3.4 Definite Integral

3.4.1 Definite Integral

 

The solution of the area problem involved the approximation to the area under the graph of a function by means of the sum
\[\sum_{i=1}^{n} f(x_i)\Delta x,\;\; \Delta x=\frac{b-a}{n}, \;\; x_i=a+i\Delta x,\;\; i=0, 1,\dots , n, \;\; n \in \mathbb{N},\]
where each product \(f(x_i)\Delta x\) represented the area of a rectangular region. Such a sum is a special type of a Riemann sum.

 

Definition 1. Let \(f\) be a 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, and let \(x_i^*\) be a random point in the \(i\)-th intervals \([x_{i-1},x_i]\). Let \(\Delta x_i\) denote the length of the \(i\)-th subinterval. The sum
\[ \sum_{i=1}^{n} f(x_i^*)\Delta x_i\]
is called a Riemann sum.

 

Remark. (1) The subintervals do not need to have the same length. We set
\[||\Delta x|| = \mathrm{length \; of \; the \; largest \; subinterval \; in \; the \; partition}.\]

 

(2) Note that it is not required that \(f(x)\geq 0\) for all \(x\in I\). If this condition is satisfied, then the Riemann sum approximates the area under the graph of \(f\) on \(I\).

 

(3) The special case when the subintervals have the same length, and \(x_i^*=x_i\), corresponds to the right-endpoint rectangular approximation: \(R(f,n)\).

 

Definition 2. Let \(f\) be a 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, and let \(x_i^*\) be a random point in the \(i\)-th intervals \([x_{i-1},x_i]\). Let \(\Delta x_i\) denote the length of the \(i\)-th subinterval. The limit, with \(n\) increasing indefinitely and \(||\Delta x|| \to 0\) at the same time,
\[ \lim_{n\to \infty, \; ||\Delta x|| \to 0}\sum_{i=1}^{n} f(x_i^*)\Delta x_i,\]
if it exists, is called the definite integral of \(f\) over the interval \(I=[a,b]\), and \(f\) is said to be integrable on \([a,b]\).

Notation. . The definite integral of \(f\) over the interval \([a,b]\) is usually denoted as follows:
\[\int_a^bf(x)\,dx.\]

A condition that guarantees that the limit of the Riemann sums exists is that \(f\) be continuous on\([a,b]\). This is the content of the following theorem that we state without proof (using the above notation).

Theorem 1. (Existence of Definite Integral). Let \(f\) be a continuous function on the closed interval \([a,b]\). Then the limit
\[\int_a^bf(x)\,dx \;\; \underset{\mathrm{Def.}}{=} \;\;
\lim_{n\to \infty, \; ||\Delta x|| \to 0} \sum_{i=1}^{n} f(x_i^*)\Delta x_i\]
exists and is independent of the partition and of the sample points.

Thus, in particular, if \(x_i^*=x_i=a+i\Delta x\) with \(\displaystyle \Delta x = \frac{b-a}{n}\), then we can compute the definite integral of a continuous function using right-endpoint Riemann sums:
\begin{equation}\label{ch03-04-eq01}
\int_a^bf(x)\,dx = \lim_{n\to \infty} \sum_{i=1}^{n} f(x_i)\Delta x\tag{3.3}.
\end{equation}
Thus, from Example 1 in 3.3.3 we have
\[\int_{-1}^2 (x^3-x^2+2)\, dx =
\lim_{n\to \infty}\sum _{i=1}^n \left(x_i^3-x_i^2+2\right)\Delta x
= \frac{27}{4}.\]

Example 1. Evaluate \(\displaystyle \int_1^3 x(x-2)\,dx\).

Solution. We can use right-endpoint approximations (see (3.3)).

 

Step 1. Partition of the interval. \(\displaystyle \Delta x= \frac{2}{n}\), \(\displaystyle x_i=1+\frac{2i}{n}\), \(i = 0, 1, \dots, n\).

