MA005: Calculus I

Every continuous function has an antiderivative, even those nondifferentiable functions with "corners" such as absolute value.

The Fundamental Theorem of Calculus (Part 1)

If $f$ is continuous and $\mathrm{A}(x)=\int_{\mathrm{a}}^{x} \mathrm{f}(t) \mathrm{dt}$

then $\frac{\mathrm{d}}{\mathrm{dx}}\left(\int_{\mathrm{a}}^{x} \mathrm{f}(t) \mathrm{dt}\right)=\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{A}(x)=\mathrm{f}(x)$. $\mathrm{A}(x)$ is an antiderivative of $\mathrm{f}(x)$.

Proof: Assume $f$ is a continuous function and let $\mathrm{A}(x)=\int_{\mathrm{a}}^{x} \mathrm{f}(t) \mathrm{dt}$. By the definition of derivative of $A$,

$\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{A}(x)=\lim _{h \rightarrow 0} \frac{A(x+h)-A(x)}{h}=\lim _{h \rightarrow 0} \frac{1}{h}\left\{\int_{a}^{x+h} \mathrm{f}(\mathrm{t}) \mathrm{dt}-\int_{a}^{x} \mathrm{f}(\mathrm{t}) \mathrm{dt}\right\}=\lim _{h \rightarrow 0} \frac{1}{h} \int_{x}^{x+h} \mathrm{f}(\mathrm{t}) \mathrm{dt}$

By Property 6 of definite integrals (Section 4.3), for $h > 0$

$\{\min \text { of } \mathrm{f} \text { on }[x, x+\mathrm{h}]\} \cdot \mathrm{h} \leq \int_{x}^{x+\mathrm{h}} \mathrm{f}(t) \mathrm{dt} \leq\{\max \text { of } \mathrm{f} \text { on }[x, x+\mathrm{h}]\} \cdot \mathrm{h} . \text { (Fig. 1) }$

Dividing each part of the inequality by $h$, we have that $\frac{1}{\mathrm{~h}} \int_{x}^{x+\mathrm{h}} \mathrm{f}(t) \mathrm{dt}$ is between the minimum and the maximum of $f$ on the interval $[x, x+\mathrm{h}]$. The function $f$ is continuous (by the hypothesis) and the interval $[x, x+\mathrm{h}]$ is shrinking (since h approaches 0), so $\lim _{h \rightarrow 0}\{\text { min of f on }[x, x+\mathrm{h}]\}=\mathrm{f}(x)$ and $\lim _{h \rightarrow 0}\{\max \text { of fon }[x, x+\mathrm{h}]\}=\mathrm{f}(x)$. Therefore, $\frac{1}{\mathrm{~h}}^{x+\mathrm{h}} \int_{\mathrm{f}}^{\mathrm{f}}(t) \mathrm{dt}$ is stuck between two quantities (Fig. 2) which both approach $\mathrm{f}(x)$.

Then $\frac{1}{\mathrm{~h}} \int_{x}^{x+\mathrm{h}} \mathrm{f}(t) \mathrm{dt}$ must also approach $\mathrm{f}(x)$, and $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{A}(x)=\lim _{\mathrm{h} \rightarrow 0} \frac{1}{\mathrm{~h}} \int_{\mathrm{x}}^{\mathrm{x}+\mathrm{h}} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{f}(x)$.

Example 1: $\mathrm{A}(x)=\int_{0}^{x} \mathrm{f}(t) \mathrm{d} \mathrm{t}$ for $f$ in Fig. 3. Evaluate $\mathrm{A}(x)$ and $\mathrm{A}^{\prime}(x)$ for $x=1,2,3 \text { and } 4$.

Solution: $\begin{gathered}\mathrm{A}(1)=\int_{0}^{1} \mathrm{f}(t) \mathrm{dt}=1 / 2, \mathrm{~A}(2)=\int_{0}^{2} \mathrm{f}(t) \mathrm{dt}=1, \mathrm{~A}(3)=\int_{0}^{3} \mathrm{f}(t) \mathrm{dt}=1 / 2 \\\end{gathered}$

$\mathrm{A}(4)=\int_{0}^{4} \mathrm{f}(t) \mathrm{dt}=-1 / 2$. Since $f$ is continuous, $A^{\prime}(x)=f(x)$ so

$A^{\prime}(1)=f(1)=1, A^{\prime}(2)=f(2)=0, A^{\prime}(3)=f(3)=-1, A^{\prime}(4)=f(4)=-1$

Practice 1: $\mathrm{A}(x)=\int_{0}^{x} \mathrm{f}(t) \mathrm{dt}$ for $f$ in Fig. 4. Evaluate $\mathrm{A}(x)$ and $\mathrm{A}^{\prime}(x)$ for $x=1,2,3$ and $4$.

Example 2: $\mathrm{A}(x)=\int_{0}^{x} \mathrm{f}(t) \mathrm{dt}$ for the function $f$ shown in Fig. 5.

For which value of $x$ is $\mathrm{A}(x)$ maximum?

For which $x$ is the rate of change of $A$ maximum?

Solution: Since $A$ is differentiable, the only critical points are where $\mathrm{A}^{\prime}(x)=0$ or at endpoints. $\mathrm{A}^{\prime}(x)=\mathrm{f}(x)=0 \text { at } x=3$ and $A$ has a maximum at $x=3$. Notice that the values of $\mathrm{A}(x)$ as $x$ goes from 0 to 3 and then the $A$ values decrease. The rate of change of $\mathrm{A}(x)$ is $A^{\prime}(x)=f(x)$, and $f(x)$ appears to have a maximum at $x=2$ so the rate of change of $\mathrm{A}(x)$ is maximum when $x=2$. Near $x=2$, a slight increase in the value of $x$ yields the maximum increase in the value of $\mathrm{A}(x)$.