Solved Examples

Important Formulas and Concepts

1. Area under a simple curve

• If the curve is given by $y = f(x)$, then the geometrical area between the curve and the x-axis from $x=a$ to $x=b$ is $$\text{Area} = \int_a^b |f(x)|\,dx$$ If $f(x) \ge 0$ on $[a,b]$, this reduces to $\int_a^b f(x)\,dx$.
• If the curve is given by $x = g(y)$, then the geometrical area between the curve and the y-axis from $y=c$ to $y=d$ is $$\text{Area} = \int_c^d |g(y)|\,dy$$

2. Area between two curves

• If $y=f(x)$ is the upper curve and $y=g(x)$ is the lower curve on $[a,b]$, then $$\text{Area} = \int_a^b [f(x)-g(x)]\,dx$$
• If the region is easier to describe using $y$, and $x=f(y)$ is the right curve while $x=g(y)$ is the left curve on $[c,d]$, then $$\text{Area} = \int_c^d [f(y)-g(y)]\,dy$$

3. Standard Area Formulas

• Area of the circle $x^2+y^2=a^2$ is $\pi a^2$.
• Area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi ab$.
• Area bounded by the parabola $y^2=4ax$ and the line $y=mx$ is $$\frac{8a^2}{3m^3}.$$ These formulas are useful for checking final answers, but in descriptive solutions the integral method should still be shown clearly.

Example 1: Basic Area under Parabola

Find the area of the region bounded by the curve $y^2 = x$ and the lines $x = 1$, $x = 4$, and the x-axis in the first quadrant.

Solution: Step 1: Rewrite the curve in usable form. Since we are in the first quadrant, $y \ge 0$, so from $y^2=x$ we get $$y = \sqrt{x}.$$

Step 2: Identify the limits. The vertical boundaries are $x=1$ and $x=4$.

Step 3: Set up the area integral. Because the curve lies above the x-axis in this interval, $$\text{Area} = \int_1^4 y\,dx = \int_1^4 \sqrt{x}\,dx = \int_1^4 x^{1/2}dx.$$

Step 4: Integrate. $$\int x^{1/2}dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}.$$ So, $$\text{Area} = \left[\frac{2}{3}x^{3/2}\right]_1^4.$$

Step 5: Evaluate the limits. $$= \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8-1) = \frac{14}{3}.$$

Answer: $14/3$ square units.

Example 2: Integration with Respect to the y-axis

Find the area of the region bounded by $x^2 = 4y$, $y = 2$, $y = 4$ and the y-axis in the first quadrant.

Solution: Step 1: Express $x$ in terms of $y$. From $$x^2 = 4y,$$ and since we are in the first quadrant where $x \ge 0$, $$x = 2\sqrt{y}.$$

Step 2: Identify the horizontal limits. The region extends from $y=2$ to $y=4$.

Step 3: Set up the area integral. The left boundary is the y-axis ($x=0$), and the right boundary is $x=2\sqrt{y}$, so $$\text{Area} = \int_2^4 x\,dy = \int_2^4 2\sqrt{y}\,dy.$$

Step 4: Integrate. $$\text{Area} = 2\int_2^4 y^{1/2}dy = 2\left[\frac{2}{3}y^{3/2}\right]_2^4 = \frac{4}{3}(4^{3/2} - 2^{3/2}).$$

Step 5: Simplify. $$4^{3/2}=8, \qquad 2^{3/2}=2\sqrt2.$$ Hence, $$\text{Area} = \frac{4}{3}(8-2\sqrt2)=\frac{32-8\sqrt2}{3}.$$

Answer: $\frac{32 - 8\sqrt{2}}{3}$ square units.

Example 3: Area of Ellipse

Find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

Solution: Step 1: Use symmetry. The ellipse is symmetric about both coordinate axes. So total area is four times the area in the first quadrant.

