1.Answer: Set $y = 0 \implies 4 - x^2 = 0 \implies x^2 = 4 \implies x = \pm 2$.
2.Answer: $y = (x-1)^2$. The vertex is $(1, 0)$.
3.Answer: Complete the square: $(x-3)^2 + y^2 = 9$. Center is $(3, 0)$ and radius is $3$.
4.Answer: $x^2 = x \implies x(x-1) = 0 \implies x=0, 1$. Points: $(0, 0)$ and $(1, 1)$.
5.Answer: $(x^2)^2 = x \implies x^4 - x = 0 \implies x(x^3-1) = 0 \implies x=0, 1$. Points: $(0, 0)$ and $(1, 1)$.
6.Answer: On the x-axis. Coordinates are $(2, 0)$.
7.Answer: Set $x = 0 \implies y^2 = 4 \implies y = \pm 2$. Points are $(0, 2)$ and $(0, -2)$.
8.Answer: $\sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4}$.
9.Answer: Set $x = 0 \implies y = e^0 = 1$. The y-intercept is $(0, 1)$.
10.Answer: $x^2 = 2x \implies x^2 - 2x = 0 \implies x(x-2) = 0 \implies x = 0$ and $x = 2$.
11.Answer: $\int_1^3 x dx = \left[ \frac{x^2}{2} \right]_1^3 = \frac{9}{2} - \frac{1}{2} = 4$ sq units.
12.Answer: $\int_0^2 x^2 dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}$ sq units.
13.Answer: $\int_0^1 x^3 dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} - 0 = \frac{1}{4}$ sq units.
14.Answer: $\int_0^\pi \sin x dx = \left[ -\cos x \right]_0^\pi = -\cos(\pi) - (-\cos 0) = -(-1) + 1 = 2$ sq units.
15.Answer: $\int_1^2 y^2 dy = \left[ \frac{y^3}{3} \right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ sq units.
16.Answer: $\int_1^{e^2} \frac{1}{x} dx = [\ln|x|]_1^{e^2} = \ln(e^2) - \ln(1) = 2 - 0 = 2$ sq units.
17.Answer: $\int_0^2 e^x dx = [e^x]_0^2 = e^2 - e^0 = e^2 - 1$ sq units.
18.Answer: $\int_0^{\pi/2} \cos x dx = [\sin x]_0^{\pi/2} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1$ sq unit.
19.Answer: $\int_1^2 3x^2 dx = [x^3]_1^2 = 8 - 1 = 7$ sq units.
20.Answer: $\int_0^9 x^{1/2} dx = [\frac{2}{3}x^{3/2}]_0^9 = \frac{2}{3}(27) = 18$ sq units.
21.Answer: Intersects at $x=\pm 2$. Area = $\int_{-2}^2 (4 - x^2) dx = 2 \int_0^2 (4 - x^2) dx = 2 [4x - \frac{x^3}{3}]_0^2 = 2(8 - \frac{8}{3}) = \frac{32}{3}$ sq units.
22.Answer: Area = $2 \int_0^1 \sqrt{x} dx = 2 [\frac{2}{3}x^{3/2}]_0^1 = \frac{4}{3}$ sq units.
23.Answer: Forms a triangle with vertices $(0,0), (-2,2), (2,2)$. Base = 4, Height = 2. Area = $\frac{1}{2}(4)(2) = 4$ sq units. (Or $\int_{-2}^2 (2 - |x|) dx$).
24.Answer: Intersects when $x^2+1 = x+1 \implies x^2-x=0 \implies x=0, 1$. Integral: $\int_0^1 ( (x+1) - (x^2+1) ) dx = \int_0^1 (x - x^2) dx$.
25.Answer: $[\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ sq units.
26.Answer: Symmetric. Area = $2 \int_0^2 x dy = 2 \int_0^2 2\sqrt{y} dy = 4 [\frac{2}{3}y^{3/2}]_0^2 = \frac{8}{3}(2\sqrt{2}) = \frac{16\sqrt{2}}{3}$ sq units.
27.Answer: Curve is $y=x^2$ ($x \ge 0$) and $y=-x^2$ ($x < 0$). By symmetry, Area = $2 \int_0^1 x^2 dx = 2 [\frac{x^3}{3}]_0^1 = \frac{2}{3}$ sq units.