 

Step 2. \(f(x_i)\Delta x\).
\[f(x_i)= x_i^2-2x_i = \left(1+\frac{2i}{n}\right )^2-2\left (1+\frac{2i}{n}\right )
=\frac{4i^2}{n^2}-1, \;\; \mathrm{and} \;\; f(x_i)\Delta x
= \left ( \frac{4i^2}{n^2}-1 \right ) \frac{2}{n}.\]

Step 3. Riemann sum.
\begin{align*}
\sum_{i=1}^n f(x_i)\Delta x &= \sum_{i=1}^n \left ( \frac{4i^2}{n^2}-1 \right ) \frac{2}{n}
\underset{?}{=}
\frac{2}{n} \left ( \frac{4}{n^2}\sum_{i=1}^ni^2 -\sum_{i=1}^n1\right )=\\
&=\frac{2}{n} \left ( \frac{4}{n^2}\frac{n(n+1)(2n+1)}{6} -n\right )
= 2 \left ( \frac{2}{3}\frac{(n+1)(2n+1)}{n^2} -1 \right ).
\end{align*}

 

Step 4. Limit of the Riemann sums as \(n\to \infty\).
\begin{align*}
\lim_{n\to \infty}\sum_{i=1}^n f(x_i)\Delta x
&= \lim_{n\to \infty} 2 \left ( \frac{2}{3}\frac{(n+1)(2n+1)}{n^2} -1 \right )=\\
& =\lim_{n\to \infty} 2 \left ( \frac{2}{3}\left (1+\frac{1}{n}\right ) \left (2+\frac{1}{n}\right ) -1 \right )
\underset{?}{=} 2\left (\frac{2}{3}(1)(2)-1\right )=\frac{2}{3}.
\end{align*}

Remark. (1) The graph of \(f(x)= x(x-2)\), \(x \in [1,3]\) is a parabola as shown in Figure 3.9(a). And we have,
\[
f(x_i)\leq 0, \;\; \mathrm{for}\;\; x_i\in[1,2], \;\;\; f(x_i)\geq 0, \;\; \mathrm{for}\;\; x_i\in[2,3],\]
thus, the contribution to the Riemann sum of the terms \( f(x_i)\Delta x \;\; \mathrm{for}\;\; x_i\in[1,2] \) is negative. It seems natural, in this case, to assign a “negative area” to the region below the \(x\)-axis, while to the region above the \(x\)-axis we continue to assign its (positive) area (see Figure 3.9(b)).

(2) In fact, we have the following:
\[
\int_1^2 x(x-2)\,dx =-\frac{2}{3}, \;\;\;
\int_2^3 x(x-2)\,dx =\frac{4}{3}, \;\;\;
\Longrightarrow
\int_1^3 x(x-2)\,dx =-\frac{2}{3}+\frac{4}{3}=\frac{2}{3}.\]
See Exercise # 1 in 3.4.3.

Example 2. Given \(f(x) =|2x-4|-2\), \(x\in [1,5]\), (a) sketch its graph; (b) evaluate \(\displaystyle \int_1^5 f(x)\,dx\) by interpreting it as a sum of positive and negative areas, and using elementary geometry to evaluate the area of each region.

Solution. First of all, we rewrite \(f\) in terms of a piecewise function.
\[|2x-4|-2
=\left\{\renewcommand{\arraystretch}{1.2}
\begin{array}{ll}
(2x-4)-2 &\text{if }\; 2x-4\geq 0\\
-(2x-4)-2 &\text{if }\; 2x-4 < 0 \end{array} \right. =\left\{\renewcommand{\arraystretch}{1.2} \begin{array}{ll} 2x-6&\text{if }\; x\geq 2\\ -2x+2 &\text{if }\; x < 2 \end{array}\right.. \] Thus, the graph of \(f\) consists of two lines as in Figure 3.9(c). It follows that in order to compute \(\displaystyle \int_1^5 f(x)\,dx\), we only need to compute the area of the two triangles shown in the figure. Assign a negative area to the triangle below the \(x\)-axis as in Figure 3.9(d), and then add their areas.

• The height of the triangle below the \(x\)-axis is \(f(2)=|2(2)-4|-2=-2\). Thus, its area is \(\displaystyle \frac{1}{2}(2)(-2)=-2\).
• The second triangle has height \(f(5)=|2(5)-4|-2= 4\). Thus, its area is \(\displaystyle \frac{1}{2}(2)(4)=4\).

