APPLICATION OF INTEGRALS

Chapter 8

Application of Integrals

Big idea: a curved shape has no ready-made area formula, but a thin strip of it is almost a rectangle. Integration adds up all the thin strips, and the sum is the exact area.

You already know area formulas for rectangles, triangles and circles. But real shapes are often curved. A designer wants glass for an arch-shaped window. The window is $4$ units wide, its bottom edge is flat, and its top edge is the curve $y = 4 - x^2$ for $-2 \le x \le 2$. How much glass is needed?

The arch fits inside a rectangle $4$ wide and $4$ tall (area $16$) and it contains a triangle with the same base and height (area $8$). So the glass area is between $8$ and $16$. We want the exact number.

In Chapter 7 you learned that the definite integral adds up thin strips. Here $\int_{-2}^{2}(4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \frac{32}{3} \approx 10.67$. That is the answer to the window problem. This chapter turns that idea into a method for finding areas of regions with curved edges.

Let $y = f(x)$ be a curve that stays above the $x$-axis between $x = a$ and $x = b$. We want the area $A$ of the region between the curve, the $x$-axis and the vertical lines $x = a$ and $x = b$.

Take a very thin vertical strip at position $x$, with width $dx$. Its height is $y = f(x)$, so its area is $dA = y\,dx$. This is called the elementary area. The region is made of all such strips from $x = a$ to $x = b$, so we add them up:

$$A = \int_a^b y\,dx = \int_a^b f(x)\,dx.$$

Sometimes it is easier to use horizontal strips. If the curve is written as $x = g(y)$, the region between the curve, the $y$-axis and the horizontal lines $y = c$ and $y = d$ has elementary area $dA = x\,dy$, and

$$A = \int_c^d x\,dy = \int_c^d g(y)\,dy.$$

Find the area under $y = x^2 + 1$ from $x = 0$ to $x = 3$. The curve is always above the axis, so $A = \int_0^3 (x^2 + 1)\,dx = \left[\frac{x^3}{3} + x\right]_0^3 = 9 + 3 = 12$ square units.

Find the area under $y = e^x$ from $x = 0$ to $x = 2$. $A = \int_0^2 e^x\,dx = \Big[e^x\Big]_0^2 = e^2 - 1 \approx 6.3891$.

Take the rectangle $0 \le x \le 2$, $0 \le y \le 4$ (area $8$). The parabola $y = x^2$ joins its corners $(0, 0)$ and $(2, 4)$. It cuts the rectangle into two regions.

• Below the curve, use vertical strips: $\int_0^2 x^2\,dx = \frac{8}{3} \approx 2.6667$.
• Above the curve, the region lies between the curve $x = \sqrt{y}$ and the $y$-axis. Use horizontal strips: $\int_0^4 \sqrt{y}\,dy = \frac{16}{3} \approx 5.3333$.
• The two areas add to $\frac{8}{3} + \frac{16}{3} = 8$, the area of the whole rectangle. This is a good check.

A direct y-example. The area between $x = y^2$, the $y$-axis and the lines $y = 1$ and $y = 3$ is $\int_1^3 y^2\,dy = \left[\frac{y^3}{3}\right]_1^3 = \frac{26}{3} \approx 8.6667$.

The formula $\int_a^b y\,dx$ counts area below the axis as negative. If the curve crosses the $x$-axis, the positive and negative parts can cancel, and the integral no longer equals the area.

Example 4. Find the area enclosed between $y = x^3 - 4x$ and the $x$-axis, for $-2 \le x \le 2$. The curve meets the axis where $x(x - 2)(x + 2) = 0$, at $x = -2, 0, 2$. At $x = -1$ the curve has height $3$ (above), and at $x = 1$ it has height $-3$ (below).

• On $[-2, 0]$: $\int_{-2}^{0}(x^3 - 4x)\,dx = 4$.
• On $[0, 2]$: $\int_{0}^{2}(x^3 - 4x)\,dx = -4$. The minus sign says "below the axis".
• The whole integral is $0$, because the parts cancel. But the area is $|4| + |-4| = 8$.