Step 2: Express $y$ in terms of $x$. From $$\frac{x^2}{16} + \frac{y^2}{9} = 1,$$ we get $$\frac{y^2}{9} = 1 - \frac{x^2}{16} = \frac{16-x^2}{16},$$ so $$y^2 = \frac{9}{16}(16-x^2).$$ In the first quadrant, $$y = \frac{3}{4}\sqrt{16-x^2}.$$

Step 3: Set up the area integral. The x-limits in the first quadrant are from $0$ to $4$. Thus, $$\text{Total Area} = 4\int_0^4 \frac{3}{4}\sqrt{16-x^2}\,dx = 3\int_0^4 \sqrt{16-x^2}\,dx.$$

Step 4: Use the standard integral. $$\int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right).$$ Here $a=4$, so $$\text{Area} = 3\left[ \frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_0^4.$$

Step 5: Evaluate. At $x=4$, $$\frac{4}{2}\sqrt{0} + 8\sin^{-1}(1) = 0 + 8\cdot\frac{\pi}{2} = 4\pi.$$ At $x=0$, the value is $0$. Thus, $$\text{Area} = 3(4\pi)=12\pi.$$

Answer: $12\pi$ square units.

Example 4: Area of Circle

Find the area of the region bounded by the circle $x^2 + y^2 = 4$.

Solution: Step 1: Identify the radius. The equation is of the form $x^2+y^2=r^2$, so $$r^2=4 \implies r=2.$$

Step 2: Use symmetry. The circle is symmetric about both axes. The area in the first quadrant is $$\int_0^2 \sqrt{4-x^2}\,dx,$$ so the total area is $$4\int_0^2 \sqrt{4-x^2}\,dx.$$

Step 3: Apply the standard formula. $$\text{Total Area} = 4\left[ \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_0^2.$$

Step 4: Evaluate. At $x=2$, $$\frac{2}{2}\sqrt{0} + 2\sin^{-1}(1) = 0 + 2\cdot\frac{\pi}{2} = \pi.$$ At $x=0$, the value is $0$. Therefore, $$\text{Total Area} = 4\pi.$$

Answer: $4\pi$ square units.

Example 5: Trigonometric Curve Area

Find the area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$.

Solution: Step 1: Determine where the curve changes sign. The curve $y=\cos x$ crosses the x-axis at $$x=\frac{\pi}{2}.$$ It is positive on $\left[0,\frac{\pi}{2}\right]$ and negative on $\left[\frac{\pi}{2},\pi\right]$.

Step 2: Split the integral. Since geometrical area is always positive, $$\text{Area} = \int_0^{\pi/2} \cos x\,dx + \left|\int_{\pi/2}^{\pi} \cos x\,dx\right|.$$

Step 3: Evaluate the first part. $$\int_0^{\pi/2} \cos x\,dx = [\sin x]_0^{\pi/2} = 1-0=1.$$

Step 4: Evaluate the second part. $$\int_{\pi/2}^{\pi} \cos x\,dx = [\sin x]_{\pi/2}^{\pi} = 0-1 = -1.$$ So its area contribution is $|-1|=1$.

Step 5: Add both parts. $$\text{Total Area} = 1+1=2.$$

Answer: $2$ square units.

Example 6: Area bounded by Curve and Line

Find the area of the region bounded by the curve $y = x^2$ and the line $y = 4$.

Solution: Step 1: Find the points of intersection. Set $$x^2 = 4 \implies x = \pm 2.$$

Step 2: Determine upper and lower curves. Between $x=-2$ and $x=2$, the horizontal line $y=4$ lies above the parabola $y=x^2$.

Step 3: Set up the integral. $$\text{Area} = \int_{-2}^2 (4-x^2)\,dx.$$ Because the region is symmetric about the y-axis, $$\text{Area} = 2\int_0^2 (4-x^2)dx.$$

Step 4: Integrate. $$= 2\left[4x - \frac{x^3}{3}\right]_0^2 = 2\left(8 - \frac{8}{3}\right).$$

Step 5: Simplify. $$= 2\cdot\frac{16}{3} = \frac{32}{3}.$$

Answer: $32/3$ square units.

