Model Answer: Area Between Curves and Volume of Revolution

Part 1: Definition of Area Between Two Curves (1 mark)

The area between two curves is the region bounded between two curves in a plane over a common interval. If $y = f(x)$ and $y = g(x)$ with $f(x) \ge g(x)$ on $[a,b]$, then:

$$A = \int_a^b [f(x) - g(x)] , dx$$

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Part 2: Area Enclosed by $y = 2 - x^2$ and $y = -x$ (3 marks)

Step 1: Intersection points

$$2 - x^2 = -x \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0$$

So $x = -1$ and $x = 2$.

Step 2: Upper curve

At $x = 0$: parabola gives $2$, line gives $0$. So parabola is above on $[-1, 2]$.

Step 3: Integrate

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

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

At $x = 2$: $;4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$

At $x = -1$: $;-2 + \frac{1}{2} + \frac{1}{3} = -2 + \frac{5}{6} = -\frac{7}{6}$

$$A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \text{ square units}$$

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Part 3: Definition of Volume Integral

A volume integral is the integral of a function over a three-dimensional region $V$, giving the volume (or accumulated quantity) of the solid:

$$V = \iiint_V dV$$

For a solid of revolution using the washer method:

$$V = \pi \int_a^b \Big([R]^2 - [r]^2\Big),dx$$

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Part 4: Volume of Solid Revolved about $y = 1$ (6 marks)

Given: curve $y^2 = x$ (i.e. $y = \sqrt{x}$), lines $y = 1$ and $x = 4$; revolve about $y = 1$.

Step 1: Identify the region

The region lies between the curve $y = \sqrt{x}$ (upper) and the line $y = 1$ (lower), from where they meet to $x = 4$.

• $y = \sqrt{x}$ meets $y = 1$ at $x = 1$.
• Right boundary $x = 4$, where $y = \sqrt{4} = 2$.

So $x$ runs from $1$ to $4$, with $\sqrt{x} \ge 1$ on this interval.

Step 2: Disk method about $y = 1$

Revolving about the horizontal line $y = 1$ with vertical strips gives disks of radius

$$R(x) = \sqrt{x} - 1$$

(The region touches $y = 1$ so the inner radius is zero.)

$$V = \pi \int_1^4 (\sqrt{x} - 1)^2 , dx$$

Step 3: Expand and integrate

$$(\sqrt{x}-1)^2 = x - 2\sqrt{x} + 1$$

$$V = \pi \int_1^4 \left(x - 2x^{1/2} + 1\right) dx = \pi \left[\frac{x^2}{2} - \frac{4}{3}x^{3/2} + x\right]_1^4$$

At $x = 4$: $;\frac{16}{2} - \frac{4}{3}(8) + 4 = 8 - \frac{32}{3} + 4 = 12 - \frac{32}{3} = \frac{4}{3}$

At $x = 1$: $;\frac{1}{2} - \frac{4}{3} + 1 = \frac{3}{2} - \frac{4}{3} = \frac{9 - 8}{6} = \frac{1}{6}$

$$V = \pi\left(\frac{4}{3} - \frac{1}{6}\right) = \pi\left(\frac{8}{6} - \frac{1}{6}\right) = \frac{7\pi}{6} \text{ cubic units}$$

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Common mistake: setting the solid up as horizontal washers with outer radius $4$ and inner radius $y^2$, that is $\int_1^2 (16 - y^4),dy$. That set-up does not represent revolving the given region about $y = 1$. The region bounded by $y^2 = x$, $y = 1$ and $x = 4$ revolved about $y = 1$ gives disks of radius $\sqrt{x} - 1$, yielding $\frac{7\pi}{6}$, not $\frac{49\pi}{5}$.

Final Answers:

• Area $= \dfrac{9}{2}$ square units
• Volume $= \dfrac{7\pi}{6}$ cubic units

Model Answer: Newton-Raphson Method and Box Optimization

Part 1: Newton-Raphson Method (2 marks)

Definition: The Newton-Raphson method is an iterative numerical technique for finding the roots (zeros) of a real-valued function $f(x) = 0$. Starting from an initial guess $x_0$, it uses the tangent line to the curve to generate successively better approximations to the root. It exhibits quadratic convergence when the initial guess is sufficiently close to the root and $f'(x) \neq 0$.

Formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \quad n = 0, 1, 2, \dots$$

The iteration is repeated until $|x_{n+1} - x_n|$ is smaller than a chosen tolerance.

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Part 2: Box Volume Optimization (8 marks)

Given Data

• Square tin sheet: side length $= 12$ inch
• Cut congruent squares of side $x$ from each corner
• Fold up to form an open-top box

Step 1: Box Dimensions

• Length: $12 - 2x$
• Width: $12 - 2x$
• Height: $x$

Valid domain: $0 < x < 6$.

Step 2: Volume Function

$$V(x) = x(12 - 2x)^2 = x(144 - 48x + 4x^2)$$ $$V(x) = 4x^3 - 48x^2 + 144x$$

Step 3: First Derivative

$$V'(x) = 12x^2 - 96x + 144$$

Set $V'(x) = 0$: $$12x^2 - 96x + 144 = 0 \implies x^2 - 8x + 12 = 0$$

Step 4: Solve the Quadratic

$$x = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2}$$ $$x = 6 \quad \text{or} \quad x = 2$$

Since $x = 6$ gives zero-dimension box (invalid), the candidate is $x = 2$.

