Definition 1.6.1. Integral function.
Let \(f\) be integrable on the interval \([a,b]\text{.}\) The function \(F\) defined on \(I\) as \(F(x)=\int_a^xf(t)\, dt\) is called an integral function.
Section 1.6 Fundamental theorem (cont.)
In this section we focus on FTC I. One peculiarity of this theorem is the function \(F(x)\) defined as
\begin{equation}
F(x)=\int_a^x f(t)\, dt\text{.}\tag{1.6.1}
\end{equation}
This is called an integral function, as the values \(F(x)\) are defined by the integral expression \(\int_a^x f(t)\, dt\text{.}\)
Examining the formula (1.6.1), we note that the input of \(F\) is an the upper limit of integration of the integral. As we vary \(x\text{,}\) the only thing that changes about the integral is the interval over which we are integrating; and this causes the values \(F(x)\) to vary with \(x\text{.}\) Using the signed area interpretation of the integral allows us to visualize the integral function (1.6.1) . In keeping with the notation in (1.6.1), we will label the horizontal axis as the \(t\)-axis. For any \(x\geq a\text{,}\) define \(\mathcal{R}_x\) to be the region in the plane that lies between the graph of \(f\) and the \(t\)-axis, and between the vertical lines \(t=a\) and \(t=x\text{;}\) as we increase \(x\text{,}\) imagine sliding the vertical line \(t=x\) to the right, creating a growing region \(\mathcal{R}_x\text{;}\) the integral function \(F(x)=\int_a^x f(t)\, dt\) computes the signed area of this growing region!
Interactive example 1.6.1. Integral function.
Example 1.6.3. Integral functions.
Let \(F(x)=\displaystyle\int_1^x \frac{1}{t^2}\, dt\text{.}\) Make a table of values of \(F(x)\) for \(x=1, 2, 3, 4, 5\text{.}\) Explain graphically what \(F(b)\) is for any \(b\geq 1\text{.}\)
Solution.
For any \(b\geq 1\) the value \(F(b)=\int_1^b f(t)\, dt\) is equal to the signed area of the region between the graph of \(f\) and the horizontal axis from \(x=1\) to \(x=b\text{.}\) In this case, since \(f(x)=1/x^2\text{,}\) the signed area is equal to the area.
\begin{equation*}
F(b)=\int_1^b \frac{1}{t^2}\, dt=-\frac{1}{b}-(-1)=\frac{b-1}{b}\text{.}
\end{equation*}
We thus have the following table of values
\begin{equation*}
\begin{array}{c|c|c|c|c|c}
x\amp 1\amp 2\amp 3\amp 4\amp 5 \\
\hline
F(x) \amp 0\amp 1/2 \amp 2/3\amp 3/4 \amp 4/5
\end{array}\text{.}
\end{equation*}
Note that \(F(b)\to\infty\) as \(b\to\infty\text{.}\) This suggests that the region
\begin{equation*}
\mathcal{R}=\{(x,y)\colon x\geq 1, 0\leq y\leq f(x)\}
\end{equation*}
of all points lying between the graph of \(f\) and the \(x\)-axis and to the right of the line \(x=1\) has area 1. We’ll return to this when we discuss improper integrals.
Now that we have a better grip on integral functions \(F\) of the form (1.6.1)), let’s look at what FTC I says about them: namely, that they are differentiable, and satisfy
\begin{equation*}
F'(x)=f(x)
\end{equation*}
for all \(x\in [a,b]\text{.}\) The corollary below follows immediately, answering a question we posed when discussing antiderivatives.
Corollary 1.6.4. Continuous functions have antiderivatives.
If \(f\) is continuous on the interval \(I=[a,b]\text{,}\) then \(f\) has an antiderivative on \(I\text{.}\)
Proof.
Let \(F\) be the integral function defined as \(F(x)=\int_a^x f(t)\, dt\text{.}\) According to FTC I, we have \(F'(x)=f(x)\) for all \(x\in I\text{.}\)
Let’s incorporate the result of FTC I into the geometric interpretation of \(F(x)=\int_a^x f(t)\, dt\) discussed above. Since \(F(x)\) computes the signed area of the variable region \(\mathcal{R}_x\text{,}\) its derivative \(F'(x)\) computes the rate of change of this signed area with respect to \(x\text{.}\) By FTC I, we have \(F'(x)=f(x)\text{.}\) Thus the rate of change of the signed area of \(\mathcal{R}_x\) at a give point \(x\) is just \(f(x)\text{,}\) the height of the graph of \(f\) at \(x\text{.}\)
Theorem 1.6.5. FTC I: geometric.