Example 7: Parabola and Oblique Line

Find the area of the region bounded by the parabola $y^2 = 4x$ and the line $y = 2x$.

Solution: Step 1: Find the intersection points. Substitute $y=2x$ into $y^2=4x$: $$(2x)^2 = 4x \implies 4x^2 - 4x = 0 \implies 4x(x-1)=0.$$ Thus, $$x=0,1.$$ The corresponding points are $(0,0)$ and $(1,2)$.

Step 2: Rewrite the parabola. In the first quadrant, $$y = \sqrt{4x} = 2\sqrt{x}.$$

Step 3: Decide upper and lower curves. For $0<x<1$, we have $\sqrt{x} > x$, so $$2\sqrt{x} > 2x.$$ Hence, the parabola lies above the line.

Step 4: Set up the integral. $$\text{Area} = \int_0^1 (2\sqrt{x} - 2x)dx = 2\int_0^1 (x^{1/2} - x)dx.$$

Step 5: Integrate. $$= 2\left[\frac{2}{3}x^{3/2} - \frac{x^2}{2}\right]_0^1 = 2\left(\frac{2}{3}-\frac{1}{2}\right).$$

Step 6: Simplify. $$= 2\cdot\frac{1}{6} = \frac{1}{3}.$$

Answer: $1/3$ square units.

Example 8: Area between Two Curves

Find the area of the region bounded by the curves $y = x^2$ and $y = x^3$ in the first quadrant.

Solution: Step 1: Find the points of intersection. Set $$x^2 = x^3 \implies x^2(x-1)=0.$$ So, $$x=0,1.$$

Step 2: Determine the upper curve. For $0<x<1$, we have $$x^2 > x^3.$$ So the upper curve is $y=x^2$ and the lower curve is $y=x^3$.

Step 3: Set up the integral. $$\text{Area} = \int_0^1 (x^2 - x^3)dx.$$

Step 4: Integrate. $$= \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = \frac{1}{3} - \frac{1}{4}.$$

Step 5: Simplify. $$= \frac{4-3}{12} = \frac{1}{12}.$$

Answer: $1/12$ square units.

Example 9: Modulus Function Area

Find the area bounded by the curve $y = |x|$ and $y = 2 - x^2$.

Solution: Step 1: Use symmetry. Both $y=|x|$ and $y=2-x^2$ are symmetric about the y-axis, so the enclosed region is symmetric about the y-axis. Thus, $$\text{Total Area} = 2 \times \text{(area in the first quadrant)}.$$

Step 2: Work in the first quadrant. For $x\ge 0$, we have $$|x| = x.$$ Now find intersections with the parabola: $$x = 2-x^2 \implies x^2+x-2=0 \implies (x+2)(x-1)=0.$$ The valid first-quadrant intersection is $$x=1.$$

Step 3: Identify upper and lower curves. On $0\le x \le 1$, the parabola $y=2-x^2$ lies above the line $y=x$.

Step 4: Set up the integral. $$\text{Total Area} = 2\int_0^1 [(2-x^2)-x]dx.$$

Step 5: Integrate. $$= 2\left[2x - \frac{x^3}{3} - \frac{x^2}{2}\right]_0^1 = 2\left(2 - \frac{1}{3} - \frac{1}{2}\right).$$

Step 6: Simplify. $$= 2\left(\frac{12-2-3}{6}\right)=2\cdot\frac{7}{6}=\frac{7}{3}.$$

Answer: $7/3$ square units.

Example 10: Two Intersecting Parabolas

Find the area of the region bounded by $y^2 = 4ax$ and $x^2 = 4ay$ ($a > 0$).