Rule. To find the area between a curve and the $x$-axis: (1) find where the curve crosses the axis; (2) integrate separately on each piece; (3) take the absolute value of each piece; (4) add them.

The circle $x^2 + y^2 = a^2$ is symmetric about both axes, so its area is $4$ times the area of the quarter in the first quadrant. In that quarter, $y = \sqrt{a^2 - x^2}$ and $x$ runs from $0$ to $a$. From the formula in Section 7.6.2, $\int_0^a \sqrt{a^2 - x^2}\,dx = \frac{\pi a^2}{4}$. So the area is $4 \cdot \frac{\pi a^2}{4} = \pi a^2$.

For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the quarter in the first quadrant has $y = \frac{b}{a}\sqrt{a^2 - x^2}$. Then the quarter area is $\frac{b}{a}\cdot\frac{\pi a^2}{4} = \frac{\pi a b}{4}$, and the whole ellipse has area $\pi a b$.

Check with numbers. For $a = 5$: the quarter area is $\int_0^5\sqrt{25 - x^2}\,dx = 19.635$, and $4 \times 19.635 = 78.5398 = 25\pi$. For the ellipse with $a = 4$, $b = 3$: the quarter area is $9.4248$, and the whole ellipse is $37.6991 = 12\pi$. Both were also found by adding thousands of thin strips by computer.

1. Find the area under $y = x^3$ from $x = 0$ to $x = 2$.
2. Find the area under $y = \dfrac{1}{x}$ from $x = 1$ to $x = e^2$.
3. Find the area between $y = x^2 - 4x$ and the $x$-axis from $x = 0$ to $x = 4$.
4. Find the area under $y = 4 - x^2$ from $x = 0$ to $x = 2$ (half of the arch).

Answers

1. $\left[\frac{x^4}{4}\right]_0^2 = 4$.
2. $\Big[\ln x\Big]_1^{e^2} = 2$.
3. The curve is below the axis on $(0, 4)$ because $x(x - 4) < 0$. The integral is $-10.6667 = -\frac{32}{3}$, so the area is $\frac{32}{3} \approx 10.6667$.
4. $\left[4x - \frac{x^3}{3}\right]_0^2 = \frac{16}{3} \approx 5.3333$, exactly half of $\frac{32}{3}$.

1. Find the area under $y = x^3$ from $x = 0$ to $x = 2$.
2. Find the area under $y = \dfrac{1}{x}$ from $x = 1$ to $x = e^2$.
3. Find the area under $y = \cos 2x$ from $x = 0$ to $x = \dfrac{\pi}{4}$.
4. Find the area enclosed between $y = x^2 - 4x$ and the $x$-axis.
5. Find the total area between $y = x(x - 1)(x - 2)$ and the $x$-axis, from $x = 0$ to $x = 2$.
6. Find the area between the curve $x = y^2 + 1$, the $y$-axis and the lines $y = 0$ and $y = 3$.
7. Find the area enclosed by the circle $x^2 + y^2 = 36$.
8. Find the area enclosed by the ellipse $9x^2 + 25y^2 = 225$.
9. Find the area between $y = |x - 1|$ and the $x$-axis from $x = 0$ to $x = 3$. Check it with triangles.
10. A runner's speed is $v(t) = 6t - t^2$ metres per second for $0 \le t \le 6$. The area under the speed graph is the distance run. How far does the runner go?
11. Find the area under $y = \dfrac{1}{1 + x^2}$ from $x = 0$ to $x = 1$.

