Chap 7. APPLICATIONS OF INTEGRATION

# Chap 7. APPLICATIONS OF INTEGRATION ## Objectifs et prérequis This chapter focuses on applying integration to solve geometric and physical problems. Students will learn to compute areas bounded by curves, volumes of solids generated by revolution, and centroids of plane regions. Mastery of definite integration techniques (polynomials, trigonometric, exponential functions) and graphing skills is essential. Familiarity with the Fundamental Theorem of Calculus and basic area under a curve concepts from Form Four is required. ## Introduction Integration extends beyond finding antiderivatives; it quantifies accumulation. In Additional Mathematics, we apply definite integrals to measure area, volume, and center of mass. These applications bridge pure calculus with real-world modeling in physics, engineering, and design. The chapter progresses from one-dimensional area (between curves) to two-dimensional solids of revolution, then to centroids—each building on the idea of summing infinitesimal elements. ## I. Area under a curve and between two curves ### Definition and principle The definite integral $\int_a^b f(x) \, dx$ gives the net signed area between the curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$. When $f(x) \geq 0$ on $[a,b]$, this equals the actual area. For area between two curves $y = f(x)$ (top) and $y = g(x)$ (bottom) where $f(x) \geq g(x)$ on $[a,b]$, the area is $\int_a^b [f(x) - g(x)] \, dx$. > **Key formula**: Area between $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is $A = \int_a^b |f(x) - g(x)| \, dx$, but when order is known, $A = \int_a^b [f(x) - g(x)] \, dx$ if $f(x) \geq g(x)$. ### Properties - Area is always non-negative; if curves cross, split the integral at intersection points. - The order of subtraction matters: top function minus bottom function. - Symmetry can simplify calculations (e.g., even functions over $[-a,a]$). ### Example 1: Area under a single curve Find the area under $y = x^2$ from $x = 0$ to $x = 2$. **Solution**: Since $x^2 \geq 0$ on $[0,2]$, $$ A = \int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3} \text{ square units}. $$ ### Example 2: Area between two curves Find the area between $y = x^2$ and $y = 2x$ from $x = 0$ to $x = 2$. **Solution**: First, find intersection: $x^2 = 2x \Rightarrow x^2 - 2x = 0 \Rightarrow x(x-2)=0 \Rightarrow x=0,2$. On $[0,2]$, $2x \geq x^2$ (test $x=1$: $2 > 1$), so top function is $y=2x$, bottom is $y=x^2$. $$ A = \int_0^2 (2x - x^2) \, dx = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = \left(4 - \frac{8}{3}\right) - 0 = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \text{ square units}. $$ ### Example 3: Area requiring splitting Find the area between $y = \sin x$ and $y = \cos x$ from $x = 0$ to $x = \frac{\pi}{2}$. **Solution**: Intersection: $\sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}$ in $[0, \frac{\pi}{2}]$. On $[0, \frac{\pi}{4}]$, $\cos x \geq \sin x$; on $[\frac{\pi}{4}, \frac{\pi}{2}]$, $\sin x \geq \cos x$. $$ A = \int_0^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx $$ First integral: $$ \int (\cos x - \sin x) dx = \sin x + \cos x \quad \Rightarrow \quad [\sin x + \cos x]_0^{\pi/4} = \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - (0 + 1) = \sqrt{2} - 1 $$ Second integral: $$ \int (\sin x - \cos x) dx = -\cos x - \sin x \quad \Rightarrow \quad [-\cos x - \sin x]_{\pi/4}^{\pi/2} = \left(-0 - 1\right) - \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = -1 + \sqrt{2} $$ Total: $A = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2$ square units. ### Exercises d'application corrigés **Exercise 1**: Find the area bounded by $y = x^3$, $y = 0$, $x = 1$, and $x = 2$. **Solution**: $y = x^3 \geq 0$ on $[1,2]$, so $$ A = \int_1^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_1^2 = \frac{16}{4} - \frac{1}{4} = 4 - 0.25 = \frac{15}{4} \text{ square units}. $$ **Exercise 2**: Find the area between $y = e^x$ and $y = x$ from $x = 0$ to $x = 1$. **Solution**: On $[0,1]$, $e^x \geq x$ (since $e^0=1>0$, $e^1\approx2.7>1$, and $e^x$ grows faster), so $$ A = \int_0^1 (e^x - x) \, dx = \left[ e^x - \frac{x^2}{2} \right]_0^1 = \left(e - \frac{1}{2}\right) - (1 - 0) = e - 0.5 - 1 = e - 1.5 \text{ square units}. $$ **Exercise 3**: Find the area enclosed by $y = x^2$ and $y = 4$. **Solution**: Intersection: $x^2 = 4 \Rightarrow x = \pm 2$. Top: $y=4$, bottom: $y=x^2$, symmetric about $y$-axis. $$ A = 2 \int_0^2 (4 - x^2) \, dx = 2 \left[ 4x - \frac{x^3}{3} \right]_0^2 = 2 \left(8 - \frac{8}{3}\right) = 2 \left(\frac{24}{3} - \frac{8}{3}\right) = 2 \cdot \frac{16}{3} = \frac{32}{3} \text{ square units}. $$ ## II. Solids of revolution ### Definition and principle A solid of revolution is formed by rotating a plane region around a line (axis). The volume can be computed using the disk method (when rotating around an axis and the region touches the axis) or the washer method (when there is a gap, creating a hollow solid). > **Disk method (about $x$-axis)**: If $y = f(x) \geq 0$ on $[a,b]$, volume $V = \pi \int_a^b [f(x)]^2 \, dx$. > **Washer method (about $x$-axis)**: If region between $y = f(x)$ (top) and $y = g(x)$ (bottom) is rotated, $V = \pi \int_a^b \left( [f(x)]^2 - [g(x)]^2 \right) dx$, provided $f(x) \geq g(x) \geq 0$. ### Properties - Axis of rotation determines variable of integration ($dx$ for horizontal axis, $dy$ for vertical). - Methods generalize to rotation about $y$-axis or other lines via substitution. - Volume is always non-negative; ensure correct outer and inner radii. ### Example 1: Disk method about $x$-axis Find the volume of the solid obtained by rotating the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ about the $x$-axis. **Solution**: Here, $f(x) = \sqrt{x} \geq 0$, so $$ V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \left( \frac{16}{2} - 0 \right) = 8\pi \text{ cubic units}. $$ ### Example 2: Washer method about $x$-axis Find the volume of the solid obtained by rotating the region between $y = x^2$ and $y = x$ from $x = 0$ to $x = 1$ about the $x$-axis. **Solution**: Intersection: $x^2 = x \Rightarrow x=0,1$. On $[0,1]$, $x \geq x^2$, so top: $f(x)=x$, bottom: $g(x)=x^2$. $$ V = \pi \int_0^1 \left[ (x)^2 - (x^2)^2 \right] dx = \pi \int_0^1 (x^2 - x^4) \, dx = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1 = \pi \left( \frac{1}{3} - \frac{1}{5} \right) = \pi \left( \frac{5-3}{15} \right) = \frac{2\pi}{15} \text{ cubic units}. $$ ### Example 3: Rotation about $y$-axis Find the volume of the solid obtained by rotating the region under $y = x^2$ from $x = 0$ to $x = 2$ about the $y$-axis. **Solution**: Use the shell method (alternative to washer when integrating with respect to $x$ for $y$-axis rotation): $$ V = 2\pi \int_a^b (\text{radius})(\text{height}) \, dx = 2\pi \int_0^2 x \cdot (x^2) \, dx = 2\pi \int_0^2 x^3 \, dx = 2\pi \left[ \frac{x^4}{4} \right]_0^2 = 2\pi \cdot \frac{16}{4} = 2\pi \cdot 4 = 8\pi \text{ cubic units}. $$ *(Note: Shell method: radius = distance from $y$-axis = $x$, height = $f(x) = x^2$)* ### Exercises d'application corrigés **Exercise 1**: Find the volume of the solid generated by rotating $y = \sin x$ from $x = 0$ to $x = \pi$ about the $x$-axis. **Solution**: $y = \sin x \geq 0$ on $[0,\pi]$, so $$ V = \pi \int_0^\pi (\sin x)^2 \, dx = \pi \int_0^\pi \frac{1 - \cos 2x}{2} \, dx = \frac{\pi}{2} \int_0^\pi (1 - \cos 2x) \, dx = \frac{\pi}{2} \left[ x - \frac{\sin 2x}{2} \right]_0^\pi $$ $$ = \frac{\pi}{2} \left( \left(\pi - 0\right) - (0 - 0) \right) = \frac{\pi}{2} \cdot \pi = \frac{\pi^2}{2} \text{ cubic units}. $$ **Exercise 2**: Find the volume of the solid obtained by rotating the region bounded by $y = x^2$, $y = 0$, $x = 2$ about the $y$-axis. **Solution**: Use shell method: radius = $x$, height = $x^2$, limits $x=0$ to $2$. $$ V = 2\pi \int_0^2 x \cdot x^2 \, dx = 2\pi \int_0^2 x^3 \, dx = 2\pi \left[ \frac{x^4}{4} \right]_0^2 = 2\pi \cdot \frac{16}{4} = 8\pi \text{ cubic units}. $$ **Exercise 3**: Find the volume of the solid obtained by rotating the region between $y = \sqrt{x}$ and $y = x$ from $x = 0$ to $x = 1$ about the $x$-axis. **Solution**: Intersection: $\sqrt{x} = x \Rightarrow x = x^2 \Rightarrow x^2 - x = 0 \Rightarrow x(x-1)=0 \Rightarrow x=0,1$. On $[0,1]$, $\sqrt{x} \geq x$ (since $x \leq 1$), so outer radius $R = \sqrt{x}$, inner radius $r = x$. $$ V = \pi \int_0^1 \left[ (\sqrt{x})^2 - (x)^2 \right] dx = \pi \int_0^1 (x - x^2) \, dx = \pi \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \pi \left( \frac{1}{2} - \frac{1}{3} \right) = \pi \cdot \frac{1}{6} = \frac{\pi}{6} \text{ cubic units}. $$ ## III. Centroid ### Definition and principle The centroid (or center of mass of a uniform lamina) of a plane region is the point $(\bar{x}, \bar{y})$ where the region would balance. For a region bounded by curves, $$ \bar{x} = \frac{1}{A} \int_a^b x [f(x) - g(x)] \, dx, \quad \bar{y} = \frac{1}{A} \int_a^b \frac{1}{2} \left[ (f(x))^2 - (g(x))^2 \right] dx, $$ where $A$ is the area of the region, $f(x)$ is the upper function, $g(x)$ the lower function on $[a,b]$. > **Key formulas**: > $$ > \bar{x} = \frac{\int_a^b x [f(x) - g(x)] \, dx}{\int_a^b [f(x) - g(x)] \, dx}, \quad > \bar{y} = \frac{\int_a^b \frac{1}{2} \left[ (f(x))^2 - (g(x))^2 \right] dx}{\int_a^b [f(x) - g(x)] \, dx} > $$ ### Properties - Centroid lies within the region for convex shapes; may lie outside for non-convex. - Symmetry: if region is symmetric about a line, centroid lies on that line. - Formulas derive from moment equilibrium: $M_x = \rho \int y \, dA$, $M_y = \rho \int x \, dA$, with $\bar{x} = M_y / (ρA)$, $\bar{y} = M_x / (ρA)$, and $\rho$ constant cancels. ### Example 1: Centroid of area under a curve Find the centroid of the region under $y = x^2$ from $x = 0$ to $x = 2$. **Solution**: First, area $A = \int_0^2 x^2 \, dx = \frac{8}{3}$ (from earlier). $$ \bar{x} = \frac{1}{A} \int_0^2 x \cdot x^2 \, dx = \frac{3}{8} \int_0^2 x^3 \, dx = \frac{3}{8} \left[ \frac{x^4}{4} \right]_0^2 = \frac{3}{8} \cdot \frac{16}{4} = \frac{3}{8} \cdot 4 = \frac{12}{8} = \frac{3}{2} $$ $$ \bar{y} = \frac{1}{A} \int_0^2 \frac{1}{2} (x^2)^2 \, dx = \frac{3}{8} \cdot \frac{1}{2} \int_0^2 x^4 \, dx = \frac{3}{16} \left[ \frac{x^5}{5} \right]_0^2 = \frac{3}{16} \cdot \frac{32}{5} = \frac{3 \cdot 32}{16 \cdot 5} = \frac{96}{80} = \frac{6}{5} $$ So centroid is $\left( \frac{3}{2}, \frac{6}{5} \right)$. ### Example 2: Centroid between two