Hence,
\[\int_1^5 f(x)\,dx = -2+4 = 2.\]

3.4.2 Basic Properties of Definite Integrals

 

Limits of Integration

 

Let \(f\) be an integrable function on the closed interval \([a,b]\). Then
\begin{equation}\label{ch03-04-eq02}
\int_a^bf(x)\,dx = -\int_b^af(x)\,dx, \;\; \text{and} \;\; \int_a^af(x)\,dx=0\tag{3.4}.
\end{equation}

Proof.
\[\int_a^bf(x)\,dx = \lim_{n\to \infty} \sum_{i=1}^{n} f(x_i)\Delta x \;\; \text{with} \;\; \Delta x = \frac{b-a}{n},\]
it follows that
\[\int_b^af(x)\,dx = \lim_{n\to \infty} \sum_{i=1}^{n} f(x_i)\Delta x \;\; \text{with} \;\; \Delta x = \frac{a-b}{n}.\]
Since \(a-b=-(b-a)\), the result follows.

 

For the second property, note that in this case \(\displaystyle \Delta x = \frac{a-a}{n}=0\). Thus, right-endpoint Riemann sums are zero.

□

 

Algebraic Operations Involving Definite Integrals

 

An immediate consequence of Theorem 1 in 2.4.1 is the following

Theorem 2. Let \(f\) and \(g\) be integrable functions on the closed interval \([a,b]\). Then

1. 1. \(f+g\) is integrable, and \(\displaystyle \int_a^b(f+g)(x)\,dx = \int_a^bf(x)\,dx +\int_a^bg(x)\,dx\).

 

1. 1. \(f-g\) is integrable, and \(\displaystyle \int_a^b(f-g)(x)\,dx = \int_a^bf(x)\,dx -\int_a^bg(x)\,dx\).

 

1. \(kf\), \(k\in \mathbb{R}\), is integrable, and \(\displaystyle \int_a^bkf(x)\,dx =k \int_a^bf(x)\,dx \).

Proof. 1. Using the algebraic operations involving limits (Theorem 1 in 2.4.1}), and (3.3), we have
\begin{align*}
\int_a^b(f+g)(x)\,dx
&= \lim_{n\to \infty} \sum_{i=1}^{n} (f(x_i) +g(x_i))\Delta x
\underset{(*)}{=} \lim_{n\to \infty} \left(\sum_{i=1}^{n} f(x_i)\Delta x +\sum_{i=1}^{n}g(x_i))\Delta x\right)=\\
&=
\left(\lim_{n\to \infty}\sum_{i=1}^{n} f(x_i)\Delta x\right) +
\left(\lim_{n\to \infty}\sum_{i=1}^{n}g(x_i))\Delta x\right)
=\int_a^bf(x)\,dx +\int_a^bg(x)\,dx
\end{align*}
\((*)\) See Proposition 1 in 3.2.2.

Similar arguments prove the other two parts.

□

Corollary 1. (Linearity of the Definite Integral). Let \(c_1\) and \(c_2\) be arbitrary constants, and \(f_1\) and \( f_2\) be integrable functions on \([a,b]\), then
\begin{equation}\label{ch03-04-eq03}
\int_a^b(c_1f_2+c_2f_2)(x)\,dx= c_1\int_a^bf_1(x)\,dx. + c_2\int_a^bf_2(x)\,dx\tag{3.5}
\end{equation}

 

Naturally, the above rules can be extended to a finite sum involving more than two terms, that is, let\(c_1,\dots,c_k\) be arbitrary constants, and \(f_1,\dots, f_k\) be integrable functions on \([a,b]\), then
\begin{equation}\label{ch03-04-eq04}
\int_a^b\sum_{i=1}^{k}c_kf_k(x)\,dx= \sum_{i=1}^{k}c_k\int_a^bf_k(x)\,dx\tag{3.6}.
\end{equation}

 

Additive Property with Respect to the Interval of Integration

 

Assume that \(f\) is integrable on \([a,b]\), and that \(f(x)\geq 0\) for all \(x\in [a.b]\). Let \(a < c < b\). Then, when interpreting \(\displaystyle \int_a^bf(x)\,dx\) as the area enclosed by the graph of \(f\) and the \(x\)-axis, the following is intuitively true (see Figure 3.10 (a)):

\begin{equation}\label{ch03-04-eq05}
\int_a^cf(x)\,dx + \int_c^bf(x)\,dx =\int_a^bf(x)\,dx\tag{3.7} .
\end{equation}