Answers

1. $\int_0^2 x^3\,dx = \left[\frac{x^4}{4}\right]_0^2 = 4$ square units.
2. $\Big[\ln x\Big]_1^{e^2} = 2 - 0 = 2$.
3. $\left[\frac{\sin 2x}{2}\right]_0^{\pi/4} = \frac{1}{2} = 0.5$.
4. The curve meets the axis at $x = 0$ and $x = 4$ and lies below it in between. $\int_0^4 (x^2 - 4x)\,dx = -10.6667 = -\frac{32}{3}$, so the area is $\frac{32}{3} \approx 10.6667$.
5. The curve crosses the axis at $0$, $1$, $2$. On $[0, 1]$ it is above the axis (at $x = 0.5$ it equals $0.375$) and the integral is $0.25 = \frac{1}{4}$. On $[1, 2]$ it is below and the integral is $-0.25$. Total area $= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.
6. Use horizontal strips: $\int_0^3 (y^2 + 1)\,dy = \left[\frac{y^3}{3} + y\right]_0^3 = 12$.
7. The area of a circle of radius $a$ is $\pi a^2$. Here $a = 6$, so the area is $36\pi \approx 113.1$.
8. Divide by $225$: $\frac{x^2}{25} + \frac{y^2}{9} = 1$, so $a = 5$ and $b = 3$. The area is $\pi a b = 15\pi \approx 47.12$.
9. Split at $x = 1$. On $[0, 1]$, $|x - 1| = 1 - x$ and the integral is $0.5$. On $[1, 3]$, $|x - 1| = x - 1$ and the integral is $2$. Total $= 2.5$. Triangles: $\frac{1}{2}(1)(1) + \frac{1}{2}(2)(2) = 2.5$.
10. The speed is never negative on $[0, 6]$. $\int_0^6 (6t - t^2)\,dt = \left[3t^2 - \frac{t^3}{3}\right]_0^6 = 36$ metres.
11. $\Big[\tan^{-1}x\Big]_0^1 = \frac{\pi}{4} \approx 0.7854$.

Find the area between the line $y = 2x - 3$, the $x$-axis and the lines $x = 0$ and $x = 4$. The line meets the axis where $2x - 3 = 0$, at $x = 1.5$. Below the axis on $[0, 1.5]$ the integral is $-2.25$. Above the axis on $[1.5, 4]$ it is $6.25$. The area is $|-2.25| + 6.25 = 8.5$.

Check with triangles: the left triangle has base $1.5$ and height $3$, area $2.25$. The right triangle has base $2.5$ and height $5$, area $6.25$. Together $8.5$.

Find the area between $y = \cos x$ and the $x$-axis from $x = 0$ to $x = 2\pi$. Split where $\cos x = 0$, at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. The three pieces have integrals $1$, $-2$ and $1$. Their absolute values add to $1 + 2 + 1 = 4$. (The plain integral from $0$ to $2\pi$ is $0$.)

Find $\int_{-6}^{0}|x + 3|\,dx$. The graph of $y = |x + 3|$ is a V with its point at $x = -3$. Each half is a triangle with base $3$ and height $3$. The left half has integral $4.5$ and the right half has $4.5$, so the total is $9$.

The height of a hill, in metres, is $y = 4x - x^2$ for $0 \le x \le 4$ (with $x$ in metres along the ground). The area of the hill's side view is $\int_0^4 (4x - x^2)\,dx = \left[2x^2 - \frac{x^3}{3}\right]_0^4 = \frac{32}{3} \approx 10.6667$ square metres.

A cyclist's velocity is $v(t) = t^2 - 4t + 3$ metres per second for $0 \le t \le 4$. The velocity is zero at $t = 1$ and $t = 3$. It is positive on $[0, 1]$ and $[3, 4]$ (moving forward) and negative on $[1, 3]$ (moving back). The three integrals are $1.3333$, $-1.3333$ and $1.3333$.

• Displacement (net change of position): $\int_0^4 v\,dt = 1.3333 = \frac{4}{3}$ m.
• Distance travelled (total path length): add the absolute values: $\frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ m.

1. Find the area under $y = x^4$ from $x = 1$ to $x = 3$.
2. Find $\int_{-1}^{1} x|x|\,dx$ and the area between $y = x|x|$, the $x$-axis and the lines $x = -1$, $x = 1$.
3. Find the area between $y = x^3$, the $x$-axis and the lines $x = -2$ and $x = 1$.
4. Find the area between $y = \sin x$ and the $x$-axis from $x = 0$ to $x = 2\pi$.
5. Find the area between the curve $x = 4 - y^2$ and the $y$-axis.
6. Find the area of the circle $x^2 + y^2 = 4$ and of the ellipse $\frac{x^2}{4} + y^2 = 1$.
7. A cyclist has velocity $v(t) = t^2 - 4t + 3$ m/s, as in Example 5. Find the distance travelled and the displacement from $t = 0$ to $t = 4$.
8. Find the area under $y = \sqrt{x}$ from $x = 0$ to $x = 9$.