curves Find the centroid of the region between $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$. **Solution**: Intersection: $x = x^2 \Rightarrow x=0,1$. On $[0,1]$, $x \geq x^2$, so $f(x)=x$, $g(x)=x^2$. Area: $A = \int_0^1 (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$. $$ \bar{x} = \frac{1}{A} \int_0^1 x (x - x^2) \, dx = 6 \int_0^1 (x^2 - x^3) \, dx = 6 \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = 6 \left( \frac{1}{3} - \frac{1}{4} \right) = 6 \cdot \frac{1}{12} = \frac{6}{12} = \frac{1}{2} $$ $$ \bar{y} = \frac{1}{A} \int_0^1 \frac{1}{2} \left[ (x)^2 - (x^2)^2 \right] dx = 6 \cdot \frac{1}{2} \int_0^1 (x^2 - x^4) \, dx = 3 \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1 = 3 \left( \frac{1}{3} - \frac{1}{5} \right) = 3 \cdot \frac{2}{15} = \frac{6}{15} = \frac{2}{5} $$ So centroid is $\left( \frac{1}{2}, \frac{2}{5} \right)$. ### Example 3: Centroid using symmetry Find the centroid of the region bounded by $y = 4 - x^2$ and $y = 0$. **Solution**: Intersection: $4 - x^2 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. Region symmetric about $y$-axis, so $\bar{x} = 0$. Area: $A = \int_{-2}^2 (4 - x^2) \, dx = 2 \int_0^2 (4 - x^2) \, dx = 2 \left[ 4x - \frac{x^3}{3} \right]_0^2 = 2 \left(8 - \frac{8}{3}\right) = 2 \cdot \frac{16}{3} = \frac{32}{3}$. $$ \bar{y} = \frac{1}{A} \int_{-2}^2 \frac{1}{2} (4 - x^2)^2 \, dx = \frac{3}{32} \cdot \frac{1}{2} \int_{-2}^2 (16 - 8x^2 + x^4) \, dx = \frac{3}{64} \cdot 2 \int_0^2 (16 - 8x^2 + x^4) \, dx \quad (\text{even function}) $$ $$ = \frac{6}{64} \int_0^2 (16 - 8x^2 + x^4) \, dx = \frac{3}{32} \left[ 16x - \frac{8x^3}{3} + \frac{x^5}{5} \right]_0^2 = \frac{3}{32} \left( 32 - \frac{64}{3} + \frac{32}{5} \right) $$ Compute inside: $32 = \frac{480}{15}$, $\frac{64}{3} = \frac{320}{15}$, $\frac{32}{5} = \frac{96}{15}$, so $\frac{480 - 320 + 96}{15} = \frac{256}{15}$ Thus, $$ \bar{y} = \frac{3}{32} \cdot \frac{256}{15} = \frac{3 \cdot 256}{32 \cdot 15} = \frac{768}{480} = \frac{8}{5} = 1.6 $$ So centroid is $\left( 0, \frac{8}{5} \right)$. ### Exercises d'application corrigés **Exercise 1**: Find the centroid of the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$. **Solution**: Area: $A = \int_0^4 \sqrt{x} \, dx = \int_0^4 x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^4 = \frac{2}{3} \cdot 8 = \frac{16}{3}$. $$ \bar{x} = \frac{1}{A} \int_0^4 x \cdot \sqrt{x} \, dx = \frac{3}{16} \int_0^4 x^{3/2} \, dx = \frac{3}{16} \left[ \frac{2}{5} x^{5/2} \right]_0^4 = \frac{3}{16} \cdot \frac{2}{5} \cdot (4^{5/2}) = \frac{6}{80} \cdot (2^5) = \frac{3}{40} \cdot 32 = \frac{96}{40} = \frac{12}{5} = 2.4 $$ $$ \bar{y} = \frac{1}{A} \int_0^4 \frac{1}{2} (\sqrt{x})^2 \, dx = \frac{3}{16} \cdot \frac{1}{2} \int_0^4 x \, dx = \frac{3}{32} \left[ \frac{x^2}{2} \right]_0^4 = \frac{3}{32} \cdot \frac{16}{2} = \frac{3}{32} \cdot 8 = \frac{24}{32} = \frac{3}{4} $$ Centroid: $\left( \frac{12}{5}, \frac{3}{4} \right)$. **Exercise 2**: Find the centroid of the region between $y = 2x$ and $y = x^2$ from $x = 0$ to $x = 2$. **Solution**: Intersection: $2x = x^2 \Rightarrow x^2 - 2x = 0 \Rightarrow x(x-2)=0 \Rightarrow x=0,2$. On $[0,2]$, $2x \geq x^2$, so $f(x)=2x$, $g(x)=x^2$. Area: $A = \int_0^2 (2x - x^2) \, dx = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$. $$ \bar{x} = \frac{1}{A} \int_0^2 x (2x - x^2) \, dx = \frac{3}{4} \int_0^2 (2x^2 - x^3) \, dx = \frac{3}{4} \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^2 = \frac{3}{4} \left( \frac{16}{3} - 