(a)

$$ \int \left( 4x^{3} - 3e^{2x} + \frac{5}{x} \right) dx $$

$$ = \frac{4x^{4}}{4} - \frac{3e^{2x}}{2} + 5\ln|x| + C $$

$$ = x^{4} - \frac{3}{2}e^{2x} + 5\ln|x| + C $$

(b)

$$ \int \frac{6x^{2} + 2}{x^{3} + x},dx $$

Let $u = x^{3} + x$, then $du = (3x^{2} + 1),dx$.

Note that $6x^{2} + 2 = 2(3x^{2} + 1)$, so:

$$ \int \frac{2,du}{u} = 2\ln|x^{3} + x| + C $$

(c)

$$ \int_{1}^{2} \frac{x}{\sqrt{2x^{2} + 1}},dx $$

Let $u = 2x^{2} + 1$, then $du = 4x,dx$ and $x,dx = \frac{du}{4}$.

When $x = 1$, $u = 3$. When $x = 2$, $u = 9$.

$$ \int_{3}^{9} \frac{1}{\sqrt{u}} \cdot \frac{du}{4} $$

$$ = \frac{1}{4} \int_{3}^{9} u^{-1/2},du $$

$$ = \frac{1}{4} \Big[ 2u^{1/2} \Big]_{3}^{9} $$

$$ = \frac{1}{2} \big( 3 - \sqrt{3} \big) $$

(a)

$$ \int 3x^{2}\sqrt{x^{3} + 4},dx $$

Let $u = x^{3} + 4$, then $du = 3x^{2},dx$.

$$ \int \sqrt{u},du = \frac{2}{3}u^{3/2} + C $$

$$ = \frac{2}{3}(x^{3} + 4)^{3/2} + C $$

(b)

$$ \int_{-1}^{2} |x|,dx $$

Split at $x = 0$:

$$ = \int_{-1}^{0} (-x),dx + \int_{0}^{2} x,dx $$

$$ = \Big[ -\frac{x^{2}}{2} \Big]{-1}^{0} + \Big[ \frac{x^{2}}{2} \Big]{0}^{2} $$

$$ = \Big( 0 + \frac{1}{2} \Big) + \big( 2 - 0 \big) $$

$$ = \frac{5}{2} $$

(c)

$y = 4 - x^{2}$ cuts the $x$-axis at $x = \pm 2$.

$$ V = \pi \int_{-2}^{2} (4 - x^{2})^{2},dx $$

$$ = \pi \int_{-2}^{2} \big( 16 - 8x^{2} + x^{4} \big),dx $$

$$ = \pi \Big[ 16x - \frac{8x^{3}}{3} + \frac{x^{5}}{5} \Big]_{-2}^{2} $$

$$ = \pi \Bigg[ \Big( 32 - \frac{64}{3} + \frac{32}{5} \Big) - \Big( -32 + \frac{64}{3} - \frac{32}{5} \Big) \Bigg] $$

$$ = \pi \Bigg[ 64 - \frac{128}{3} + \frac{64}{5} \Bigg] $$

$$ = \pi \Bigg[ \frac{960 - 640 + 192}{15} \Bigg] $$

$$ = \frac{512\pi}{15} $$

(a)

$$ \text{Area} = \int_{0}^{\pi} \sin x,dx $$

$$ = \Big[ -\cos x \Big]_{0}^{\pi} $$

$$ = (-\cos\pi) - (-\cos 0) $$

$$ = 1 + 1 = 2 $$

(b)

Intersection: $x^{2} = 2x + 3$

$$ x^{2} - 2x - 3 = 0 \implies (x-3)(x+1) = 0 \implies x = -1,; 3 $$

$$ \text{Area} = \int_{-1}^{3} \big[ (2x+3) - x^{2} \big],dx $$

$$ = \Big[ x^{2} + 3x - \frac{x^{3}}{3} \Big]_{-1}^{3} $$

$$ = (9 + 9 - 9) - \Big( 1 - 3 + \frac{1}{3} \Big) $$

$$ = 9 - \Big( -\frac{5}{3} \Big) $$

$$ = 9 + \frac{5}{3} = \frac{32}{3} $$

(a)

$$ \text{Area} = \int_{0}^{4} \sqrt{x},dx $$

$$ = \Big[ \frac{2}{3}x^{3/2} \Big]_{0}^{4} $$

$$ = \frac{2}{3}(8 - 0) = \frac{16}{3} $$

(b)

$$ \text{Area} = \int_{0}^{\pi/4} (\cos x - \sin x),dx $$

$$ = \Big[ \sin x + \cos x \Big]_{0}^{\pi/4} $$

$$ = \Big( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \Big) - (0 + 1) $$

$$ = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 $$

(a)

