1.Answer: Substitute $y = x$ into $x^2 + y^2 = 8 \implies x^2 + x^2 = 8 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$. Points: $(2, 2)$ and $(-2, -2)$.
2.Answer: Substitute $x = 3$ into $y^2 = 4x \implies y^2 = 12 \implies y = \pm 2\sqrt{3}$. Points: $(3, 2\sqrt{3})$ and $(3, -2\sqrt{3})$.
3.Answer: $x^2 = 8 - x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$. When $x = \pm 2$, $y = 4$. Points: $(2, 4)$ and $(-2, 4)$.
4.Answer: Set $y = 0 \implies x^2 - 4x + 3 = 0 \implies (x-1)(x-3) = 0 \implies x = 1$ and $x = 3$.
5.Answer: $x^2 + 6x - 16 = 0 \implies (x+8)(x-2) = 0 \implies x = -8, 2$. Since $x = -8$ gives $y^2 = -48$ (no real $y$), we use $x = 2$. $y^2 = 12 \implies y = \pm 2\sqrt{3}$. Points: $(2, 2\sqrt{3}), (2, -2\sqrt{3})$.
6.Answer: $e^x = e^{-x} \implies e^{2x} = 1 \implies 2x = 0 \implies x = 0$. When $x = 0$, $y = 1$. Point: $(0, 1)$.
7.Answer: The line intercepts the axes at $(3,0)$ and $(0,2)$. It forms a right triangle. Area = $\frac{1}{2} \times 3 \times 2 = 3$ sq units.
8.Answer: Solving $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and $y = 2(1 - \frac{x}{3})$. Substitution yields points $(3, 0)$ and $(0, 2)$.
9.Answer: $\tan x = 1 \implies x = \frac{\pi}{4}$. Point: $(\frac{\pi}{4}, 1)$.
10.Answer: $y + y^2 = 2 \implies y^2 + y - 2 = 0 \implies (y+2)(y-1) = 0 \implies y = -2, 1$. Since $y = x^2 \ge 0$, $y = 1$. Then $x = \pm 1$. Points: $(1, 1)$ and $(-1, 1)$.
11.Answer: $x^2 = (x-2)^2 \implies x^2 = x^2 - 4x + 4 \implies 4x = 4 \implies x = 1$. Then $1^2 + y^2 = 4 \implies y^2 = 3 \implies y = \pm \sqrt{3}$. Points: $(1, \sqrt{3})$ and $(1, -\sqrt{3})$.
12.Answer: $\cos x = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5\pi}{3}$.
13.Answer: $x^2 - (x-1)^2 = 1 \implies x^2 - (x^2 - 2x + 1) = 1 \implies 2x - 1 = 1 \implies 2x = 2 \implies x = 1$. Then $y = 0$. Point: $(1, 0)$.
14.Answer: $x^2 + 1 = 2x + 1 \implies x^2 - 2x = 0 \implies x(x-2) = 0 \implies x = 0, 2$. Points: $(0, 1)$ and $(2, 5)$.
15.Answer: $2\sin x = 2\sin x \cos x \implies 2\sin x (1 - \cos x) = 0 \implies \sin x = 0$ or $\cos x = 1$. In $[0, \pi]$, $x = 0, \pi$. Points: $(0, 0)$ and $(\pi, 0)$.
16.Answer: Area = $|\int_{-2}^0 x^3 dx| + \int_0^1 x^3 dx = |[\frac{x^4}{4}]_{-2}^0| + [\frac{x^4}{4}]_0^1 = |0 - 4| + \frac{1}{4} = 4 + \frac{1}{4} = \frac{17}{4}$ sq units.
17.Answer: Area = $\int_0^{\pi/4} \sec^2 x dx = [\tan x]_0^{\pi/4} = 1 - 0 = 1$ sq unit.
18.Answer: Intersects x-axis at $x = \pm 2$. Area = $\int_{-2}^2 (4 - x^2) dx = [4x - \frac{x^3}{3}]_{-2}^2 = (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$ sq units.
19.Answer: Intersects y-axis at $y = \pm 2$. Area = $\int_{-2}^2 (4 - y^2) dy = 2\int_0^2 (4 - y^2) dy = 2[4y - \frac{y^3}{3}]_0^2 = 2(8 - \frac{8}{3}) = \frac{32}{3}$ sq units.
20.Answer: Area = $\int_0^\pi \sin x dx + |\int_\pi^{2\pi} \sin x dx| = [-\cos x]_0^\pi + |[-\cos x]_\pi^{2\pi}| = (1+1) + |(-1-1)| = 2 + 2 = 4$ sq units.