Answers

1. $\left[\frac{x^5}{5}\right]_1^3 = \frac{243 - 1}{5} = 48.4$.
2. The function is odd, so the integral is $0$. For the area, $|x|x| = x^2$ and the area is $\int_{-1}^{1} x^2\,dx = \frac{2}{3} \approx 0.6667$.
3. On $[-2, 0]$ the curve is below the axis, area $4$. On $[0, 1]$ it is above, area $0.25$. Total $= 4 + \frac{1}{4} = \frac{17}{4} = 4.25$.
4. Each hump has area $2$, so the total is $4$. (The plain integral is $0$.)
5. The curve meets the $y$-axis ($x = 0$) at $y = \pm 2$. Use horizontal strips: $\int_{-2}^{2}(4 - y^2)\,dy = \left[4y - \frac{y^3}{3}\right]_{-2}^{2} = \frac{32}{3} \approx 10.6667$.
6. Circle: $\pi a^2 = 4\pi \approx 12.57$. Ellipse: $a = 2$, $b = 1$, so $\pi a b = 2\pi \approx 6.28$.
7. Distance $= 4$ m (add the absolute values of the three parts). Displacement $= \frac{4}{3} \approx 1.3333$ m.
8. $\int_0^9 x^{1/2}\,dx = \left[\frac{2}{3}x^{3/2}\right]_0^9 = \frac{2}{3}\cdot 27 = 18$.

• Area with vertical strips. If $y = f(x) \ge 0$ on $[a, b]$, the area between the curve, the $x$-axis and the lines $x = a$, $x = b$ is $A = \int_a^b y\,dx = \int_a^b f(x)\,dx$.
• Area with horizontal strips. If $x = g(y)$, the area between the curve, the $y$-axis and the lines $y = c$, $y = d$ is $A = \int_c^d x\,dy = \int_c^d g(y)\,dy$.
• Elementary area. A thin strip has area (length) $\times$ (thickness): $dA = y\,dx$ or $dA = x\,dy$.
• If the curve crosses the axis, the integral is signed (area below the axis counts as negative). Split at the crossing points and add the absolute values to get the area.
• Circle and ellipse. The circle $x^2 + y^2 = a^2$ has area $\pi a^2$. The ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ has area $\pi a b$. (Four times the first-quadrant area.)
• Distance and displacement. For a velocity $v(t)$, $\int v\,dt$ is the displacement and $\int |v|\,dt$ is the distance.

The ideas behind integration are much older than the symbols. The story has three stages.

• Ancient Greece: exhaustion. Greek mathematicians found areas and volumes of curved shapes by fitting simple shapes inside them, again and again, until almost nothing was left over. This is called the method of exhaustion. It was developed especially by Eudoxus (4th century BCE) and Archimedes (3rd century BCE).
• The 17th century: a system. Several mathematicians, including Kepler, Cavalieri, Fermat and Wallis, found clever ways to calculate areas. In 1665 Isaac Newton began his work on calculus. He treated the antiderivative as the reverse of finding a tangent to a curve.
• Newton and Leibniz. In the 1680s Gottfried Wilhelm Leibniz published his own version of calculus. He thought of an area as a sum of infinitely many tiny pieces, and he chose the stretched "S" sign $\int$ for "sum". In 1696, after a suggestion by Johann Bernoulli, he renamed the subject "integral calculus". Newton and Leibniz worked separately and in different ways, but their results were practically the same. Leibniz saw most clearly that the antiderivative and the definite integral are connected.
• The 19th century: firm foundations. Early in the 1800s, Augustin-Louis Cauchy put integration on a firm footing using the idea of a limit, which is how you met it in Chapter 7.

The key discovery, that differentiation and integration undo each other, is credited to Newton and Leibniz.