                                                                                (b)[/caption]

One way to see that this is valid using Riemann sums is as follows: consider a partition of the interval \([a.b]\) so that one of the partition points is precisely \(c\), say \(c=x_p\). Then
\[
\sum_{i=1}^nf(x_i)\Delta x_i= \sum_{i=1}^pf(x_i)\Delta x_i + \sum_{i=p+1}^nf(x_i)\Delta x_i.\]
It is clear that these sums are a Riemann sums for \(\displaystyle \int_a^cf(x)\,dx \) and \(\displaystyle \int_c^bf(x)\,dx \), respectively. Hence the result.

 

Basic Integration Formulas

 

Let \(c\in \mathbb{R}\), and \(a < b\). Then
\[
\int_a^bc\,dx = c(b-a),\; \;
\int_a^bx\,dx = \frac{1}{2}(b^2 – a^2), \;\;
\int_a^bx^2\,dx = \frac{1}{3}(b^3 – a^3), \;\;
\int_a^bx^3\,dx = \frac{1}{4}\left (b^4 – a^4\right ).
\]
Proof. Here is an outline of the proof for the last formula. The proofs of these formulas are left as an exercise (see # 8 in 3.4.3).

Using a right-endpoint approximation, with \(a=0 < b\), \(\displaystyle \Delta x = \frac{b-0}{n}=\frac{b}{n}\), and \(\displaystyle x_i=\frac{bi}{n}\) we have
\begin{align*}
\int_0^bx^3\,dx &=
\lim_{n\to \infty}\sum_{i=1}^{n} f\left (x_i\right )\Delta x =
\lim_{n\to \infty}\sum_{i=1}^{n} \left (\frac{bi}{n}\right )^3\frac{b}{n} =
\lim_{n\to \infty}\left (\frac{b}{n}\right )^4\sum_{i=1}^{n} i^3 =\\
&=\lim_{n\to \infty}\left (\frac{b}{n}\right )^4\left (\frac{n(n+1)}{2}\right )^2
=\lim_{n\to \infty} \frac{b^4}{4} \frac{n^2(n+1)^2}{n^4} \underset{?}{=}\frac{b^4}{4}.
\end{align*}
This proves that the formula is valid for \(a=0\).

Assume that \(a < 0 < b\), then
\[\int_a^bx^3\,dx \underset{?}{=} \int_a^0x^3\,dx+ \int_0^bx^3\,dx = -\int_0^ax^3\,dx +\int_0^bx^3\,dx = -\frac{a^4}{4}+\frac{b^4}{4} = \frac{1}{4}(b^4-a^4).\]
Assume that \(0 < a < b\), then
\[\int_0^bx^3\,dx \underset{?}{=} \int_0^ax^3\,dx+ \int_a^bx^3\,dx =\frac{a^4}{4} +\int_a^bx^3\,dx \Longrightarrow \int_a^bx^3\,dx = \frac{b^4}{4} -\frac{a^4}{4}= \frac{1}{4}(b^4-a^4).\]
The case \(b < 0\) can be proven in a similar fashion.

□

Remark. It is possible to prove the following general formula:
\begin{equation}\label{ch03-04-eq06}
\int_a^bx^n\,dx = \frac{1}{n+1}\left (b^{n+1} – a^{n+1}\right ),\;\;\;a < b, n\in \mathbb{N}\tag{3.8}.
\end{equation}
However, we postpone its proof until we have a more efficient method to compute definite integrals, that is, The Fundamental Theorem of Calculus..

 

Remark. (3.6) and (3.8) can be used to obtain the definite integral of any polynomial function.
\begin{equation}\label{ch03-04-eq07}
\int_a^b\sum_{i=0}^{n}c_ix^i\,dx =\sum_{i=0}^{n} \frac{c_i}{i+1}\left (b^{i+1} – a^{i+1}\right )\tag{3.9}.
\end{equation}

3.4.4 Net Accumulate Area

Definition 3. Let \(f\) be a continuous function defined on the interval \([a,b]\), and \(c \in [a,b]\). The net accumulated area starting at \(\pmb{c}\) is the function
\[F(x)= \int_c^xf(t)\,dt, \;\;\; x\in [a,b].\]

Observation. (1) Note that 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.

