{"id":382,"date":"2025-02-13T19:44:29","date_gmt":"2025-02-13T19:44:29","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-definite-integral-learn-it-2\/"},"modified":"2025-02-13T19:44:29","modified_gmt":"2025-02-13T19:44:29","slug":"the-definite-integral-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-definite-integral-learn-it-2\/","title":{"raw":"The Definite Integral: Learn It 2","rendered":"The Definite Integral: Learn It 2"},"content":{"raw":"\n

Area and the Definite Integral<\/h2>\n

When we defined the definite integral, we lifted the requirement that [latex]f(x)[\/latex] be nonnegative. But how do we interpret \u201cthe area under the curve\u201d when [latex]f(x)[\/latex] is negative?<\/p>\n

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Net Signed Area<\/h3>\n

Let us return to the Riemann sum.<\/p>\n

Consider, for example, the function [latex]f(x)=2-2x^2[\/latex] (shown in Figure 2) on the interval [latex][0,2][\/latex]. Use [latex]n=8[\/latex] and choose [latex]x_i^*[\/latex] as the left endpoint of each interval. Construct a rectangle on each subinterval of height [latex]f(x_i^*)[\/latex] and width [latex]\\Delta x[\/latex].<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]\"A Figure 2. For a function that is partly negative, the Riemann sum is the area of the rectangles above the [latex]x[\/latex]-axis minus the area of the rectangles below the [latex]x[\/latex]-axis.[\/caption]\n\n

When [latex]f(x_i^*)[\/latex] is positive, the product [latex]f(x_i^*) \\Delta x[\/latex] represents the area of the rectangle, as before. When [latex]f(x_i^*)[\/latex] is negative, however, the product [latex]f(x_i^*) \\Delta x[\/latex] represents the negative<\/em> of the area of the rectangle.<\/p>\n

The Riemann sum then becomes<\/p>\n

[latex]\\displaystyle\\sum_{i=1}^{8}f(x_i^*) \\Delta x =[\/latex] (Area of rectangles above the [latex]x[\/latex]-axis) [latex]-[\/latex] (Area of rectangles below the [latex]x[\/latex]-axis)<\/div>\n
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 <\/div>\n<\/div>\n

Taking the limit as [latex]n\\to \\infty[\/latex], the Riemann sum approaches the area between the curve above the [latex]x[\/latex]-axis and the [latex]x[\/latex]-axis, minus the area between the curve below the [latex]x[\/latex]-axis and the [latex]x[\/latex]-axis, as shown in Figure 3.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]\"A Figure 3. In the limit, the definite integral equals area [latex]A_1[\/latex] minus area [latex]A_2[\/latex], or the net signed area.[\/caption]\n\n

Then,<\/p>\n

[latex]\\begin{array}{ll} \\displaystyle\\int_0^2 f(x) dx & =\\underset{n\\to \\infty}{\\lim} \\displaystyle\\sum_{i=1}^{n} f(c_i) \\Delta x \\\\ & =A_1-A_2 \\end{array}[\/latex]<\/div>\n

The quantity [latex]A_1-A_2[\/latex] is called the net signed area<\/strong>.<\/p>\n

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Notice that net signed area can be positive, negative, or zero. If the area above the [latex]x[\/latex]-axis is larger, the net signed area is positive. If the area below the [latex]x[\/latex]-axis is larger, the net signed area is negative. If the areas above and below the [latex]x[\/latex]-axis are equal, the net signed area is zero.<\/p>\n<\/section>\n

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Find the net signed area between the curve of the function [latex]f(x)=2x[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][-3,3][\/latex].
\n[reveal-answer q=\"fs-id1170571712242\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1170571712242\"]<\/p>\n

The function produces a straight line that forms two triangles: one from [latex]x=-3[\/latex] to [latex]x=0[\/latex] and the other from [latex]x=0[\/latex] to [latex]x=3[\/latex] (Figure 4). Using the geometric formula for the area of a triangle, [latex]A=\\frac{1}{2}bh[\/latex], the area of triangle [latex]A_1[\/latex], above the axis, is<\/p>\n

[latex]A_1=\\frac{1}{2}3(6)=9[\/latex],<\/p>\n

where [latex]3[\/latex] is the base and [latex]2(3)=6[\/latex] is the height. The area of triangle [latex]A_2[\/latex], below the axis, is<\/p>\n

[latex]A_2=\\frac{1}{2}(3)(6)=9[\/latex],<\/div>\n

where [latex]3[\/latex] is the base and [latex]6[\/latex] is the height. Thus, the net area is<\/p>\n

[latex]\\displaystyle\\int_{-3}^3 2x dx=A_1-A_2=9-9=0[\/latex].<\/span><\/p>\n

\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]\"A Figure 4. The area above the curve and below the [latex]x[\/latex]-axis equals the area below the curve and above the [latex]x[\/latex]-axis.[\/caption]\n<\/div>\nAnalysis<\/strong>\n

If [latex]A_1[\/latex] is the area above the [latex]x[\/latex]-axis and [latex]A_2[\/latex] is the area below the [latex]x[\/latex]-axis, then the net area is [latex]A_1-A_2[\/latex]. Since the areas of the two triangles are equal, the net area is zero.<\/p>\n

Watch the following video to see the worked solution to this example.<\/p>\n