Solution: Step 1: Rewrite the second curve. From $$x^2 = 4ay,$$ we get $$y = \frac{x^2}{4a}.$$ Also, from $$y^2 = 4ax,$$ we get in the first quadrant $$y = 2\sqrt{ax}.$$

Step 2: Find the intersection points. Substitute $y = \frac{x^2}{4a}$ into $y^2 = 4ax$: $$\left(\frac{x^2}{4a}\right)^2 = 4ax$$ $$\frac{x^4}{16a^2} = 4ax$$ $$x^4 = 64a^3x$$ $$x(x^3 - 64a^3)=0.$$ So, $$x=0, 4a.$$

Step 3: Set up the area integral. The upper curve is $y=2\sqrt{ax}$ and the lower curve is $y=\frac{x^2}{4a}$, so $$\text{Area} = \int_0^{4a} \left(2\sqrt{a}x^{1/2} - \frac{x^2}{4a}\right)dx.$$

Step 4: Integrate. $$\text{Area} = \left[\frac{4\sqrt{a}}{3}x^{3/2} - \frac{x^3}{12a}\right]_0^{4a}.$$

Step 5: Evaluate at $x=4a$. Since $$(4a)^{3/2} = 8a\sqrt{a},$$ we get $$\frac{4\sqrt{a}}{3}(8a\sqrt{a}) - \frac{(4a)^3}{12a} = \frac{32a^2}{3} - \frac{64a^3}{12a}.$$ Now, $$\frac{64a^3}{12a} = \frac{16a^2}{3}.$$ Therefore, $$\text{Area} = \frac{32a^2}{3} - \frac{16a^2}{3} = \frac{16a^2}{3}.$$

Answer: $16a^2/3$ square units.

Example 11: Circle and a Line Cutting It

Find the area of the smaller region bounded by the circle $x^2 + y^2 = 4$ and the line $x + y = 2$.

Solution: Step 1: Find the intersection points. From the line, $$y = 2-x.$$ Substitute into the circle: $$x^2 + (2-x)^2 = 4$$ $$x^2 + 4 - 4x + x^2 = 4$$ $$2x^2 - 4x = 0$$ $$2x(x-2)=0.$$ So, $$x=0,2.$$ The points are $(0,2)$ and $(2,0)$.

Step 2: Identify upper and lower curves. In the first quadrant, the upper boundary is the circle $$y = \sqrt{4-x^2},$$ and the lower boundary is the line $$y = 2-x.$$

Step 3: Set up the area integral. $$\text{Area} = \int_0^2 \left(\sqrt{4-x^2} - (2-x)\right)dx.$$

Step 4: Integrate. $$\text{Area} = \left[ \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) - 2x + \frac{x^2}{2} \right]_0^2.$$

Step 5: Evaluate. At $x=2$, $$0 + 2\sin^{-1}(1) - 4 + 2 = 2\cdot\frac{\pi}{2} - 2 = \pi - 2.$$ At $x=0$, the value is $0$. Thus, $$\text{Area} = \pi - 2.$$

Answer: $\pi - 2$ square units.

Example 12: Ellipse and a Line

Find the area of the smaller region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the line $\frac{x}{a} + \frac{y}{b} = 1$.

Solution: Step 1: Find the intersection points. The line cuts the axes at $(a,0)$ and $(0,b)$. Both these points also satisfy the ellipse, so these are the endpoints of the smaller enclosed region.

Step 2: Write both curves in the form $y=$ function of $x$. From the ellipse, $$y = \frac{b}{a}\sqrt{a^2-x^2}$$ in the first quadrant. From the line, $$\frac{y}{b} = 1 - \frac{x}{a} \implies y = \frac{b}{a}(a-x).$$

Step 3: Set up the integral. The ellipse lies above the line between $x=0$ and $x=a$, so $$\text{Area} = \int_0^a \left[\frac{b}{a}\sqrt{a^2-x^2} - \frac{b}{a}(a-x)\right]dx.$$ Factor out $\frac{b}{a}$: $$\text{Area} = \frac{b}{a}\int_0^a \left(\sqrt{a^2-x^2} - a + x\right)dx.$$

Step 4: Integrate. $$\text{Area} = \frac{b}{a}\left[ \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) - ax + \frac{x^2}{2} \right]_0^a.$$

Step 5: Evaluate. At $x=a$, $$0 + \frac{a^2}{2}\cdot\frac{\pi}{2} - a^2 + \frac{a^2}{2} = \frac{\pi a^2}{4} - \frac{a^2}{2}.$$ At $x=0$, the value is $0$. So, $$\text{Area} = \frac{b}{a}\left(\frac{\pi a^2}{4} - \frac{a^2}{2}\right).$$ Simplify: $$= ab\left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{ab}{4}(\pi - 2).$$

Answer: $\frac{ab}{4}(\pi - 2)$ square units.

