle \frac{25}{4}}-{\displaystyle \frac{25}{2}}\right)={\displaystyle \frac{49}{4}}\text{.}}$

The area under the curve on the interval ${\displaystyle \left[-4,1\right]}$ is the sum of these two areas:

${\displaystyle {\displaystyle \frac{9}{4}}+{\displaystyle \frac{49}{4}}={\displaystyle \frac{29}{2}},}$

as we found in Example 6.26.

However, if we consider the definite integral on the interval ${\displaystyle \left[-4,1\right]\text{,}}$ we see that

${\displaystyle {\displaystyle {\int }_{-4}^{1}2x+5dx=x^2+5x{|}_{-4}^1}=1+5-\left(16-20\right)=10.}$

This result is not the same as the area under the curve on the interval ${\displaystyle \left[-4,1\right]\text{.}}$ Rather, it is the area under the curve on ${\displaystyle \left[-5∕2,1\right]\text{,}}$ minus the area under the curve on ${\displaystyle \left[-4,-5∕2\right]\text{.}}$ And as we can see,

${\displaystyle {\displaystyle \frac{49}{4}}-{\displaystyle \frac{9}{4}}=10\text{.}}$

Example 6.32. If we look at the graph of the cosine function on the interval ${\displaystyle \left[0,\pi \right]\text{,}}$ we understand why the definite integral on this interval is zero: on this interval, the area above the $x$-axis is equal to the area below the $x$-axis.

${\displaystyle {\int }_0^{\pi }\cos xdx=\sin x{|}_0^{\pi }=sin\left(\pi \right)-sin\left(0\right)=0\text{.}}$

Example 6.33. Let us evaluate ${\displaystyle {\int }_{-1}^4\sqrt{x+4}+xdx\text{:}}$

${\displaystyle \begin{array}{ll}{\displaystyle {\int }_{-1}^4\sqrt{x+4}}+xdx\hfill & ={\displaystyle \frac{2{\left(x+4\right)}^{3/2}}{3}}+{{\displaystyle \frac{x^2}{2}}|}_{-1}^4\hfill \\ \hfill & ={\displaystyle \frac{2{\left(8\right)}^{3/2}}{3}}+{\displaystyle \frac{4^2}{2}}-\left({\displaystyle \frac{2{\left(3\right)}^{3/2}}{3}}+{\displaystyle \frac{1}{2}}\right)\hfill \\ \hfill & ={\displaystyle \frac{32\sqrt{2}}{3}}-2\sqrt{3}+{\displaystyle \frac{15}{2}}\text{.}\hfill \end{array}}$

Now let us evaluate ${\displaystyle {\int }_0^{\pi ∕4}{sec}^{2}x-sin\left(2x\right)dx}$

${\displaystyle \begin{array}{ll}{\displaystyle {\int }_0^{\pi /4}{\sec }^{2}x-\sin \left(2x\right)}dx\hfill & =\tan x+{{\displaystyle \frac{\cos \left(2x\right)}{2}}|}_0^{\pi /4}\hfill \\ \hfill & =\tan \left({\displaystyle \frac{\pi }{4}}\right)+{\displaystyle \frac{\cos \left({\displaystyle \frac{\pi }{2}}\right)}{2}}-\left(\tan \left(0\right)+{\displaystyle \frac{\cos \left(0\right)}{2}}\right)\hfill \\ \hfill & =1-{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{2}}\text{.}\hfill \end{array}}$

When we use the $u$-identification or the $u$-substitution to evaluate definite integrals, we have two choices to deal with the limits of integration.

1. We can change the limits of integration according to the function ${\displaystyle u\text{.}}$
2. We can find the antiderivative in terms of the original variable, and then evaluate at the original limits of integration.

Example 6.34. Let us use the $u$-substitution to find the antiderivative of the integrand function

${\displaystyle {\int }_{-1}^{4}x\sqrt[3]{x+1}dx\text{.}}$

Set ${\displaystyle u=x+1\text{,}}$ then ${\displaystyle du=dx}$ and ${\displaystyle x=u-1\text{.}}$

1. We can change the limits of integration according to the function $u$. So, if $x = -1$, $u = 0$ and if $x = 4$, $u = 5$. These are the limits of integration according to the variable $u$, and the integral is

   \begin{align*} \int_0^5 (u - 1)(u)^{1/3}\,du & = \int_0^5 u^{4/3} - u^{1/3}\;du \\ & = \dfrac{3u^{7/3}}{7} - \dfrac{3u^{4/3}}{4} \Biggm|_0^5\\ & = \dfrac{3(5^{7/3})}{7} - \dfrac{3(5^{4/3})}{4}\\ & = 5^{4/3}\left(\dfrac{15}{7} - \dfrac{3}{4}\right)\\ & = \dfrac{39\sqrt[3]{5^4}}{28} \end{align*}

2. We can find the antiderivative in terms of the original variable (in this case, $x$), and then we can evaluate it at the original limits of integration.