(2) Let \(a \leq c_1 < c_2\leq b\), and \(F_1\) and \(F_2\) be the accumulated areas, respectively, starting at \(c_1\) and \(c_2\). Then, \(F_1\) differs from \(F_2\) by the constant \(\displaystyle \int_{c_1}^{c_2}f(t)\,dt\). This follows from
\[F_1(x)= \int_{c_1}^xf(t)\,dt \underset{(\ref{ch03-04-eq05})}{=} \int_{c_1}^{c_2}f(t)\,dt + \int_{c_2}^xf(t)\,dt = \int_{c_1}^{c_2}f(t)\,dt + F_2(x), \;\;\; x\in [a,b].\]

Example 1. Let \(\displaystyle f(x)=|2x+2| – 2x – 5\), \(x\in[-4,3]\), use elementary geometry to find an expression for the net accumulated area starting at \(-4\).

Solution. First, we rewrite \(f\) as a piecewise function:
\[f(x) = \left \{
\renewcommand{\arraystretch}{1.5}
\begin{array}{ll}
2x+2 – 2x – 5 & x \geq -1\\
-(2x+2) – 2x – 5 & x < -1
\end{array}
\right .
= \left \{
\renewcommand{\arraystretch}{1.5}
\begin{array}{ll}
-3 & x \geq -1\\
– 4x – 7 & x < -1
\end{array}
\right ..\]
It follows that the \(x\)-intercept is \(\displaystyle \left (-\frac{7}{4}, 0\right )\), and the graph of \(f\) is as in Figure 3.11 (a).

To compute the net accumulate area, we consider three cases:

 

(1) \(\displaystyle -4 \leq x \leq -\frac{7}{4}\). In this case, the accumulated area is the area of a trapezoid, as shown in Figure 3.11 (b). Thus, its area is given by
\[F(x) = \frac{1}{2}\left (f(-4)+f(x)\right )(x+4) = \frac{1}{2}(9 +(-4x-7))(x+4) = (1-2x)(x+4) = -2 x^2-7 x+4.\]

 

(2) \(\displaystyle -\frac{7}{4} < x \leq -1\). In this case, the accumulated are is the area of the triangle above the interval \(\displaystyle \left [-4,-\frac{7}{4}\right ]\), minus the area of the triangle below the interval \(\displaystyle \left [-\frac{7}{4}, x\right ]\) (draw the accumulated area). Namely
\[F(x) = F\left (-\frac{7}{4}\right ) + \frac{1}{2}f(x) \left (\frac{7}{4}+x\right ) \underset{?}{=}\frac{81}{8}- \frac{1}{2}(4x+7) \left (\frac{7}{4}+x\right ) \underset{?}{=} -2x^2-7 x+4.\]
Note that in this case \(f(x)= -4x-7<0\).

 

(3) \(\displaystyle -1 < x \leq3\). In this case, the accumulated are is the area of the triangle above the interval \(\displaystyle \left [-4,-\frac{7}{4}\right ]\), minus the area of the triangle below the interval \(\displaystyle \left [-\frac{7}{4}, -1\right ]\), minus the area of the rectangle below the interval \([-1,x]\) (draw the accumulated area). Namely
\[F(x) = F\left (-1\right ) -3(x+1) \underset{?}{=}9-3(x+1) = -3x+6.\]

In conclusion, we have that the accumulated area is given by
\[F(x) = \left \{\renewcommand{\arraystretch}{1.25}
\begin{array}{ll}
-2x^2-7 x+4& -4\leq x\leq -1\\
-3x+6 & -1 < x < 3
\end{array}
\right ..\]

3.4.6 Definite Integrals of Roots of any Order

 

Here we evaluate the definite integral of a radical. In this context, the partition of the interval will involve intervals of \textit{different lengths}. In the example the computation is done for \(f(x)=\sqrt[3]{x}\). Naturally, the calculations can be extended for other roots. However, as you will see, this is not an efficient way to evaluate such integrals. There is, indeed, a systematic way to compute definite integrals. That is precisely the content of \textit{The Fundamental Theorem of Calculus}, as we shall learn later on.

Example 1. Show that \(\displaystyle \int_0^b\sqrt[3]{x}\,dx =\frac{3}{4}b^{4/3}\), \(b>0\).