Example 13: Area bounded by y-axis Integration

Find the area bounded by the curve $x = y^2$ and the line $x = y + 2$.

Solution: Step 1: Find intersections. Set $$y^2 = y+2 \implies y^2-y-2=0 \implies (y-2)(y+1)=0.$$ Thus, $$y=-1,2.$$

Step 2: Determine right and left curves. At $y=0$, $$x=y^2=0, \qquad x=y+2=2.$$ So the line $x=y+2$ lies to the right of the parabola $x=y^2$.

Step 3: Set up the area integral. $$\text{Area} = \int_{-1}^2 [(y+2)-y^2]dy.$$

Step 4: Integrate. $$= \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^2.$$

Step 5: Evaluate. At $y=2$, $$\frac{4}{2} + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}.$$ At $y=-1$, $$\frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6}.$$ Hence, $$\text{Area} = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{27}{6} = \frac{9}{2}.$$

Answer: $9/2$ square units.

Example 14: Integrating with Fractional Powers

Find the area of the region bounded by $y = \sqrt{x}$ and $y = x$.

Solution: Step 1: Find the points of intersection. Set $$\sqrt{x} = x.$$ Squaring gives $$x = x^2 \implies x(x-1)=0.$$ Thus, $$x=0,1.$$

Step 2: Determine upper and lower curves. For $0<x<1$, $$\sqrt{x} > x.$$ So the upper curve is $y=\sqrt{x}$ and the lower curve is $y=x$.

Step 3: Set up the integral. $$\text{Area} = \int_0^1 (\sqrt{x} - x)dx.$$

Step 4: Integrate. $$= \left[\frac{2}{3}x^{3/2} - \frac{x^2}{2}\right]_0^1 = \frac{2}{3} - \frac{1}{2}.$$

Step 5: Simplify. $$= \frac{4-3}{6} = \frac{1}{6}.$$

Answer: $1/6$ square units.

Example 15: Area Enclosed by a Triangle using Integration

Find the area of the triangle whose vertices are $A(1,0), B(2,2)$ and $C(3,1)$.

Solution: Step 1: Find the equations of the sides. For $AB$: Slope $=\frac{2-0}{2-1}=2$, so $$y = 2(x-1) = 2x-2.$$ For $BC$: Slope $=\frac{1-2}{3-2}=-1$, so $$y-2 = -(x-2) \implies y=-x+4.$$ For $CA$: Slope $=\frac{1-0}{3-1}=\frac12$, so $$y = \frac12(x-1).$$

Step 2: Split the area into two vertical parts. From $x=1$ to $x=2$, the upper side is $AB$ and the lower side is $CA$. From $x=2$ to $x=3$, the upper side is $BC$ and the lower side is $CA$. Thus, $$\text{Area} = \int_1^2 (y_{AB} - y_{CA})dx + \int_2^3 (y_{BC} - y_{CA})dx.$$

Step 3: First part. $$y_{AB} - y_{CA} = (2x-2) - \frac12(x-1) = \frac32(x-1).$$ So, $$\int_1^2 \frac32(x-1)dx = \frac32\left[\frac{(x-1)^2}{2}\right]_1^2 = \frac34.$$

Step 4: Second part. $$y_{BC} - y_{CA} = (-x+4) - \frac12(x-1) = -\frac32x + \frac92.$$ Hence, $$\int_2^3 \left(-\frac32x + \frac92\right)dx = \left[-\frac34x^2 + \frac92x\right]_2^3 = \frac34.$$

Step 5: Add both parts. $$\text{Area} = \frac34 + \frac34 = \frac32.$$

Answer: $3/2$ square units.

