Example 3: Area between $y = 3x + 2$⁠, the $x$-axis, $x = -1$ and $x = 1$

Basics you need

Solution

1. Find where the line crosses the $x$-axis, and sketch

   On the $x$-axis $y = 0$⁠: $3x + 2 = 0$⁠, so $x = -\frac23$⁠. Call this point $A\left(-\frac23, 0\right)$⁠. It lies between $-1$ and 1.

   At $x = -1$⁠, $y = -3 + 2 = -1$⁠: the point $B(-1, -1)$⁠, below the axis. At $x = 1$⁠, $y = 5$⁠: the point $D(1, 5)$⁠. Let $C(-1, 0)$ and $E(1, 0)$ be the feet of the ordinates.

   Since the slope 3 is positive, the line rises: $3x + 2 < 0$ for $x < -\frac23$ and $3x + 2 > 0$ for $x > -\frac23$⁠.

   The region consists of two pieces: triangle $ACB$ between $x = -1$ and $x = -\frac23$⁠, below the axis, and triangle $ADE$ between $x = -\frac23$ and $x = 1$⁠, above it. Knowing where the sign changes tells us where to split.
   A curve meets the $x$-axis where $y = 0$⁠. Basics: reading the edges from the equation

2. Split the area at the crossing

   $$\begin{aligned} \text{Area} &= \text{area } ACBA + \text{area } ADEA \\ &= \left|\int_{-1}^{-2/3}(3x + 2)\,dx\right| \\ &\quad + \int_{-2/3}^{1}(3x + 2)\,dx. \end{aligned}$$
   On $\left[-1, -\frac23\right]$ the integrand is $\le 0$⁠, so that integral is negative and only its absolute value is the area of $ACBA$⁠. On $\left[-\frac23, 1\right]$ the integrand is $\ge 0$⁠, so that integral is the area of $ADEA$ as it stands.
   Remark, Section 8.2: $A = |A_1| + A_2$⁠. Lesson: below the axis, take the absolute value

3. Find an antiderivative

   $$F(x) = \int(3x + 2)\,dx = \frac{3x^2}{2} + 2x.$$

   Check: $F'(x) = \frac{3\cdot 2x}{2} + 2 = 3x + 2$⁠.

   The power formula $\int x^n\,dx = \frac{x^{n+1}}{n+1}$ with $n = 1$ for $3x$⁠, and $\int 2\,dx = 2x$⁠. No constant is needed for a definite integral; it cancels in $F(b) - F(a)$⁠.
   Formula (i), Section 7.2. Basics: a definite integral is a number

4. The part below the axis

   $F\left(-\frac23\right) = \frac{3}{2}\cdot\frac49 - \frac43 = \frac23 - \frac43 = -\frac23$ and $F(-1) = \frac32 - 2 = -\frac12$⁠. So

   $$\begin{aligned} \int_{-1}^{-2/3}(3x + 2)\,dx &= -\frac23 - \left(-\frac12\right) \\ &= -\frac16, \\ \text{area } ACBA &= \left|-\frac16\right| = \frac16. \end{aligned}$$
   The integral is negative, as expected for a region below the axis. Its size $\frac16$ is the area.
   Below the axis, the integral is minus the area. Basics: area below the axis

5. The part above the axis

   $F(1) = \frac32 + 2 = \frac72$⁠, so

   $$\begin{aligned} \int_{-2/3}^{1}(3x + 2)\,dx &= \frac72 - \left(-\frac23\right) \\ &= \frac{21}{6} + \frac{4}{6} = \frac{25}{6}. \end{aligned}$$
   On this interval the line is on or above the axis, so the integral is positive and equals the area of $ADEA$⁠.
   Theorem 2, Section 7.8.3: $F(\text{upper}) - F(\text{lower})$⁠. Basics: a definite integral is a number

6. Add the two parts

   $$\text{Area} = \frac16 + \frac{25}{6} = \frac{26}{6} = \frac{13}{3} \text{ sq. units}.$$
   An area counts every piece as positive, so the two pieces add. Compare the plain integral $\int_{-1}^{1}(3x + 2)\,dx = \frac72 - \left(-\frac12\right) = 4$⁠: it equals $\frac{25}{6} - \frac16$⁠, the upper piece minus the lower one, which is not the area.
   Split at every crossing, then add the sizes. Lesson: when the curve dips below the axis, Examples 3–4

7. Check with triangles

   Both pieces are right triangles. $ACB$ has legs $CA = -\frac23 - (-1) = \frac13$ and $CB = 1$⁠, so its area is $\frac12\cdot\frac13\cdot 1 = \frac16$⁠. $ADE$ has legs $AE = 1 - \left(-\frac23\right) = \frac53$ and $ED = 5$⁠, so its area is $\frac12\cdot\frac53\cdot 5 = \frac{25}{6}$⁠.

   Elementary geometry gives the same two numbers as the integrals, which confirms both the split and the arithmetic.
   Area of a triangle $= \frac12 \times$ base $\times$ height. Basics: area as a sum of strips

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