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Worked solution

Find the exact area of a rational curve

Find the exact area of the region bounded by y=xx+3y=\frac{x}{x+3}, the xx-axis, and the lines x=−1x=-1 and x=2x=2.

CJC · Lecture · Example 8

Sketch the correct branch

The curve has vertical asymptote x=−3x=-3 and crosses the axis at x=0x=0. On [−1,0][-1,0] it is negative, while on [0,2][0,2] it is positive. The asymptote is outside the required interval. Shade only the two pieces bounded by x=−1x=-1, x=2x=2 and the axis.

How did this part go?

Find an antiderivative

Rewrite xx+3=1−3x+3\frac{x}{x+3}=1-\frac3{x+3}. Therefore ∫xx+3 dx=x−3ln⁡∣x+3∣+C\int\frac{x}{x+3}\,dx=x-3\ln|x+3|+C.

How did this part go?

Add the two positive areas

The area is −∫−10xx+3 dx+∫02xx+3 dx=−[(0−3ln⁡3)−(−1−3ln⁡2)]+[(2−3ln⁡5)−(0−3ln⁡3)]-\int_{-1}^0\frac{x}{x+3}\,dx+\int_0^2\frac{x}{x+3}\,dx=-[(0-3\ln3)-(-1-3\ln2)]+[(2-3\ln5)-(0-3\ln3)]. Simplifying gives 1+3[ln⁡3−ln⁡2+ln⁡3−ln⁡5]=1+3ln⁡(9/10)1+3[\ln3-\ln2+\ln3-\ln5]=1+3\ln(9/10) square units.

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Compare your sketch

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The branch of y=x/(x+3) to the right of asymptote x=−3. Shaded negative region from −1 to 0 and positive region from 0 to 2 are added as geometric areas.
The branch of y=x/(x+3) to the right of asymptote x=−3. Shaded negative region from −1 to 0 and positive region from 0 to 2 are added as geometric areas.
The branch of y=x/(x+3) to the right of asymptote x=−3. Shaded negative region from −1 to 0 and positive region from 0 to 2 are added as geometric areas.

Common mistakes

The lower piece has negative signed integral but positive area. Change its sign before adding the two regions.

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