21.Answer: x-intercept is $x=1$. Area = $|\int_0^1 \ln x dx| + \int_1^e \ln x dx$. Wait, domain is $x>0$. Usually area with x-axis and $x=e$ implies from root $x=1$. Let's assume from $x=1$ to $e$. $\int_1^e \ln x dx = [x\ln x - x]_1^e = (e - e) - (0 - 1) = 1$ sq unit.
22.Answer: Area = $2 \int_0^3 2\sqrt{x} dx = 4 [\frac{2}{3}x^{3/2}]_0^3 = \frac{8}{3}(3\sqrt{3}) = 8\sqrt{3}$ sq units.
23.Answer: Area = $\int_{-1}^1 \frac{1}{1+x^2} dx = [\tan^{-1}x]_{-1}^1 = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}$ sq units.
24.Answer: Area = $\int_0^1 e^{2x} dx = [\frac{1}{2}e^{2x}]_0^1 = \frac{1}{2}(e^2 - 1)$ sq units.
25.Answer: Area = $|\int_{-1}^0 y^3 dy| + \int_0^2 y^3 dy = |[\frac{y^4}{4}]_{-1}^0| + [\frac{y^4}{4}]_0^2 = |0 - \frac{1}{4}| + (4 - 0) = \frac{1}{4} + 4 = \frac{17}{4}$ sq units.
26.Answer: Intersects x-axis at $x=0, 4$. Area = $|\int_0^4 (x^2 - 4x) dx| = |[\frac{x^3}{3} - 2x^2]_0^4| = |\frac{64}{3} - 32| = |-\frac{32}{3}| = \frac{32}{3}$ sq units.
27.Answer: This is the upper half of a circle of radius $a$. Area = $\frac{1}{2} \pi a^2$ sq units.
28.Answer: Area = $\int_0^{\pi/4} (\cos x - \sin x) dx = [\sin x + \cos x]_0^{\pi/4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$ sq units.
29.Answer: Area = $\int_1^2 (x^2 + 2) dx = [\frac{x^3}{3} + 2x]_1^2 = (\frac{8}{3} + 4) - (\frac{1}{3} + 2) = \frac{20}{3} - \frac{7}{3} = \frac{13}{3}$ sq units.
30.Answer: Area = $\int_0^{\pi/3} \tan x dx = [\ln|\sec x|]_0^{\pi/3} = \ln(\sec\frac{\pi}{3}) - \ln(\sec 0) = \ln(2) - 0 = \ln 2$ sq units.
31.Answer: Intersects at $x=0, 2$. Area = $\int_0^2 (2x - x^2) dx = [x^2 - \frac{x^3}{3}]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$ sq units.
32.Answer: Intersects at $x=\pm 2$. Area = $\int_{-2}^2 (4 - x^2) dx = 2[4x - \frac{x^3}{3}]_0^2 = 2(8 - \frac{8}{3}) = \frac{32}{3}$ sq units.
33.Answer: Area = $2 \int_0^2 \sqrt{8x} dx = 2\sqrt{8} [\frac{2}{3}x^{3/2}]_0^2 = 4\sqrt{2} (\frac{2}{3} \cdot 2\sqrt{2}) = \frac{32}{3}$ sq units.
34.Answer: Area = $2 \int_0^a 2\sqrt{a}\sqrt{x} dx = 4\sqrt{a} [\frac{2}{3}x^{3/2}]_0^a = \frac{8\sqrt{a}}{3} \cdot a\sqrt{a} = \frac{8a^2}{3}$ sq units.
35.Answer: Area = $2 \int_{2\sqrt{2}}^4 \sqrt{16-x^2} dx$. After using the formula and limits, area = $2[ \frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\frac{x}{4} ]_{2\sqrt{2}}^4 = 4\pi - 8$ sq units.
36.Answer: Area = $2 \int_1^2 \sqrt{4-x^2} dx = 2[ \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2} ]_1^2 = \frac{4\pi}{3} - \sqrt{3}$ sq units.
37.Answer: Intersects at $x=0, \pi/2$. (Because $\sin(\pi/2)=1$ and $\frac{2}{\pi}(\pi/2)=1$). Area = $\int_0^{\pi/2} (\sin x - \frac{2x}{\pi}) dx = [-\cos x - \frac{x^2}{\pi}]_0^{\pi/2} = (0 - \frac{\pi}{4}) - (-1 - 0) = 1 - \frac{\pi}{4}$ sq units.
38.Answer: $x^2 = 4(\frac{x+2}{4}) \implies x^2 - x - 2 = 0 \implies x = -1, 2$. Area = $\int_{-1}^2 (\frac{x+2}{4} - \frac{x^2}{4}) dx = \frac{1}{4}[\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^2 = \frac{9}{8}$ sq units.
