Areas, Integrals, and Antiderivatives

Read this section to learn about the relationship among areas, integrals, and antiderivatives. Work through practice problems 1-5.

Using Antiderivatives to Evaluate

Now we can put the ideas of areas and antiderivatives together to get a way of evaluating definite integrals that is exact and often easy.

If \mathrm{A}(x)=\int_{\mathrm{a}}^{x} \mathrm{f}(t) \mathrm{dt}\mathrm{A}(\mathrm{a})=\int_{\mathrm{a}}^{\mathrm{a}} \mathrm{f}(t) \mathrm{dt}=0, \mathrm{~A}(\mathrm{~b})=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(t) \mathrm{dt} and \mathrm{A}(\mathrm{x}) is an antiderivative of f, \mathrm{A}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x})

We also know that if \mathrm{F}(\mathrm{x}) is any antiderivative of f, then \mathrm{F}(\mathrm{x}) and \mathrm{A}(\mathrm{x}) have the same derivative so \mathrm{F}(\mathrm{x}) and \mathrm{A}(\mathrm{x}) are "parallel" and differ by a constant, F(x)=A(x)+C for all x.

Then \mathrm{F}(\mathrm{b})-\mathrm{F}(\mathrm{a})=\{\mathrm{A}(\mathrm{b})+\mathrm{C}\}-\{\mathrm{A}(\mathrm{a})+\mathrm{C}\}=\mathrm{A}(\mathrm{b})-\mathrm{A}(\mathrm{a})=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(t) \mathrm{dt}-\int_{\mathrm{a}}^{\mathrm{a}} \mathrm{f}(t) \mathrm{dt}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(t) \mathrm{dt}.

To evaluate a definite integral \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(t) \mathrm{dt}, we can find any antiderivative F of f and evaluate \mathrm{F}(\mathrm{b})-\mathrm{F}(\mathrm{a}).

This result is a special case of Part 2 of the Fundamental Theorem of Calculus, and it will be used hundreds of times in the next several chapters. The Fundamental Theorem is stated and proved in Section 4.5.

Antiderivatives and Definite Integrals

If f is a continuous, nonnegative function and and F is any antiderivative of f \left(\mathrm{F}^{\prime}(x)=\mathrm{f}(x)\right)  on the interval [a,b],

then \left\{\begin{array}{l}
    \text { area bounded between the graph } \\
    \text { of } \mathrm{f} \text { and the } x \text {-axis and } \\
    \text { vertical lines at } x=\mathrm{a} \text { and } x=\mathrm{b}
    \end{array}\right\} = \int_{\mathrm{a}}^{\mathbf{b}} \mathrm{f}(x) \mathrm{dx}=\mathbf{F}(\mathrm{b})-\mathrm{F}(\mathrm{a})

The problem of finding the exact value of a definite integral reduces to finding some (any) antiderivative F of the integrand and then evaluating \mathrm{F}(\mathrm{b})=\mathrm{F}(\mathrm{a}). Even finding one antiderivative can be difficult, and, for now, we will stick to functions which have easy antiderivatives. Later we will explore some methods for finding antiderivatives of more difficult functions.

The evaluation \mathrm{F}(\mathrm{b})=\mathrm{F}(\mathrm{a}) is represented by the symbol \left.F(x)\right|_{a} ^{b}.

Example 3: Evaluate \int_{1}^{3} x \mathrm{dx} in two ways:

By sketching the graph of y = x and geometrically finding the area.

By finding an antiderivative of \mathrm{F}(\mathrm{x}) of f and evaluating \mathrm{F}(3)-\mathrm{F}(1).


(i) The graph of y = x is shown in Fig. 6, and the shaded region has area 4.

(ii) One antiderivative of x is \mathrm{F}(x)=\frac{1}{2} x^{2} (check that \mathrm{D}\left(x^{2} / 2\right)=x), and \left.\mathrm{F}(x)\right|_{1} ^{3}=\mathrm{F}(3)-\mathrm{F}(1)=\frac{1}{2}\left(3^{2}\right)-\frac{1}{2}\left(1^{2}\right)=\frac{9}{2}-\frac{1}{2}=4 which agrees with (i).

If someone chose another antiderivative of x, say F(x)=\frac{1}{2} x^{2}+7 ( check that \mathbf{D}\left(x^{2} / 2+7\right)=x), then \left.\mathrm{F}(x)\right|_{1} ^{3}=\mathrm{F}(3)-\mathrm{F}(1)=\left\{\frac{1}{2}\left(3^{2}\right)+7\right\}-\left\{\frac{1}{2}\left(1^{2}\right)+7\right\}=\frac{23}{2}-\frac{15}{2}=4. No matter which antiderivative F is chosen, \mathrm{F}(3)-\mathrm{F}(1) equals 4.

Practice 2: Evaluate \int_{1}^{3}(x-1) \mathrm{d} x in the two ways of the previous example.

This antiderivative method is an extremely powerful way to evaluate some definite integrals, and it is used often. However, it can only be used to evaluate a definite integral of a function defined by a formula.

Example 4: Find the area between the graph of the cosine and the horizontal axis for x between 0 and \pi / 2.

Solution: The area we want (Fig. 7) is \int_{0}^{\pi / 2} \cos (x) \mathrm{d} \mathrm{x} so we need an antiderivative of \mathrm{f}(x)=\cos (x).

\mathrm{F}(x)=\sin (x) is one antiderivative of \cos (x). (Check that \mathbf{D}(\sin (x)) =\cos (x)). Then \text { area }=\int_{0}^{\pi / 2} \cos (x) \mathrm{dx}=\left.\sin (x)\right|_{0} ^{\pi / 2}=\sin (\pi / 2)-\sin (0)=1-0=1.

Practice 3: Find the area between the graph of y=3 x^{2} and the horizontal axis for x between 1 and 2.