Example 16: Two Intersecting Circles

Find the area of the region enclosed between the two circles $x^2 + y^2 = 4$ and $(x-2)^2 + y^2 = 4$.

Solution: Step 1: Find the points of intersection. Subtract the equations: $$(x-2)^2 + y^2 - (x^2+y^2)=0$$ $$x^2 - 4x + 4 - x^2 = 0 \implies -4x + 4 = 0 \implies x=1.$$ Substitute into $x^2+y^2=4$: $$1 + y^2 = 4 \implies y = \pm\sqrt3.$$

Step 2: Use symmetry. The common region is symmetric about the x-axis, so total area is twice the upper-half area.

Step 3: Set up the integral for the upper half. From $x=0$ to $x=1$, the upper boundary is the second circle: $$y = \sqrt{4-(x-2)^2}.$$ From $x=1$ to $x=2$, the upper boundary is the first circle: $$y = \sqrt{4-x^2}.$$ So, $$\text{Area} = 2\left[ \int_0^1 \sqrt{4-(x-2)^2}dx + \int_1^2 \sqrt{4-x^2}dx \right].$$

Step 4: Evaluate. Each of the two integrals equals $$\frac{2\pi}{3} - \frac{\sqrt3}{2}.$$ Therefore, $$\text{Area} = 2\left[2\left(\frac{2\pi}{3} - \frac{\sqrt3}{2}\right)\right] = \frac{8\pi}{3} - 2\sqrt3.$$

Answer: $\frac{8\pi}{3} - 2\sqrt{3}$ square units.

Example 17: Parabola and a Modulus Function

Find the area bounded by the curve $y = x^2$ and $y = |x|$.

Solution: Step 1: Find the intersections in the first quadrant. For $x\ge 0$, $|x|=x$. Then $$x^2 = x \implies x(x-1)=0,$$ so $x=0,1$.

Step 2: Use symmetry. The two curves are symmetric about the y-axis, so total area is twice the first-quadrant area.

Step 3: Determine upper and lower curves on $[0,1]$. For $0<x<1$, we have $$x > x^2.$$ Thus the upper curve is $y=x$ and the lower curve is $y=x^2$.

Step 4: Set up and evaluate the integral. $$\text{Area} = 2\int_0^1 (x-x^2)dx = 2\left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = 2\left(\frac12 - \frac13\right)=\frac13.$$

Answer: $1/3$ square units.

Example 18: Area Defined by Inequalities

Find the area of the region $\{(x, y): y^2 \le 4x,\; 4x^2 + 4y^2 \le 9\}$.

Solution: Step 1: Interpret the inequalities. The condition $$y^2 \le 4x$$ means the point lies to the right of the parabola $y^2=4x$. The condition $$4x^2 + 4y^2 \le 9$$ means the point lies inside the circle $$x^2 + y^2 \le \frac{9}{4},$$ which has radius $\frac{3}{2}$.

Step 2: Use symmetry. The region is symmetric about the x-axis, so total area is twice the upper-half area.

Step 3: Find where the parabola and circle meet in the upper half. In the upper half of the parabola, $$y = 2\sqrt{x}.$$ In the upper half of the circle, $$y = \sqrt{\frac94 - x^2}.$$ Set them equal: $$2\sqrt{x} = \sqrt{\frac94 - x^2}.$$ Squaring, $$4x = \frac94 - x^2 \implies 4x^2 + 16x - 9 = 0.$$ This gives the relevant positive root $$x = \frac12.$$

Step 4: Split the upper-half area. From $x=0$ to $x=\frac12$, the parabola lies below the circle, so the top boundary is $y=2\sqrt{x}$. From $x=\frac12$ to $x=\frac32$, the circle becomes the top boundary. Thus, $$\text{Area} = 2\left[ \int_0^{1/2} 2\sqrt{x}\,dx + \int_{1/2}^{3/2} \sqrt{\frac94 - x^2}\,dx \right].$$