Solution.

Step 1. Partition of the interval. Set
\[x_i=b\left (\frac{i}{n}\right )^3,\;\; i = 0,\, 1, \,\dots,\, n \;\;\Longrightarrow \;\;
\Delta x_i=x_i-x_{i-1} = b\left (\frac{i}{n}\right )^3-b\left (\frac{i-1}{n}\right )^3
\underset{?}{=}
\frac{3 b i^2}{n^3}-\frac{3 b i}{n^3}+\frac{b}{n^3}
.\]

 

Remark. Note that since \(0\leq i\leq n\),
\[\Delta x_i = \frac{3 b i^2}{n^3}-\frac{3 b i}{n^3}+\frac{b}{n^3}
\leq \frac{3 bn^2}{n^3}+\frac{3 b n}{n^3}+\frac{b}{n^3}
=\frac{3 b}{n}+\frac{3 b }{n^2}+\frac{b}{n^3}
\underset{n\to \infty\vphantom{\frac{1}{1^2}}}{\longrightarrow}0.\]
It follows that when computing
\[
\int_0^b\sqrt[3]{x}\,dx =\lim_{n\to \infty} \sum_{i=1}^n f(x_i)\Delta x_i,\]
the number of intervals increases to infinity, and, at the same time, the lengths of the subintervals decrease to zero.

 

Step 2. \(f(x_i)\Delta x_i\).
\[f(x_i)= \sqrt[3]{ x_i} = \sqrt[3]{ b\left (\frac{i}{n}\right )^3}
=\sqrt[3]{b}\frac{i}{n}, \]
\[f(x_i)\Delta x_i
=\sqrt[3]{b}\frac{i}{n}\left (\frac{3 b i^2}{n^3}-\frac{3 b i}{n^3}+\frac{b}{n^3}\right )
=\frac{b\sqrt[3]{b}}{n^4}\left ( 3 i^3 -3 i^2+ i\right )
.\]

 

Step 3. Riemann sum.
\begin{align*}
\sum_{i=1}^n f(x_i)\Delta x_i
&= \sum_{i=1}^n \frac{b\sqrt[3]{b}}{n^4}\left ( 3 i^3 -3 i^2+ i\right )
\underset{?}{=}
\frac{b\sqrt[3]{b}}{n^4}\left (3\sum_{i=1}^n i^3 -3\sum_{i=1}^n i^2 +
\sum_{i=1}^n i \right ) = \\
&=\frac{b\sqrt[3]{b}}{n^4}\left (3 \left (\frac{n(n+1)}{2}\right )^2
-3\frac{n(n+1)(2n+1)}{6} +
\frac{n(n+1)}{2}\right)=\\
&= b\sqrt[3]{b} \left (\frac{3 }{4}\frac{(n+1)^2}{n^2}
-\frac{1}{2}\frac{(n+1)(2n+1)}{n^3} +
\frac{1}{2}\frac{(n+1)}{n^3}\right).
\end{align*}

 

Step 4. Limit of the Riemann sums as \(n\to \infty\).
\begin{align*}
\lim_{n\to \infty}\sum_{i=1}^n f(x_i)\Delta x_i
&=\lim_{n\to \infty} b\sqrt[3]{b} \left (\frac{3 }{4}\frac{(n+1)^2}{n^2}
-\frac{1}{2}\frac{(n+1)(2n+1)}{n^3} +
\frac{1}{2}\frac{(n+1)}{n^3}\right)\\
&\underset{?}{=} \lim_{n\to \infty} b\sqrt[3]{b}\left (\frac{3}{4}\left (1+\frac{1}{n}\right )^2
-\frac{1}{2}\frac{1}{n}\left (1+\frac{1}{n}\right )\left (2+\frac{1}{n}\right ) +
\frac{1}{2}\frac{1}{n^2}\left (1+\frac{1}{n}\right )\right)\\
&\underset{?}{=} b\sqrt[3]{b}\left (\frac{3}{4}\right )=\frac{3}{4}b^{4/3}.
\end{align*}

It follows immediately from (3.6) that (explain why this is so)
\[ \int_a^b\sqrt[3]{x}\,dx =\frac{3}{4}\left (b^{4/3}-a^{4/3}\right ),\;\; 0\leq a < b.\]