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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.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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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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