Step 5: Second Derivative Test

$$V''(x) = 24x - 96$$ At $x = 2$: $$V''(2) = 48 - 96 = -48 < 0$$ Hence $x = 2$ gives a maximum.

Step 6: Maximum Volume

$$V(2) = 2(12 - 4)^2 = 2 \times 64 = 128 \text{ cubic inches}$$

Result

The squares cut from each corner should be $2$ inch $\times$ $2$ inch.

• Box dimensions: $8 \times 8 \times 2$ inch
• Maximum volume: $\boxed{128 \text{ cubic inches}}$

Model Answer: Integration

Definition of Integration

Integration is the reverse process of differentiation. It is the process of finding a function (the antiderivative) whose derivative is a given function. If $F'(x) = f(x)$, then:

$$\int f(x),dx = F(x) + C$$

where $C$ is the constant of integration. For a definite integral:

$$\int_a^b f(x),dx = [F(x)]_a^b = F(b) - F(a)$$

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Integral 1: $\int \dfrac{dx}{(x-1)(x-2)}$

Partial fractions:

$$\frac{1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$$

$$1 = A(x-2) + B(x-1)$$

• Put $x = 1$: $1 = A(-1) \Rightarrow A = -1$
• Put $x = 2$: $1 = B(1) \Rightarrow B = 1$

So:

$$\int \frac{dx}{(x-1)(x-2)} = \int\left(\frac{-1}{x-1} + \frac{1}{x-2}\right)dx$$

$$= -\ln|x-1| + \ln|x-2| + C = \ln\left|\frac{x-2}{x-1}\right| + C$$

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Integral 2: $\int_{-1}^{1} 3x^2\sqrt{x^3+1},dx$

Substitution: Let $u = x^3 + 1$, so $du = 3x^2,dx$.

Limits:

• $x = -1 \Rightarrow u = 0$
• $x = 1 \Rightarrow u = 2$

$$\int_{-1}^{1} 3x^2\sqrt{x^3+1},dx = \int_0^2 \sqrt{u},du = \left[\frac{2}{3}u^{3/2}\right]_0^2$$

$$= \frac{2}{3}(2)^{3/2} - 0 = \frac{2}{3}\cdot 2\sqrt{2} = \frac{4\sqrt{2}}{3}$$

Answer: $\displaystyle \frac{4\sqrt{2}}{3} \approx 1.886$

Both results confirmed.

Model Answer: Integral Test and Series Convergence

Given Data

• Series: $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2+4}$
• Required: (i) Define integral test [1] (ii) Test convergence/divergence [4]

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1. Definition of Integral Test (1 mark)

Integral Test: Let $f(x)$ be continuous, positive, and decreasing on the interval $[N, \infty)$ for some positive integer $N$, and let $a_n = f(n)$.

Then the infinite series $\displaystyle\sum_{n=N}^{\infty} a_n$ and the improper integral $\displaystyle\int_{N}^{\infty} f(x),dx$ either both converge or both diverge.

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2. Testing $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2+4}$ (4 marks)

Step 1: Define the corresponding function and verify conditions

Let $$f(x) = \frac{1}{x^2+4}, \quad x \ge 1$$

• Continuous: $x^2 + 4 > 0$ for all $x$, so $f$ is continuous on $[1,\infty)$. ✓
• Positive: $\dfrac{1}{x^2+4} > 0$ for all $x \ge 1$. ✓
• Decreasing: $f'(x) = -\dfrac{2x}{(x^2+4)^2} < 0$ for $x \ge 1$, so $f$ is decreasing. ✓

All conditions are satisfied; the integral test applies.

Step 2: Evaluate the improper integral

$$\int_{1}^{\infty} \frac{1}{x^2+4},dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x^2+4},dx$$

Using $\displaystyle\int \frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\frac{x}{a} + C$, with $a = 2$:

$$\int_{1}^{t} \frac{1}{x^2+4},dx = \left[\frac{1}{2}\arctan\frac{x}{2}\right]_{1}^{t} = \frac{1}{2}\arctan\frac{t}{2} - \frac{1}{2}\arctan\frac{1}{2}$$

Step 3: Take the limit

$$\lim_{t \to \infty}\left(\frac{1}{2}\arctan\frac{t}{2} - \frac{1}{2}\arctan\frac{1}{2}\right) = \frac{1}{2}\cdot\frac{\pi}{2} - \frac{1}{2}\arctan\frac{1}{2}$$

$$= \frac{\pi}{4} - \frac{1}{2}\arctan\frac{1}{2}$$

Numerically: $\dfrac{\pi}{4} \approx 0.7854$ and $\dfrac{1}{2}\arctan(0.5) \approx \dfrac{1}{2}(0.4636) = 0.2318$, giving

$$\int_{1}^{\infty} \frac{1}{x^2+4},dx \approx 0.5536 \quad \text{(finite)}$$

Step 4: Conclusion

Since the improper integral converges to a finite value, by the Integral Test the series

$$\boxed{\sum_{n=1}^{\infty} \frac{1}{n^2+4} \text{ converges.}}$$