28.Answer: Triangle vertices: $(0,0), (3,3), (-3,3)$. Base = 6, Height = 3. Area = $\frac{1}{2} \times 6 \times 3 = 9$ sq units.
29.Answer: $\int_0^2 (2x+1) dx = [x^2 + x]_0^2 = 4 + 2 = 6$ sq units.
30.Answer: Intersects at $x=0, 1$. Line is above curve. $\int_0^1 (x - x^3) dx = [\frac{x^2}{2} - \frac{x^4}{4}]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ sq units.
31.Answer: $\int_0^3 \sqrt{9-x^2} dx = [\frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\sin^{-1}(\frac{x}{3})]_0^3 = \frac{9}{2} \sin^{-1}(1) = \frac{9\pi}{4}$ sq units.
32.Answer: Total Area = $4 \times (\text{Area in Quadrant 1})$. By formula $\pi r^2 = \pi(4)^2 = 16\pi$ sq units.
33.Answer: Area in 1st Quad = $\frac{1}{4} \times (\pi a b) = \frac{1}{4} \times \pi(2)(3) = \frac{6\pi}{4} = \frac{3\pi}{2}$ sq units.
34.Answer: Area = $2 \times \int_0^3 \sqrt{4x} dx = 2 \int_0^3 2\sqrt{x} dx = 4 \int_0^3 x^{1/2} dx$.
35.Answer: $4 [\frac{2}{3}x^{3/2}]_0^3 = \frac{8}{3} (3\sqrt{3}) = 8\sqrt{3}$ sq units.
36.Answer: Here $a=5, b=4$. Total area = $\pi a b = \pi(5)(4) = 20\pi$ sq units.
37.Answer: In Q1, $x+y=1 \implies y = 1-x$. Area = $\int_0^1 (1-x) dx = [x - \frac{x^2}{2}]_0^1 = \frac{1}{2}$ sq unit.
38.Answer: Total Area = $4 \times (\text{Area in Q1}) = 4 \times \frac{1}{2} = 2$ sq units.
39.Answer: $\cos x$ is even, so symmetric about y-axis. Area = $2 \int_0^{\pi/2} \cos x dx = 2 [\sin x]_0^{\pi/2} = 2(1) = 2$ sq units.
40.Answer: Area = $(\text{Area of Circle in Q1}) - (\text{Area of Ellipse in Q1}) = \frac{\pi a^2}{4} - \frac{\pi a b}{4} = \frac{\pi a(a-b)}{4}$.
41.Answer: $4a=4 \implies a=1$, $4b=4 \implies b=1$. Area = $\frac{16(1)(1)}{3} = \frac{16}{3}$ sq units.
42.Answer: $4a=8 \implies a=2$, $4b=8 \implies b=2$. Area = $\frac{16(2)(2)}{3} = \frac{64}{3}$ sq units.
43.Answer: $4a=12 \implies a=3$. Area = $\frac{8(3)^2}{3} = \frac{8(9)}{3} = 24$ sq units.
44.Answer: $4a=16 \implies a=4$. Area = $\frac{8(4)^2}{3} = \frac{8(16)}{3} = \frac{128}{3}$ sq units.
45.Answer: Rewrite as $\frac{x^2}{9} + \frac{y^2}{4} = 1$. $a=3, b=2$. Area = $\pi(3)(2) = 6\pi$ sq units.
46.Answer: Intersects at $x=0, 4$. Area = $\int_0^4 (2\sqrt{x} - x) dx = [\frac{4}{3}x^{3/2} - \frac{x^2}{2}]_0^4 = \frac{32}{3} - 8 = \frac{8}{3}$ sq units.
47.Answer: $y^2 = 4x \implies a=1$. $y=x \implies m=1$. Area = $\frac{8(1)^2}{3(1)^3} = \frac{8}{3}$ sq units. (Matches).
48.Answer: $4a=1 \implies a=1/4$. $m=2$. Area = $\frac{8(1/4)^2}{3(2)^3} = \frac{8(1/16)}{24} = \frac{1/2}{24} = \frac{1}{48}$ sq units.
49.Answer: Area between $y=x^2$ and $y=x$. Intersects at $x=0, 1$. Area = $\int_0^1 (x - x^2) dx = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ sq units.
50.Answer: $(\text{Area of Circle}) - (\text{Area of Ellipse}) = \pi a^2 - \pi a b = \pi a(a-b)$.