Highest derivative is $\dfrac{d^{2}y}{dx^{2}}$, so:

$$ \boxed{\text{Order} = 2,; \text{Degree} = 3} $$

(b)

$$ y = Ae^{2x} + Be^{-3x} $$

$$ y' = 2Ae^{2x} - 3Be^{-3x} $$

$$ y'' = 4Ae^{2x} + 9Be^{-3x} $$

Substitute into the DE:

$$ y'' + y' - 6y $$

$$ = \big( 4Ae^{2x} + 9Be^{-3x} \big) + \big( 2Ae^{2x} - 3Be^{-3x} \big) - 6\big( Ae^{2x} + Be^{-3x} \big) $$

$$ = (4A + 2A - 6A)e^{2x} + (9B - 3B - 6B)e^{-3x} $$

$$ = 0 \quad \checkmark $$

(a)

$$ \frac{dy}{dx} = \frac{x}{y} $$

$$ y,dy = x,dx $$

$$ \frac{y^{2}}{2} = \frac{x^{2}}{2} + C $$

$$ y^{2} = x^{2} + 2C $$

$y(0) = 2$:

$$ 4 = 0 + 2C \implies C = 2 $$

$$ y^{2} = x^{2} + 4 \implies \boxed{y = \sqrt{x^{2} + 4}} $$

(b)

$$ \frac{dy}{dx} = \frac{x^{2} + 1}{y^{2}} $$

$$ y^{2},dy = (x^{2} + 1),dx $$

$$ \frac{y^{3}}{3} = \frac{x^{3}}{3} + x + C $$

$y(1) = 2$:

$$ \frac{8}{3} = \frac{1}{3} + 1 + C \implies C = \frac{4}{3} $$

$$ y^{3} = x^{3} + 3x + 4 \implies \boxed{y = \sqrt[3]{x^{3} + 3x + 4}} $$

(a)

$$ \lim_{k \to \infty} \frac{3k^{2} + 1}{2k^{2} + 5} $$

$$ = \lim_{k \to \infty} \frac{3 + 1/k^{2}}{2 + 5/k^{2}} $$

$$ = \frac{3}{2} $$

Since the limit exists and is finite, the sequence converges to $\dfrac{3}{2}$.

(b)

$$ \sum_{r=1}^{20} (3r^{2} - 2r + 1) $$

$$ = 3\sum_{r=1}^{20} r^{2} - 2\sum_{r=1}^{20} r + \sum_{r=1}^{20} 1 $$

$$ = 3 \cdot \frac{20 \cdot 21 \cdot 41}{6} - 2 \cdot \frac{20 \cdot 21}{2} + 20 $$

$$ = 8610 - 420 + 20 = \boxed{8210} $$

(a)

$$ \frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1} $$

(b)

$$ \sum_{r=1}^{n} \frac{1}{r(r+1)} $$

$$ = \sum_{r=1}^{n} \Big( \frac{1}{r} - \frac{1}{r+1} \Big) $$

Telescoping:

$$ = \Big( 1 - \frac{1}{2} \Big) + \Big( \frac{1}{2} - \frac{1}{3} \Big) + \cdots + \Big( \frac{1}{n} - \frac{1}{n+1} \Big) $$

$$ = 1 - \frac{1}{n+1} = \boxed{\frac{n}{n+1}} $$

(c)

$$ \sum_{r=1}^{\infty} \frac{1}{r(r+1)} = \lim_{n \to \infty} \frac{n}{n+1} = \boxed{1} $$

(a)

$$ (2x - 3y)^{4} $$

$$ = \binom{4}{0}(2x)^{4} + \binom{4}{1}(2x)^{3}(-3y) + \binom{4}{2}(2x)^{2}(-3y)^{2} + \binom{4}{3}(2x)(-3y)^{3} + \binom{4}{4}(-3y)^{4} $$

$$ = 16x^{4} - 96x^{3}y + 216x^{2}y^{2} - 216xy^{3} + 81y^{4} $$

(b)

General term:

$$ \binom{9}{r}(x^{2})^{9-r} \Big( \frac{2}{x} \Big)^{r} = \binom{9}{r} , 2^{r} , x^{18-3r} $$

For term independent of $x$:

$$ 18 - 3r = 0 \implies r = 6 $$

Term:

$$ \binom{9}{6} , 2^{6} = 84 \times 64 = \boxed{5376} $$

(a)

General term:

$$ \binom{8}{r}(1)^{8-r}(2x)^{r} = \binom{8}{r} , 2^{r} , x^{r} $$

Coefficient of $x^{5}$:

$$ \binom{8}{5} , 2^{5} = 56 \times 32 = \boxed{1792} $$

(b)

$$ (1.98)^{6} = (2 - 0.02)^{6} = 64(1 - 0.01)^{6} $$

$$ (1 - 0.01)^{6} = 1 - 0.06 + 0.0015 - 0.00002 + \cdots = 0.94148 $$

$$ (1.98)^{6} \approx 64 \times 0.94148 = \boxed{60.255} $$

(a)

Vertex $(2, 3)$, focus $(2, 5)$ — vertical, opening upward.

