08 3areaunder

Negative Area

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If the region being investigated is below the x-axis, then the definite integral will give a negative answer.

08.3area1.gif
  • $f(x)$ is negative, so $f(x) \times h$ will be negative

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But Area should always be positive!

  • Hence, if the curve is below the x-axis:
08.3rule1.gif

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Example 1

Find (to 3 decimal places) the area between the x-axis and the curve $y = x^2 - 3$
between the values $x = 0.5$ and $x = 1.5$

Solution

… … A sketch graph reveals curve is below axis in the domain $\big[0.5,\; 1.5\big]$

08.3eg1.GIF

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… … Definite integral is:

… … $\displaystyle{ \int\limits_{0.5}^{1.5} \,x^2-3 \; dx}$
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… … … $= \left[ \frac{1}{3}x^3 - 3x \right]_{0.5}^{1.5}$
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… … … $= \big( -3.375 \big) - \big( -1.458 \big)$

… … … $= -1.917$

08.3calc1.gif

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… … Hence Area = $+1.917$ square units.

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Reversing Terminals

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Reversing the order of the terminals (or limits) reverses the sign of the definite integral

08.3rule2.gif
  • This result can sometimes be useful when manipulating areas.

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Combining Regions

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If the curve crosses the x-axis within the domain, the definite integral over the entire domain will not give the correct area.

08.3area2.gif
  • This is because the negative section and the positive section will cancel each other out.

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  • The solution is to calculate each section separately and add the absolute value of the results:
08.3rule3.gif

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Example 2

Find (correct to 3 decimal places) the area between the function $y = 2\sin\big(x - 1.5\big)$ and the x-axis and between $x = 0$ and $x = 2$.
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Solution

… … Note: $\displaystyle{ \int\limits_0^2 2\sin\big(x-1.5\big)\; dx} = -1.614$

… … The definite integral over the domain $\big[0, \; 2 \big]$ gives Incorrect Area

08.3calc2.gif

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… … A sketch graph reveals curve has an x-intercept at $x = 1.5$
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… … Area $= \left| \displaystyle{\int\limits_0^{1.5} 2\sin(x-1.5)\, dx} \right| + \left| \displaystyle{ \int\limits_{1.5}^2 2\sin(x-1.5)\, dx} \right|$
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… … Area $= \Big| -1.859 \Big| + \Big| 0.245 \Big|$
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… … Area $= 1.859 + 0.245$
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… … Area $= 2.104$ square units

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