   ${\displaystyle \begin{array}{ll}{\displaystyle \int x\sqrt[3]{x+1}dx}\hfill & ={\displaystyle \int u^{4/3}-u^{1/3}du={\displaystyle \frac{3u^{7/3}}{7}}-{\displaystyle \frac{3u^{4/3}}{4}}+C}\hfill \\ \hfill & ={\displaystyle \frac{3{\left(x+1\right)}^{7/3}}{7}}-{\displaystyle \frac{3{\left(x+1\right)}^{4/3}}{4}}+C\text{.}\hfill \end{array}}$

   Hence,

   ${\displaystyle \begin{array}{ll}{\displaystyle {\int }_{-1}^{4}x\sqrt[3]{x+1}dx}\hfill & ={\displaystyle \frac{3{\left(x+1\right)}^{7/3}}{7}}-{{\displaystyle \frac{3{\left(x+1\right)}^{4/3}}{4}}|}_{-1}^4\hfill \\ \hfill & ={\displaystyle \frac{3{\left(5\right)}^{7/3}}{7}}-{\displaystyle \frac{3{\left(5\right)}^{4/3}}{4}}\text{=}{\displaystyle \frac{39\sqrt[3]{5^4}}{28}}\text{.}\hfill \end{array}}$

Warning

The most common mistakes in the evaluation of definite integrals are

1. evaluating the antiderivative in the wrong order.
2. Pay attention: The value of the antiderivative should be at the upper limit of integration minus at the lower limit of integration.
2. mixing up the positive and negative signs. Always use parentheses to evaluate the antiderivative to avoid this type of error.
3. incorrect use of the limits of integrations when using the $u$ identification or substitution methods.

Note: A definite integral $\color{#384877}{\displaystyle{\int_a^b f(x)\;dx}}$ is equal to $\color{#384877}{F(b) - F(a)}$ where $\color{#384877}{F'(x) = f(x)}$, only if the integrand function $\color{#384877}{f(x)}$ is continuous on the interval $\color{#384877}{[a,b]}$.

The Fundamental Theorem of Calculus cannot be applied to the evaluation of the integral

${\displaystyle {\int }_0^{\pi ∕2}\tan xdx}$

because ${\displaystyle \tan x}$ is continuous on ${\displaystyle \left[0,\pi ∕2\right)\text{,}}$ not on ${\displaystyle \left[0,\pi ∕2\right]\text{.}}$ This type of integral is called an improper integral, and its evaluation is not covered in this course.

When the author of your textbook says that a definite integral exists or does not exist, he is referring to the existence of the limit of the Riemann sum in the Fundamental Theorem of Calculus, Part 2. But this limit always exists if the integrand function is continuous on the closed interval given by the limits of integration, and this is what he wants you to check.

When we have difficulties finding the antiderivative of a function, we have the option of using the properties of the definite integral to find at least an estimate of its value.

Properties of the Definite Integral

1. ${\displaystyle {\int }_a^{b}cf\left(x\right)dx=c{\int }_a^{b}f\left(x\right)dx}$ for any constant ${\displaystyle c\text{.}}$
2. ${\displaystyle {\int }_a^{b}f\left(x\right)\pm g\left(x\right)dx={\int }_a^{b}f\left(x\right)\pm {\int }_a^{b}g\left(x\right)dx}$
3. ${\displaystyle {\int }_a^{b}mdx=m\left(b-a\right)}$
4. ${\displaystyle {\int }_a^{b}f\left(x\right)dx={\int }_a^{c}f\left(x\right)dx+{\int }_c^{b}f\left(x\right)dx}$ for any constant $c$
5. ${\displaystyle {\int }_a^{b}f\left(x\right)dx\ge 0}$ if $f$ is positive on ${\displaystyle \left[a,b\right]}$

6. If ${\displaystyle f\left(x\right)\le g\left(x\right)}$ for all $x$ in ${\displaystyle \left[a,b\right]\text{,}}$ then

   ${\displaystyle {\int }_a^{b}f\left(x\right)dx\le {\int }_a^{b}g\left(x\right)dx}$

7. If ${\displaystyle m\le f\left(x\right)\le M}$ for all $x$ in ${\displaystyle \left[a,b\right]\text{,}}$ then

   ${\displaystyle m\left(b-a\right)\le {\int }_a^{b}f\left(x\right)dx\le M\left(b-a\right)\text{.}}$

8. ${\displaystyle -{\int }_a^{b}f\left(x\right)dx={\int }_b^{a}f\left(x\right)dx\text{.}}$

Proof. Property 3 is Example 6.30. Property 5 is true because the definite integral is the area under the function $f$ on the interval ${\displaystyle \left[a,b\right]\text{.}}$

To prove Properties 1, 2, and 4, let $F$ and $G$ be the antiderivatives of $f$ and $g$ respectively.

