MA005: Calculus I

This section explores properties of functions defined as areas and examines some of the connections among areas, integrals and antiderivatives. In order to focus on the geometric meaning and connections, all of the functions in this section are nonnegative, but the results are generalized in the next section and proved true for all continuous functions. This section also introduces examples to illustrate how areas, integrals and antiderivatives can be used.

When $f$ is a continuous, nonnegative function, then the "area function"

$\mathrm{A}(x)=\int_{\mathrm{a}}^{x} \mathrm{f}(t) \mathrm{d} \mathrm{t}$ represents the area between the graph of $f$, the t–axis, and between the vertical lines at $t=a$ and $t=x$ (Fig. 1), and the derivative of $\mathrm{A}(x)$ represents the rate of change (growth) of $\mathrm{A}(x)$ . Examples 2 and 3 of Section 4.3 showed that for some functions f, the derivative of $\mathrm{A}(x)$ was equal to $f$ so $\mathrm{A}(x)$ was an antiderivative of $f$. The next theorem says the result is true for every continuous, nonnegative function $f$.

The Area Function is an Antiderivative

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

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

This result relating integrals and antiderivatives is a special case (for nonnegative functions $f$) of the Fundamental Theorem of Calculus (Part 1) which is proved in Section 4.5 . This result is important for two reasons:

(i) it says that a large collection of functions have antiderivatives, and

(ii) it leads to an easy way of exactly evaluating definite integrals.

Example 1: $\mathrm{A}(x)=\int_{1}^{x} \mathrm{f}(t) \mathrm{d} \mathrm{t}$ for the function $\mathrm{f}(t)$shown in Fig. 2.

Estimate the values of $\mathrm{A}(x)$ and $\mathrm{A}^{\prime}(x)$ for $x=2,3,4$ and $5$ and use these values to sketch the graph of $\mathrm{y}=\mathrm{A}(x)$.

Solution: Dividing the region into squares and triangles, it is easy to see that $\mathrm{A}(2)=2, \mathrm{~A}(3)=4.5, \mathrm{~A}(4)=7$ and $\mathrm{A}(5)=8.5$. Since $A^{\prime}(x)=f(x)$, we know that $A^{\prime}(2)=f(2)=2, A^{\prime}(3)=f(3)=3, A^{\prime}(4)=f(4)=2$ and $A^{\prime}(5)=f(5)=1$. The graph of $y=A(x)$ is shown in Fig. 3.

It is important to recognize that f is not differentiable at $x = 2$ and $x = 3$. However, the values of $A$ change smoothly near 2 and 3, and the function $A$ is differentiable at those points and at every other point from 1 to 5. Also, $\mathrm{f}^{\prime}(4)=-1$ ($f$ is clearly decreasing near $x = 4$ ), but $A^{\prime}(4)=f(4)=2$ is positive (the area $A$ is growing even though f is getting smaller).

Practice 1: $\mathrm{B}(x)$ is the area bounded by the horizontal axis, vertical lines at $t = 0$ and $t = x$, and the graph of $\mathrm{f}(t)$ shown in Fig. 4. Estimate the values of $\mathrm{B}(x)$ and $\mathrm{B}^{\prime}(x)$ for $x=1,2,3,4$ and $5$.

Example 2: Let $\mathrm{G}(x)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\int_{0}^{x} \sin (t) \mathrm{dt}\right)$. Evaluate $\mathrm{G}(x)$ for $x=\pi / 4, \pi / 2, \text { and } 3 \pi / 4$.

Solution: Fig. 5(a) shows the graph of $\mathrm{A}(x)=\int_{0}^{x} \sin (t) \mathrm{dt}$, and $\mathrm{G}(x)$ is the derivative of $\mathrm{A}(\mathrm{x})$. By the theorem, $\mathrm{A}^{\prime}(x)=\sin (x)$ so $\mathrm{A}^{\prime}(\pi / 4)=\sin (\pi / 4) \approx .707, \mathrm{~A}^{\prime}(\pi / 2)=\sin (\pi / 2)=1$, and $\mathrm{A}^{\prime}(3 \pi / 4)=\sin (3 \pi / 4) \approx .707$. Fig. 5(b) shows the graph of $y=A(x)$ and 5(c) is the graph of $\mathrm{y}=\mathrm{A}^{\prime}(\mathrm{x})=\mathrm{G}(\mathrm{x})$.

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-5.5-Areas-Integrals-Antiderivatives.pdf
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