Distance from vertex to focus: $a = 2$.

$$ \boxed{(x - 2)^{2} = 8(y - 3)} $$

Directrix:

$$ y = 3 - 2 = 1 $$

Latus rectum:

$$ |4a| = 8 $$

(b)

Sketch: vertex $(2,3)$, axis $x = 2$, opens upward, latus rectum endpoints $(-2,5)$ and $(6,5)$.

(a)

$$ y^{2} + 8x - 6y + 25 = 0 $$

$$ y^{2} - 6y = -8x - 25 $$

$$ (y - 3)^{2} - 9 = -8x - 25 $$

$$ (y - 3)^{2} = -8(x + 2) $$

• Vertex: $(-2, 3)$
• $4a = 8 \implies a = 2$, opens left
• Focus: $(-4, 3)$
• Directrix: $x = 0$

(b)

Focus $(-1, 2)$, directrix $x = 3$.

Vertex is midpoint:

$$ x = \frac{-1 + 3}{2} = 1 \implies \text{Vertex }(1, 2) $$

$a = |1 - (-1)| = 2$, opens left.

$$ \boxed{(y - 2)^{2} = -8(x - 1)} $$

(a)

$$ \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 $$

• Centre: $(0, 0)$
• $a = 5$, $b = 3$
• $c^{2} = a^{2} - b^{2} = 25 - 9 = 16 \implies c = 4$
• Horizontal ($a > b$)
• Vertices: $(\pm 5, 0)$
• Foci: $(\pm 4, 0)$
• Eccentricity: $e = \dfrac{c}{a} = \dfrac{4}{5}$
• Latus rectum: $\dfrac{2b^{2}}{a} = \dfrac{18}{5}$

(b)

Horizontal ellipse, centred at origin, vertices at $(\pm 5, 0)$, co-vertices at $(0, \pm 3)$.

(a)

$$ 4x^{2} + 9y^{2} - 24x + 36y + 36 = 0 $$

$$ 4(x^{2} - 6x) + 9(y^{2} + 4y) = -36 $$

$$ 4(x - 3)^{2} - 36 + 9(y + 2)^{2} - 36 = -36 $$

$$ 4(x - 3)^{2} + 9(y + 2)^{2} = 36 $$

$$ \boxed{\frac{(x - 3)^{2}}{9} + \frac{(y + 2)^{2}}{4} = 1} $$

• Centre: $(3, -2)$
• $a = 3$, $b = 2$
• $c^{2} = 9 - 4 = 5 \implies c = \sqrt{5}$
• Vertices: $(0, -2)$ and $(6, -2)$
• Foci: $(3 \pm \sqrt{5}, -2)$

(b)

Centre $(1, -2)$, focus $(1, 0)$, vertex $(1, 3)$ — vertical.

$a = 3$, $c = 2$.

$$ b^{2} = a^{2} - c^{2} = 9 - 4 = 5 $$

$$ \boxed{\frac{(x - 1)^{2}}{5} + \frac{(y + 2)^{2}}{9} = 1} $$

(a)

$$ \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 $$

• Centre: $(0, 0)$
• $a = 4$, $b = 3$
• $c^{2} = a^{2} + b^{2} = 16 + 9 = 25 \implies c = 5$
• Horizontal
• Vertices: $(\pm 4, 0)$
• Foci: $(\pm 5, 0)$
• Asymptotes: $y = \pm \dfrac{3}{4}x$
• Latus rectum: $\dfrac{2b^{2}}{a} = \dfrac{9}{2}$

(b)

Sketch: centre at origin, two branches opening left/right, vertices $(\pm 4, 0)$, asymptotes $y = \pm \frac{3}{4}x$.

(a)

$$ 9x^{2} - 4y^{2} - 36x - 32y - 64 = 0 $$

$$ 9(x^{2} - 4x) - 4(y^{2} + 8y) = 64 $$

$$ 9(x - 2)^{2} - 36 - 4(y + 4)^{2} + 64 = 64 $$

$$ 9(x - 2)^{2} - 4(y + 4)^{2} = 36 $$

$$ \boxed{\frac{(x - 2)^{2}}{4} - \frac{(y + 4)^{2}}{9} = 1} $$

• Centre: $(2, -4)$
• $a = 2$, $b = 3$
• $c^{2} = 4 + 9 = 13 \implies c = \sqrt{13}$
• Vertices: $(0, -4)$ and $(4, -4)$
• Foci: $(2 \pm \sqrt{13}, -4)$
• Asymptotes: $y = \dfrac{3}{2}x - 7$ and $y = -\dfrac{3}{2}x - 1$

(b)

Foci $(\pm 5, 0)$, vertices $(\pm 3, 0)$ — horizontal, centre at origin.

$a = 3$, $c = 5$.

$$ b^{2} = c^{2} - a^{2} = 25 - 9 = 16 $$

$$ \boxed{\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1} $$