The Fundamental Theorem of Calculus

Read this section to see the connection between derivatives and integrals. Work through practice problems 1-5.

Part 1: Antiderivatives

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).