grace77
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Can anyone explain this to me? What does if mean that the area may sometimes be negative but that the area must be positive??
Definite Integral Explained: Negative & Positive Areas
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Can anyone explain this to me? What does if mean that the area may sometimes be negative but that the area must be positive??
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The integral "area" may come out negative, but physical areas like "floor space" cannot be negative.
This means you have to be careful about when the areas under the x-axis are subtracted.
This means you have to be careful about when the areas under the x-axis are subtracted.
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What that pitiful "explanation" is trying to say is that if the function ##f## assumes both positive and negative values, then the definite integral of ##f## is the difference of two areas:
$$\int_a^b f(x) dx = A_1 - A_2$$
where ##A_1## is the area between the ##x## axis and the positive part of the function, and ##A_2## is the area between the ##x## axis and the negative part of the function.
In the picture in your attachment, area ##A_1## is approximated by the first four and last two rectangles, and ##A_2## is approximated by the middle five rectangles.
[edit]: By pitiful "explanation" I mean of course the one in your image attachment, not in the response given by Simon Bridge.
$$\int_a^b f(x) dx = A_1 - A_2$$
where ##A_1## is the area between the ##x## axis and the positive part of the function, and ##A_2## is the area between the ##x## axis and the negative part of the function.
In the picture in your attachment, area ##A_1## is approximated by the first four and last two rectangles, and ##A_2## is approximated by the middle five rectangles.
[edit]: By pitiful "explanation" I mean of course the one in your image attachment, not in the response given by Simon Bridge.
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It doesn't say that. If you read it correctly you will see that it says 'the "area" may be negative but area is always positive'. The first "area" is in quotes which means that it is using the word in a very loose sense. It is saying that the integral you get, which if it were above the x-axis would be the area, is negative so cannot be the actual area.
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@grace77: In case you are still confused about this, you should remember that the formula for the area between two curves between ##x=a## and ##x=b## is always$$
A =\int_a^b y_{upper}-y_{lower}~dx$$Students get used to working problems where ##f(x)\ge 0## on ##[a,b]## with ##\int_a^bf(x)~dx## which works since ##y_{upper}=f(x)## and ##y_{lower}=0##, the ##x## axis. But when you are calculating the area between a curve and the ##x## axis, and the curve crosses the ##x## axis, ##x## axis is now the upper curve, and on that portion ##y_{upper}-y_{lower}## becomes ##0-f(x)##, which changes the sign of the integrand on that portion. You normally break such integrals into two or more parts to work them.
A =\int_a^b y_{upper}-y_{lower}~dx$$Students get used to working problems where ##f(x)\ge 0## on ##[a,b]## with ##\int_a^bf(x)~dx## which works since ##y_{upper}=f(x)## and ##y_{lower}=0##, the ##x## axis. But when you are calculating the area between a curve and the ##x## axis, and the curve crosses the ##x## axis, ##x## axis is now the upper curve, and on that portion ##y_{upper}-y_{lower}## becomes ##0-f(x)##, which changes the sign of the integrand on that portion. You normally break such integrals into two or more parts to work them.
FilupSmith
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Having a positive or negative area indicates its direction above or below the axis.
For example:
$$ \int^{1}_{0}-x^{2}dx=\left[-\dfrac{1}{3}x^{3}\right]^{1}_{0} $$
##= -\dfrac{1}{3} ##
This indicates, that between x=0 and x=1 for x2, there is a 1/3 area2 DOWNWARDS (below the x-axis in this case)
For finding the area, for example, as mentioned above, floor-space, you take the absolute value of the integral. That is:
$$ \left|\int^{1}_{0}-x^{2}dx\right|=\left|\left[-\dfrac{1}{3}x^{3}\right]^{1}_{0}\right| $$
##=\left|-\dfrac{1}{3}\right| ##
##=\dfrac{1}{3}## units2
For example:
$$ \int^{1}_{0}-x^{2}dx=\left[-\dfrac{1}{3}x^{3}\right]^{1}_{0} $$
##= -\dfrac{1}{3} ##
This indicates, that between x=0 and x=1 for x2, there is a 1/3 area2 DOWNWARDS (below the x-axis in this case)
For finding the area, for example, as mentioned above, floor-space, you take the absolute value of the integral. That is:
$$ \left|\int^{1}_{0}-x^{2}dx\right|=\left|\left[-\dfrac{1}{3}x^{3}\right]^{1}_{0}\right| $$
##=\left|-\dfrac{1}{3}\right| ##
##=\dfrac{1}{3}## units2
TitoSmooth
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Also remember that it can be easier to do these problems with respect to x or y.
Remember to split your integral from [a, b] into smaller" pieces if there are sharp(cusps) or the curves where the area is bounded by intersect. Your book should explain. I believe there are 4 scenarios from the top of my head.
Would you like an example?
Remember to split your integral from [a, b] into smaller" pieces if there are sharp(cusps) or the curves where the area is bounded by intersect. Your book should explain. I believe there are 4 scenarios from the top of my head.
Would you like an example?
TitoSmooth
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Also use of symmetry can make the problem (computational wise) easier. And some people prefer to use the integral from b to a of THE HIGHEST FUNCTION - MINUS THE LO WEST FUNCTION
Instead of the taking the absolute value. I believe the first option is best because it allows you to get an intuitive feel for the application aspects. Such as finance, bio, Chem problems.
Instead of the taking the absolute value. I believe the first option is best because it allows you to get an intuitive feel for the application aspects. Such as finance, bio, Chem problems.