§ 6.4 Areas in the xy-Plane

Section Outline Properties of Definite Integrals Area Between Two Curves Finding the Area Between Two Curves

Properties of Definite Integrals Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #44

Area Between Two Curves Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #45

Finding the Area Between Two Curves EXAMPLE Find the area of the region between y = x2 – 3x and the x-axis (y = 0) from x = 0 to x = 4. SOLUTION Upon sketching the graphs we can see that the two graphs cross; and by setting x2 – 3x = 0, we find that they cross when x = 0 and when x = 3. Thus one graph does not always lie above the other from x = 0 to x = 4, so that we cannot directly apply our rule for finding the area between two curves. However, the difficulty is easily surmounted if we break the region into two parts, namely the area from x = 0 to x = 3 and the area from x = 3 to x = 4. For from x = 0 to x = 3, y = 0 is on top; and from x = 3 to x = 4, y = x2 – 3x is on top. Consequently, Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #46

Finding the Area Between Two Curves CONTINUED Thus the total area is 4.5 + 1.833 = 6.333. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #47

Finding the Area Between Two Curves CONTINUED y = x2 – 3x y = 0 Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #48

Finding the Area Between Two Curves EXAMPLE Write down a definite integral or sum of definite integrals that gives the area of the shaded portion of the figure. SOLUTION Since the two shaded regions are (1) disjoint and (2) have different functions on top, we will need a separate integral for each. Therefore Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #49

Finding the Area Between Two Curves CONTINUED Therefore, to represent all the shaded regions, we have Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #50

Finding the Area Between Two Curves EXAMPLE Two rockets are fired simultaneously straight up into the air. Their velocities (in meters per second) are v1(t) and v2(t), respectively, and v1(t) ≥ v2(t) for t ≥ 0. Let A denote the area of the region between the graphs of y = v1(t) and y = v2(t) for 0 ≤ t ≤ 10. What physical interpretation may be given to the value of A? SOLUTION Since v1(t) ≥ v2(t) for t ≥ 0, this suggests that the first rocket is always traveling at least as fast as the second rocket. Therefore, we have Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #51

Finding the Area Between Two Curves CONTINUED But again, since v1(t) ≥ v2(t) for t ≥ 0, we know that . So, this implies that . This means that the position of the first rocket is always at least as high (up in the air) as that of the second rocket. That is, the first rocket is always higher up than the second rocket (or at the same height). Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #52