532
Chapter 5 | Integration
6
∫
9−( x −3) 2 dx ,
3
⎦ . The formula for the area of a circle is A = πr 2 .
we want to find the area under the curve over the interval ⎡ ⎣ 3, 6 ⎤ The area of a semicircle is just one-half the area of a circle, or A = ⎛ ⎝ 1 2 ⎞
⎠ πr 2 . The shaded area in Figure 5.16
covers one-half of the semicircle, or A = ⎛ ⎝ 1 4 ⎞
⎠ πr 2 . Thus,
3 6
∫
9−( x −3) 2 = 1 4
π (3) 2
= 9 4 π ≈7.069.
Figure 5.16 The value of the integral of the function f ( x ) over the interval ⎡ ⎣ 3, 6 ⎤ ⎦ is the area of the shaded region.
Use the formula for the area of a trapezoid to evaluate ∫ 2 4
5.8
(2 x +3) dx .
Area and the Definite Integral When we defined the definite integral, we lifted the requirement that f ( x ) be nonnegative. But how do we interpret “the area under the curve” when f ( x ) is negative? Net Signed Area Let us return to the Riemann sum. Consider, for example, the function f ( x ) =2−2 x 2 (shown in Figure 5.17 ) on the interval [0, 2]. Use n =8 and choose ⎧ ⎩ ⎨ x i * } as the left endpoint of each interval. Construct a rectangle on each
subinterval of height f ⎛ ⎝ x i * ⎞ rectangle, as before. When f ⎛ ⎝ x i * ⎞
⎠ and width Δ x . When f ⎛ ⎝ x i * ⎞
⎠ is positive, the product f ⎛ ⎝ x i * ⎞
⎠ Δ x represents the area of the
⎠ is negative, however, the product f ⎛ ⎝ x i * ⎞
⎠ Δ x represents the negative of the area of the
rectangle. The Riemann sum then becomes ∑ i =1 8 f ⎛ ⎝ x i * ⎞ ⎠ Δ x = ⎛
⎝ Area of rectangles above the x -axis ⎞ ⎠ − ⎛
⎝ Area of rectangles below the x -axis ⎞ ⎠
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