Property 1

${\displaystyle \begin{array}{ll}{\displaystyle {\int }_a^{b}cf\left(x\right)dx}\hfill & =cF{\left(x\right)|}_a^b=cF\left(b\right)-cF\left(a\right)\hfill \\ \hfill & =c\left(F\left(b\right)-F\left(a\right)\right)=c{\displaystyle {\int }_a^{b}f\left(x\right)dx\text{.}}\hfill \end{array}}$

Property 2

${\displaystyle \begin{array}{ll}{\displaystyle {\int }_a^{b}f\left(x\right)\pm \text{g}\left(x\right)dx}\hfill & =F{\left(x\right)\pm G\left(x\right)|}_a^b\hfill \\ \hfill & =F\left(b\right)\pm G\left(b\right)-\left(F\left(a\right)\pm G\left(a\right)\right)\hfill \\ \hfill & =\left(F\left(b\right)-F\left(a\right)\right)\pm \left(G\left(b\right)-G\left(a\right)\right)\hfill \\ \hfill & ={\displaystyle {\int }_a^{b}f\left(x\right)\pm {\displaystyle {\int }_a^{b}g\left(x\right)dx\text{.}}}\hfill \end{array}}$

Property 4

${\displaystyle \begin{array}{ll}{\displaystyle {\int }_a^{c}f\left(x\right)dx+{\displaystyle {\int }_c^{b}f\left(x\right)dx}}\hfill & =F\left(c\right)-F\left(a\right)+F\left(b\right)-F\left(c\right)\hfill \\ \hfill & =F\left(b\right)-F\left(a\right)={\displaystyle {\int }_a^{b}f\left(x\right)dx\text{.}}\hfill \end{array}}$

Property 6

We know that ${\displaystyle g\left(x\right)-f\left(x\right)\ge 0}$ on the interval ${\displaystyle \left[a,b\right]\text{.}}$ So, by Properties 5 and 2,

${\displaystyle {\int }_a^{b}g\left(x\right)dx-{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}g\left(x\right)-f\left(x\right)dx\ge 0;}$

thus,

${\displaystyle {\int }_a^{b}f\left(x\right)dx\le {\int }_a^{b}g\left(x\right)dx\text{.}}$

Property 7

By Properties 3 and 6, we have

${\displaystyle m\left(b-a\right)={\int }_a^{b}mdx\le {\int }_a^{b}f\left(x\right)dx\le {\int }_a^{b}Mdx=M\left(b-a\right)\text{.}}$

Property 8

${\displaystyle -{\int }_a^{b}f\left(x\right)dx=-\left(F\left(b\right)-F\left(a\right)\right)=F\left(a\right)-F\left(b\right)={\int }_b^{a}f\left(x\right)dx\text{.}}$

Q.E.D

Example 6.35. We do not know how to find the antiderivative of the function ${\displaystyle x\sin x\text{,}}$ but we know that ${\displaystyle \left|\sin x\right|\le 1\text{.}}$

If $x$ is in the interval ${\displaystyle \left[0,\pi ∕3\right]\text{,}}$ then

${\displaystyle 0\le x\le {\displaystyle \frac{\pi }{3}},}$

and since ${\displaystyle -1\le \sin x\le 1\text{,}}$ we multiply this inequality by ${\displaystyle x\ge 0}$ and we get ${\displaystyle -x\le x\sin x\le x\text{.}}$

Since ${\displaystyle x\sin x\ge 0}$ on this interval, we have

${\displaystyle 0\le x\sin x\le x\text{ for any }x\text{ in }\left[0,{\displaystyle \frac{\pi }{3}}\right]\text{.}}$

By Properties 5 and 6, above,

${\displaystyle 0\le {\int }_0^{\pi ∕3}x\sin xdx\le {\int }_0^{\pi ∕3}xdx={\displaystyle \frac{{\pi }^2}{18}}\text{.}}$

That is, the value of the integral ${\displaystyle {\int }_0^{\pi ∕3}x\sin xdx}$ is in the interval ${\displaystyle \left[0,{\displaystyle \frac{{\pi }^2}{18}}\right]\text{.}}$

If $f$ is continuous on the interval ${\displaystyle \left[a,b\right]\text{,}}$ then by the Extreme Value Theorem, there are numbers $m$ and ${\displaystyle M\text{,}}$ such that ${\displaystyle m\le f\left(x\right)\le M}$ for all $x$ in ${\displaystyle \left[a,b\right]\text{.}}$ Hence, by Property 7, we can always find lower and upper bounds for a definite integral.

Example 6.36. On the interval ${\displaystyle \left[1,6\right]\text{,}}$ the function

${\displaystyle f\left(x\right)={\displaystyle \frac{1}{x}}}$

is decreasing and continuous.

Hence, it is bounded by ${\displaystyle 1}$ and ${\displaystyle {\displaystyle \frac{1}{6}}}$; that is, ${\displaystyle {\displaystyle \frac{1}{6}}\le {\displaystyle \frac{1}{x}}\le 1\text{,}}$ and by Property 7,

${\displaystyle {\displaystyle \frac{1}{6}}\left(5\right)\le {\int }_1^6{\displaystyle \frac{1}{x}}dx\le 5\text{.}}$

The value of the integral is in the interval ${\displaystyle \left[{\displaystyle \frac{5}{6}},5\right]\text{.}}$

Part 1 of the Fundamental Theorem of Calculus refers to the inverse nature of differentiation and integration. We know that the integral of the derivative function is equal to the function; that is, ${\displaystyle \int f^{\prime }\left(x\right)dx=f\left(x\right)\text{.}}$ Now we will see that the converse also holds; that is, the derivative of an integral is the integrand function

Theorem 6.12. The Fundamental Theorem of Calculus, Part 1.

Let $f(x)$ be a continuous function on the interval ${\displaystyle \left[a,b\right]\text{.}}$ For ${\displaystyle a\le x\le b\text{,}}$ we define the function

${\displaystyle g\left(x\right)={\int }_a^{x}f\left(t\right)dt\text{.}}$

Then

${\displaystyle g^{\prime }\left(x\right)={\displaystyle \frac{d}{dx}}{\int }_a^{x}f\left(t\right)dt=f\left(x\right)\text{.}}$

Proof. If $F$ is the antiderivative of ${\displaystyle f\text{,}}$ then

${\displaystyle g\left(x\right)={\int }_a^{x}f\left(t\right)dt=F\left(t\right){|}_a^x=F\left(x\right)-F\left(a\right)\text{.}}$

Therefore ${\displaystyle g^{\prime }\left(x\right)=F^{\prime }\left(x\right)=f\left(x\right)\text{.}}$

Q.E.D

If we compose the function ${\displaystyle g\left(x\right)={\int }_a^{x}f\left(t\right)dt}$ with a differentiable function ${\displaystyle h\left(x\right)\text{,}}$ then we have

${\displaystyle g\left(h\left(x\right)\right)={\int }_a^{h\left(x\right)}f\left(t\right)dt\text{.}}$

By the Chain Rule,

${\displaystyle {\displaystyle \frac{d}{dx}}{\int }_a^{h\left(x\right)}f\left(t\right)dt={\displaystyle \frac{d}{dx}}g\left(h\left(x\right)\right)=g^{\prime }\left(h\left(x\right)\right)h^{\prime }\left(x\right)=f\left(h\left(x\right)\right)h^{\prime }\left(x\right)\text{.}}$

Example 6.37.

1. ${\displaystyle {\displaystyle \frac{d}{dx}}{\int }_0^{3x^4}{\displaystyle \frac{3}{x^2+6}}dx={\displaystyle \frac{3}{{\left(3x^4\right)}^2+6}}\left(12x^3\right)={\displaystyle \frac{12x^3}{3x^8+2}}\text{.}}$
2. Observe that Part 1 of the Fundamental Theorem of Calculus holds only if the lower limit of integration is a constant. If it is not, we use Property 8 to change the limits of integration to apply this theorem.
3. \begin{align*} \dfrac{d}{dx} \int_{\sqrt{x}}^4 t^2\cos t \, dt & = - \! \int_4^{\sqrt{x}} t^2 \cos t\,dt \\ & = -((\sqrt{x})^2 \cos (\sqrt{x}))\dfrac{1}{2\sqrt{x}} \\ & = -\dfrac{\sqrt{x} \, \cos (\sqrt{x})}{2}{\text{.}} \end{align*}
3. If the upper and lower limits of integration are functions, then we use Property 4 to split the integral into two separate integrals, each of which has an arbitrarily chosen constant as a lower limit of integration.
4. \begin{align*} \dfrac{d}{dx}\int_{3x^2}^{\sin x} t^4 + 6t^3\,dt & = \dfrac{d}{dx}\int_{3x^2}^0 t^4 + 6t^3\,dt + \dfrac{d}{dx}\int_0^{\sin x} t^4 + 6t^3\,dt \\[3pt] & = -\dfrac{d}{dx}\int_0^{3x^2} t^4 + 6t^3\,dt + \dfrac{d}{dx} \int_0^{\sin x}t^4 + 6t^3\;dt \\[3pt] & = -[(3x^2)^4 + 6(3x^2)^3](6x) + (\sin^4x + 6\sin^3 x)(\cos x) \\[3pt] & = -486x^7(x^2 + 2)+(\sin x + 6)\sin^3 x\cos x \end{align*}