Step 5: Evaluate the first part. $$2\cdot\int_0^{1/2} 2\sqrt{x}\,dx = 4\left[\frac{2}{3}x^{3/2}\right]_0^{1/2} = \frac{8}{3}\left(\frac12\right)^{3/2} = \frac{2\sqrt2}{3}.$$

Step 6: Evaluate the second part. Using the standard integral for $\sqrt{a^2-x^2}$ with $a=\frac32$, $$2\int_{1/2}^{3/2} \sqrt{\frac94 - x^2}\,dx = \frac{9\pi}{8} - \frac{\sqrt2}{2} - \frac{9}{4}\sin^{-1}\left(\frac13\right).$$

Step 7: Add both contributions. $$\text{Area} = \frac{2\sqrt2}{3} + \left(\frac{9\pi}{8} - \frac{\sqrt2}{2} - \frac{9}{4}\sin^{-1}\left(\frac13\right)\right).$$ Combine the $\sqrt2$ terms: $$\frac{2\sqrt2}{3} - \frac{\sqrt2}{2} = \frac{\sqrt2}{6}.$$ Hence, $$\text{Area} = \frac{\sqrt2}{6} + \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\left(\frac13\right).$$

Answer: $\frac{\sqrt{2}}{6} + \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right)$ square units.

Example 19: Exponential Area

Find the area bounded by the curves $y = e^x$, $y = e^{-x}$ and the line $x = 1$.

Solution: Step 1: Find the point where the two curves intersect. $$e^x = e^{-x} \implies e^{2x}=1 \implies x=0.$$

Step 2: Determine upper and lower curves. For $0<x<1$, we have $$e^x > e^{-x}.$$ So the upper curve is $y=e^x$ and the lower curve is $y=e^{-x}$.

Step 3: Set up the integral. $$\text{Area} = \int_0^1 (e^x - e^{-x})dx.$$

Step 4: Integrate. $$\int (e^x - e^{-x})dx = e^x + e^{-x}.$$ Thus, $$\text{Area} = [e^x + e^{-x}]_0^1.$$

Step 5: Evaluate. $$= \left(e + \frac{1}{e}\right) - (1+1) = e + \frac{1}{e} - 2.$$

Answer: $e + 1/e - 2$ square units.

Example 20: Area with Logarithm

Find the area bounded by the curve $y = \ln x$, the x-axis, and the ordinate $x = e$.

Solution: Step 1: Find where the curve meets the x-axis. Since $y=0$ on the x-axis, $$\ln x = 0 \implies x=1.$$ So the region lies between $x=1$ and $x=e$.

Step 2: Set up the integral. In this interval, $\ln x > 0$, so $$\text{Area} = \int_1^e \ln x\,dx.$$

Step 3: Integrate by parts. We know $$\int \ln x\,dx = x\ln x - x.$$ Thus, $$\text{Area} = [x\ln x - x]_1^e.$$

Step 4: Evaluate. At $x=e$, $$e\ln e - e = e - e = 0.$$ At $x=1$, $$1\ln1 - 1 = 0 -1 = -1.$$ Therefore, $$\text{Area} = 0 - (-1) = 1.$$

Answer: $1$ square unit.

Example 21: Overlapping Regions

Find the area of the region bounded by $0 \le y \le x^2+1$, $0 \le y \le x+1$, $0 \le x \le 2$.

Solution: Step 1: Understand the region. The allowed y-values must satisfy both $$y \le x^2+1 \quad \text{and} \quad y \le x+1.$$ So the actual top boundary is the smaller of the two curves.

Step 2: Find where the two curves intersect. $$x^2+1 = x+1 \implies x^2-x=0 \implies x(x-1)=0.$$ So they meet at $$x=0,1.$$

Step 3: Determine the top boundary on each interval.

• For $0<x<1$, we have $x^2 < x$, so $$x^2+1 < x+1,$$ hence the top boundary is $y=x^2+1$.
• For $1<x<2$, we have $x < x^2$, so $$x+1 < x^2+1,$$ hence the top boundary is $y=x+1$.