4 \right) = \frac{3}{4} \left( \frac{16}{3} - \frac{12}{3} \right) = \frac{3}{4} \cdot \frac{4}{3} = 1 $$ $$ \bar{y} = \frac{1}{A} \int_0^2 \frac{1}{2} \left[ (2x)^2 - (x^2)^2 \right] dx = \frac{3}{4} \cdot \frac{1}{2} \int_0^2 (4x^2 - x^4) \, dx = \frac{3}{8} \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_0^2 = \frac{3}{8} \left( \frac{32}{3} - \frac{32}{5} \right) $$ $$ = \frac{3}{8} \cdot 32 \left( \frac{1}{3} - \frac{1}{5} \right) = 12 \left( \frac{5-3}{15} \right) = 12 \cdot \frac{2}{15} = \frac{24}{15} = \frac{8}{5} = 1.6 $$ Centroid: $\left( 1, \frac{8}{5} \right)$. **Exercise 3**: Find the centroid of the region bounded by $y = e^x$, $y = 0$, $x = 0$, $x = 1$. **Solution**: Area: $A = \int_0^1 e^x \, dx = [e^x]_0^1 = e - 1$. $$ \bar{x} = \frac{1}{A} \int_0^1 x e^x \, dx $$ Integrate by parts: $u = x$, $dv = e^x dx$ → $du = dx$, $v = e^x$ $$ \int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C $$ So $$ \int_0^1 x e^x dx = \left[ x e^x - e^x \right]_0^1 = (e - e) - (0 - 1) = 0 + 1 = 1 $$ Thus, $\bar{x} = \frac{1}{e-1}$. $$ \bar{y} = \frac{1}{A} \int_0^1 \frac{1}{2} (e^x)^2 \, dx = \frac{1}{2(e-1)} \int_0^1 e^{2x} dx = \frac{1}{2(e-1)} \left[ \frac{e^{2x}}{2} \right]_0^1 = \frac{1}{4(e-1)} (e^2 - 1) $$ Note: $e^2 - 1 = (e-1)(e+1)$, so $$ \bar{y} = \frac{1}{4(e-1)} \cdot (e-1)(e+1) = \frac{e+1}{4} $$ Centroid: $\left( \frac{1}{e-1}, \frac{e+1}{4} \right)$. ## Synthèse du chapitre - **Area under/between curves**: Definite integral of function or difference of functions; requires identifying top/bottom and splitting at intersections. - **Solids of revolution**: Disk/washer method for rotation about $x$- or $y$-axis; shell method as alternative for $y$-axis; volume depends on squared radii. - **Centroid**: Balance point of lamina; formulas involve first moments divided by area; symmetry simplifies $\bar{x}$ or $\bar{y}$. - **Connections**: All rely on partitioning region into infinitesimal elements (strips, disks, rings) and summing via integration. - **Common pitfalls**: Incorrect order of functions (top minus bottom), forgetting $\pi$ in volume, misidentifying radius in washer/shell, miscalculating area before centroid. ## Glossaire - **Definite integral**: Integral with limits, yielding a numerical value representing net area or accumulation. - **Area between curves**: Region enclosed by two functions; computed as integral of upper minus lower function. - **Solid of revolution**: 3D object formed by rotating a plane region around a line. - **Disk method**: Volume technique for solids with no hole; uses cross-sectional disks perpendicular to axis. - **Washer method**: Volume technique for solids with a hole; uses washers (disks with holes) perpendicular to axis. - **Shell method**: Volume technique using cylindrical shells parallel to axis of rotation; alternative to washer/disk. - **Centroid**: Geometric center of a plane region; point where area is evenly distributed. - **First moment**: Integral of distance times area element; used to compute centroid via $M_y = \int x \, dA$, $M_x = \int y \, dA$. - **Uniform lamina**: Thin plate of constant density; centroid equals center of mass. - **Axis of rotation**: Line about which a region is rotated to form a solid of revolution. - **Radius (outer/inner)**: Distance from axis of rotation to outer/inner boundary of region; critical in washer method. - **Moment about axis**: Tendency of area to rotate about an axis; $M_x = \int y \, dA$, $M_y = \int x \, dA$.