39.Answer: Intersects at $x=0, 1$. Area = $\int_0^1 (\sqrt{x} - x) dx = [\frac{2}{3}x^{3/2} - \frac{x^2}{2}]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$ sq units.
40.Answer: $2 - x^2 = -x \implies x^2 - x - 2 = 0 \implies x = -1, 2$. Area = $\int_{-1}^2 (2 - x^2 + x) dx = [2x - \frac{x^3}{3} + \frac{x^2}{2}]_{-1}^2 = \frac{9}{2}$ sq units.
41.Answer: Intersects at $x=-1, 0, 1$. By symmetry, Area = $2 \int_0^1 (x - x^3) dx = 2 [\frac{x^2}{2} - \frac{x^4}{4}]_0^1 = 2(\frac{1}{4}) = \frac{1}{2}$ sq units.
42.Answer: $x^2 - x = 2x \implies x^2 - 3x = 0 \implies x=0, 3$. Area = $\int_0^3 (2x - (x^2-x)) dx = \int_0^3 (3x - x^2) dx = [\frac{3x^2}{2} - \frac{x^3}{3}]_0^3 = \frac{27}{2} - 9 = \frac{9}{2}$ sq units.
43.Answer: Region forms a sector of the circle with angle $45^\circ$ ($\pi/4$). Area = $\frac{1}{8} \pi r^2 = \frac{1}{8} \pi (32) = 4\pi$ sq units.
44.Answer: Curves intersect at $x=0$. Area = $\int_0^1 (e^x - e^{-x}) dx = [e^x + e^{-x}]_0^1 = (e + e^{-1}) - (1 + 1) = e + \frac{1}{e} - 2$ sq units.
45.Answer: $y^2 = 4(\frac{y+4}{2}) \implies y^2 - 2y - 8 = 0 \implies y = -2, 4$. Integrate w.r.t $y$: Area = $\int_{-2}^4 (\frac{y+4}{2} - \frac{y^2}{4}) dy = 9$ sq units.
46.Answer: Formula $\frac{16ab}{3}$. Here $4a=4 \implies a=1$ and $4b=4 \implies b=1$. Area = $\frac{16(1)(1)}{3} = \frac{16}{3}$ sq units.
47.Answer: Intersects at $0, 1$. Area = $\int_0^1 (\sqrt{x} - x^2) dx = [\frac{2}{3}x^{3/2} - \frac{x^3}{3}]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ sq units.
48.Answer: $x^2 = 8 - x^2 \implies 2x^2 = 8 \implies x = \pm 2$. Area = $\int_{-2}^2 ((8-x^2) - x^2) dx = \int_{-2}^2 (8 - 2x^2) dx = [8x - \frac{2x^3}{3}]_{-2}^2 = \frac{64}{3}$ sq units.
49.Answer: Intersects at $x=1, y=\pm\sqrt{3}$. Area is $2 \times [ \int_0^1 \sqrt{4-(x-2)^2} dx + \int_1^2 \sqrt{4-x^2} dx ] = \frac{8\pi}{3} - 2\sqrt{3}$ sq units.
50.Answer: $\int_0^{\pi/4} (\cos x - \sin x) dx = [\sin x + \cos x]_0^{\pi/4} = \sqrt{2} - 1$ sq units.
51.Answer: $x^2 + 6x - 16 = 0 \implies x=2, -8$. Valid is $x=2$. Area = $2 [ \int_0^2 \sqrt{6x} dx + \int_2^4 \sqrt{16-x^2} dx ] = \frac{4\pi}{3} + \frac{32\sqrt{3}}{3}$ sq units (approx, based on exact integration).
52.Answer: Intersects at $x=\pm 1$. Area = $2 \int_0^1 (x - x^2) dx = 2 [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = 2(\frac{1}{6}) = \frac{1}{3}$ sq units.
53.Answer: Curves cross at $x=0$. Area = $\int_{-1}^0 (2^{-x} - 2^x) dx + \int_0^1 (2^x - 2^{-x}) dx = \frac{1}{\ln 2} - \frac{1}{2\ln 2} \dots = \frac{1}{\ln 2}$ sq units.
54.Answer: Represents area in first quadrant between circle $x^2+y^2=16$ and line $x+y=4$. Area = $\frac{1}{4}\pi(16) - \frac{1}{2}(4)(4) = 4\pi - 8$ sq units.
55.Answer: $x^2 = x+2 \implies x^2 - x - 2 = 0 \implies x=-1, 2$. Area = $\int_{-1}^2 (x+2 - x^2) dx = \frac{9}{2}$ sq units.