Step 4: Set up the integral. $$\text{Area} = \int_0^1 (x^2+1)dx + \int_1^2 (x+1)dx.$$

Step 5: Evaluate the first part. $$\int_0^1 (x^2+1)dx = \left[\frac{x^3}{3} + x\right]_0^1 = \frac13 + 1 = \frac43.$$

Step 6: Evaluate the second part. $$\int_1^2 (x+1)dx = \left[\frac{x^2}{2} + x\right]_1^2 = (2+2) - \left(\frac12 + 1\right) = 4 - \frac32 = \frac52.$$

Step 7: Add. $$\text{Area} = \frac43 + \frac52 = \frac{8+15}{6} = \frac{23}{6}.$$

Answer: $23/6$ square units.

Example 22: Absolute Value Equations

Find the area enclosed by the graph of $|x| + |y| = 1$.

Solution: Step 1: Interpret the graph. The equation $|x|+|y|=1$ represents a diamond-shaped figure with vertices $$(1,0), (0,1), (-1,0), (0,-1).$$

Step 2: Use symmetry. The figure is symmetric in all four quadrants. In the first quadrant, the equation becomes $$x+y=1 \implies y = 1-x.$$ The first-quadrant area is the area under this line from $x=0$ to $x=1$.

Step 3: Set up and evaluate the integral. $$\text{Total Area} = 4\int_0^1 (1-x)dx = 4\left[x - \frac{x^2}{2}\right]_0^1 = 4\left(1 - \frac12\right) = 4\cdot\frac12 = 2.$$

Answer: $2$ square units.

Example 23: Area bounded by Tangent

Find the area bounded by the curve $y = x^2$, the tangent to this curve at $(1,1)$, and the x-axis.

Solution: Step 1: Find the equation of the tangent. For $y=x^2$, $$\frac{dy}{dx} = 2x.$$ At $x=1$, slope $m=2$. Using point-slope form at $(1,1)$, $$y-1 = 2(x-1) \implies y = 2x-1.$$

Step 2: Find where the tangent meets the x-axis. Set $y=0$: $$2x-1=0 \implies x = \frac12.$$

Step 3: Visualize the region. The enclosed region consists of:

• the area under the parabola from $x=0$ to $x=1$
• minus the area under the tangent from $x=\frac12$ to $x=1$

So, $$\text{Area} = \int_0^1 x^2dx - \int_{1/2}^1 (2x-1)dx.$$

Step 4: Evaluate both integrals. $$\int_0^1 x^2dx = \left[\frac{x^3}{3}\right]_0^1 = \frac13.$$ And, $$\int_{1/2}^1 (2x-1)dx = [x^2 - x]_{1/2}^1 = (1-1) - \left(\frac14 - \frac12\right) = \frac14.$$

Step 5: Subtract. $$\text{Area} = \frac13 - \frac14 = \frac{1}{12}.$$

Answer: $1/12$ square units.

Example 24: Area involving $x|x|$

Find the area bounded by $y = x|x|$ and $y = x$.

Solution: Step 1: Rewrite the piecewise form. $$x|x| = \begin{cases} x^2, & x \ge 0 \\ -x^2, & x < 0 \end{cases}$$

Step 2: Find points of intersection. For $x\ge0$: $$x^2 = x \implies x=0,1.$$ For $x<0$: $$-x^2 = x \implies x^2 + x = 0 \implies x(x+1)=0,$$ so the valid negative root is $x=-1$. Hence intersections are at $x=-1,0,1$.

Step 3: Use symmetry. The enclosed regions in the first and third quadrants have equal area. So total area is twice the area on $[0,1]$.

Step 4: Compare the curves on $[0,1]$. There, $x|x|=x^2$, and since $x>x^2$ for $0<x<1$, the line $y=x$ is above $y=x^2$.

Step 5: Set up and evaluate the integral. $$\text{Area} = 2\int_0^1 (x-x^2)dx = 2\left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = 2\left(\frac12 - \frac13\right)=\frac13.$$

Answer: $1/3$ square units.