Let \(f\) be continuous on the interval \(I=[a,b]\) and define \(F(x)=\int_a^x f(t)\, dt\text{.}\) In what follows we label the horizontal axis in the plane as the \(t\)-axis.
- For all \(x\in I\text{,}\) let \(\mathcal{R}_x\) be the region lying between the graph of \(f\) and the \(t\)-axis, and between the vertical lines \(t=a\) and \(t=x\text{.}\) We have\begin{equation} F(x)=\operatorname{sgn\, area} \mathcal{R}_x\tag{1.6.2} \end{equation}for all \(x\in I\text{.}\)
- At any \(x\in I\) the rate of change of \(\operatorname{sgn \, area} \mathcal{R}_x\) is given by \(F'(x)=f(x)\text{.}\) In other words, the rate of change of the signed area is equal to the (directed) height of the graph of \(f\) at \(x\text{.}\)
We can make more interesting integral functions by choosing the upper and/or lower limit of integration to be a function of \(x\text{.}\) When all the functions are “nice enough”, we can combine FTC I with the chain rule and integral properties to compute the integral function’s derivative.
Example 1.6.6. FTC I+chain rule.
For each \(F(x)\) defined below, use FTC I along with integral properties and/or the chain rule to compute \(\displaystyle F'(x)=\frac{d}{dx}F(x)\text{.}\)
- \(\displaystyle \displaystyle F(x)=\int_{x}^5 \sqrt{t+1}\, dt\)
- \(\displaystyle \displaystyle F(x)=\int_{-2}^{\sin x}\cos(u^2) \, du\)
- \(\displaystyle \displaystyle F(x)=\int_{4x}^{\sqrt{x^2+1}}\sin(s^2)\, ds\)
Solution.
The idea for all of these examples is to express the given “complicated” integral function in terms of “simpler” integral function \(G(x)\) to which we can directly apply FTC I.
- Letting \(\displaystyle G(x)=\int_5^x \sqrt{t+1}\, dt\text{,}\) we have\begin{align*} F(x) \amp=\int_{x}^5 \sqrt{t+1}\, dt \\ \amp = -\int_5^x\sqrt{t+1}\, dt\\ \amp = -G(x) \end{align*}and hence\begin{align*} F'(x) \amp= -G'(x) \\ \amp= -\sqrt{x+1} \amp (\knowl{./knowl/xref/th_FTC_I.html}{\text{FTC I}}: G'(x)=\sqrt{x+1})\text{.} \end{align*}
- Letting \(\displaystyle G(x)=\int_{-2}^x \cos(u^2)\, du\text{,}\) we have\begin{align*} F(x) \amp = \int_{-2}^{\sin x} \cos(u^2)\, du\\ \amp = G(\sin x) \end{align*}and thus\begin{align*} F'(x) \amp =\frac{d}{dx} G(\sin x)\\ \amp = G'(\sin x)\cdot (\sin x)' \amp \text{(chain rule)}\\ \amp = \cos(\sin^2 x)\cdot \cos x \amp (\knowl{./knowl/xref/th_FTC_I.html}{\text{FTC I}}: G'(x)=\cos(x^2)) \end{align*}
- Letting \(\displaystyle G(x)=\int_0^x \sin(s^2)\, ds\text{,}\) we have\begin{align*} F(x) \amp = \int_{4x}^{\sqrt{x^2+1}}\sin(s^2)\, ds\\ \amp = \int_{4x}^{0}\sin(s^2)\, ds+\int_{0}^{\sqrt{x^2+1}}\sin(s^2)\, ds\\ \amp =-\int_{0}^{4x}\sin(s^2)\, ds+\int_{0}^{\sqrt{x^2+1}}\sin(s^2)\, ds\\ \amp =-G(4x)+G(\sqrt{x^2+1})\text{.} \end{align*}It follows that\begin{align*} F'(x) \amp = -\frac{d}{dx}G(4x)+\frac{d}{dx}G(\sqrt{x^2+1})\\ \amp = -G'(4x)\cdot 4+G'(\sqrt{x^2+1})(\sqrt{x^2+1})' \amp \text{(chain rule)}\\ \amp = -4G'(4x)+\frac{x\cdot G'(\sqrt{x^2+1})}{\sqrt{x^2+1}}\\ \amp =-4\sin(16x^2)+\frac{x\sin(x^2+1)}{\sqrt{x^2+1}} \amp (\knowl{./knowl/xref/th_FTC_I.html}{\text{FTC I}}: G'(x)=\sin(x^2)) \end{align*}