56.Answer: Intersection at $y=x$. The area consists of 8 identical slices. Standard evaluation gives Area = $12\sin^{-1}(2/3)$ or similar complex term depending on exact bounds.
57.Answer: $\int_0^{\ln 2} (e^x - e^{-x}) dx = [e^x + e^{-x}]_0^{\ln 2} = (2 + \frac{1}{2}) - (1 + 1) = \frac{1}{2}$ sq units.
58.Answer: Curves intersect at $x=3/2$. Area = $\int_1^{1.5} \ln x dx + \int_{1.5}^2 \ln(3-x) dx = 3\ln(1.5) - 1$ sq units.
59.Answer: Intersect at $x=\pm 1$. Area = $2 \int_0^1 (\frac{2}{x^2+1} - x^2) dx = 2[2\tan^{-1}x - \frac{x^3}{3}]_0^1 = 2(\frac{\pi}{2} - \frac{1}{3}) = \pi - \frac{2}{3}$ sq units.
60.Answer: $x^2 + 4x - 5 = 0 \implies x=1, -5$. Valid $x=1$. Area = $2[\int_0^1 2\sqrt{x} dx + \int_1^{\sqrt{5}} \sqrt{5-x^2} dx]$ which requires standard circle area formula evaluation.
61.Answer: Break at $x=-1$. $\int_{-3}^{-1} -(x+1) dx + \int_{-1}^1 (x+1) dx = [\frac{-x^2}{2} - x]_{-3}^{-1} + [\frac{x^2}{2} + x]_{-1}^1 = 2 + 2 = 4$ sq units.
62.Answer: Represents a square with vertices $(2,0), (0,2), (-2,0), (0,-2)$. Diagonals = 4. Area = $\frac{1}{2} \times 4 \times 4 = 8$ sq units.
63.Answer: Plotting shows a quadrilateral. Intersections at $x=2$ (where $x-1 = 3-x$) and $x=-1$ (where $1-x = 3+x$). Area = $\int_{-1}^1 (3+x - (1-x)) dx + \int_1^2 (3-x - (x-1)) dx = 4$ sq units.
64.Answer: Intersects $y=3$ at $x=\pm 2$. $\int_{-2}^2 3 dx - \int_{-2}^2 |x^2-1| dx = 12 - 2[ \int_0^1 (1-x^2)dx + \int_1^2 (x^2-1)dx ] = 12 - 2[2/3 + 4/3] = 8$ sq units.
65.Answer: $4 \int_0^a \frac{b}{a}\sqrt{a^2-x^2} dx = \frac{4b}{a} [ \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(x/a) ]_0^a = \frac{4b}{a} \frac{a^2\pi}{4} = \pi a b$ sq units.
66.Answer: Forms a rectangle with vertices $(1,2), (-1,2), (-1,-2), (1,-2)$. Width = 2, Height = 4. Area = $2 \times 4 = 8$ sq units.
67.Answer: Symmetric across x and y axis. Area = $4 \int_0^1 (1-x^2) dx = 4[x - x^3/3]_0^1 = 4(2/3) = 8/3$ sq units.
68.Answer: $(x-a)^2 + y^2 \le a^2$. This is a circle of radius $a$. Area = $\pi a^2$ sq units.
69.Answer: The region is bounded by $x=0$, $y=\cos x$, $y=\sin x$. Intersect at $x=\pi/4$. Area = $\int_0^{\pi/4} (\cos x - \sin x) dx = \sqrt{2} - 1$ sq units.
70.Answer: Area = $2 \int_0^2 x^2 dx = 2 [x^3/3]_0^2 = 16/3$ sq units.
71.Answer: Symmetric. $2 \int_0^1 (x - x^2) dx = 2[1/2 - 1/3] = 1/3$ sq units.
72.Answer: Region forms a sector of the circle $x^2+y^2=2$ between $y=x$ and $y=-x$. Angle is $90^\circ$ ($\pi/2$). Area = $\frac{1}{4} \pi (\sqrt{2})^2 = \pi/2$ sq units.
73.Answer: $2 \int_0^\pi \sin x dx = 2[-\cos x]_0^\pi = 2(2) = 4$ sq units.
74.Answer: Intersects where $2-x = 4-x^2 \implies x^2-x-2=0 \implies x=-1, 2$. Area = $\int_{-1}^2 (4-x^2 - (2-x)) dx = \int_{-1}^2 (2+x-x^2) dx = 9/2$ sq units.
75.Answer: Symmetric. $2 \int_0^2 |x-1| dx = 2 [ \int_0^1 (1-x) dx + \int_1^2 (x-1) dx ] = 2[1/2 + 1/2] = 2$ sq units.
