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Use the formula for the area of a trapezoid to evaluate 2 4 ( 2 x + 3 ) d x .

18 square units

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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 [link] ) 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 * ) and width Δ x . When f ( x i * ) is positive, the product f ( x i * ) Δ x represents the area of the rectangle, as before. When f ( x i * ) 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 )
A graph of a downward opening parabola over [-1, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). Eight rectangles are drawn evenly over [0,2] with heights determined by the value of the function at the left endpoints of each.
For a function that is partly negative, the Riemann sum is the area of the rectangles above the x -axis less the area of the rectangles below the x -axis.

Taking the limit as n , the Riemann sum approaches the area between the curve above the x -axis and the x -axis, less the area between the curve below the x -axis and the x -axis, as shown in [link] . Then,

0 2 f ( x ) d x = lim n i = 1 n f ( c i ) Δ x = A 1 A 2 .

The quantity A 1 A 2 is called the net signed area    .

A graph of a downward opening parabola over [-2, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). The area in quadrant one under the curve is shaded blue and labeled A1. The area in quadrant four above the curve and to the left of x=2 is shaded blue and labeled A2.
In the limit, the definite integral equals area A 1 less area A 2 , or the net signed area.

Notice that net signed area can be positive, negative, or zero. If the area above the x -axis is larger, the net signed area is positive. If the area below the x -axis is larger, the net signed area is negative. If the areas above and below the x -axis are equal, the net signed area is zero.

Finding the net signed area

Find the net signed area between the curve of the function f ( x ) = 2 x and the x -axis over the interval [ −3 , 3 ] .

The function produces a straight line that forms two triangles: one from x = −3 to x = 0 and the other from x = 0 to x = 3 ( [link] ). Using the geometric formula for the area of a triangle, A = 1 2 b h , the area of triangle A 1 , above the axis, is

A 1 = 1 2 3 ( 6 ) = 9 ,

where 3 is the base and 2 ( 3 ) = 6 is the height. The area of triangle A 2 , below the axis, is

A 2 = 1 2 ( 3 ) ( 6 ) = 9 ,

where 3 is the base and 6 is the height. Thus, the net area is

−3 3 2 x d x = A 1 A 2 = 9 9 = 0 .
A graph of an increasing line over [-6, 6] going through the origin and (-3, -6) and (3,6). The area under the line in quadrant one over [0,3] is shaded blue and labeled A1, and the area above the line in quadrant three over [-3,0] is shaded blue and labeled A2.
The area above the curve and below the x -axis equals the area below the curve and above the x -axis.

Analysis

If A 1 is the area above the x -axis and A 2 is the area below the x -axis, then the net area is A 1 A 2 . Since the areas of the two triangles are equal, the net area is zero.

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Find the net signed area of f ( x ) = x 2 over the interval [ 0 , 6 ] , illustrated in the following image.

A graph of an increasing line going through (-2,-4), (0,-2), (2,0), (4,2) and (6,4). The area above the curve in quadrant four is shaded blue and labeled A2, and the area under the curve and to the left of x=6 in quadrant one is shaded and labeled A1.

6

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Total area

One application of the definite integral is finding displacement when given a velocity function. If v ( t ) represents the velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original position. This is a very important application of the definite integral, and we examine it in more detail later in the chapter. For now, we’re just going to look at some basics to get a feel for how this works by studying constant velocities.

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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