Evaluating Definite Integrals

Evaluating definite integrals this way can be quite tedious because of the complexity of the calculations. Later in this chapter we develop techniques for evaluating definite integrals without taking limits of Riemann sums. However, for now, we can rely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by using geometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we can then discuss what to do in the case of a curve of a function dropping below the x-axis.

Example 5.8

Using Geometric Formulas to Calculate Definite Integrals

Use the formula for the area of a circle to evaluate \(\int_{3}^{6} \sqrt{9 - \left(\right. x - 3 \left.\right)^{2}} d x .\)

Solution

The function describes a semicircle with radius 3. To find

\[\int_{3}^{6} \sqrt{9 - \left(\right. x - 3 \left.\right)^{2}} d x ,\]

we want to find the area under the curve over the interval \(\left[\right. 3 , 6 \left]\right. .\) The formula for the area of a circle is \(A = \pi r^{2} .\) The area of a semicircle is just one-half the area of a circle, or \(A = \left(\right. \frac{1}{2} \left.\right) \pi r^{2} .\) The shaded area in Figure 5.16 covers one-half of the semicircle, or \(A = \left(\right. \frac{1}{4} \left.\right) \pi r^{2} .\) Thus,

\[\begin{aligned} \\ \\ \int_{3}^{6} \sqrt{9 - \left(\right. x - 3 \left.\right)^{2}} & = \frac{1}{4} \pi \left(\left(\right. 3 \left.\right)\right)^{2} \\ & = \frac{9}{4} \pi \\ & \approx 7.069. \end{aligned}\]

Figure 5.16 The value of the integral of the function \(f \left(\right. x \left.\right)\) over the interval \(\left[\right. 3 , 6 \left]\right.\) is the area of the shaded region.