Calculus

Limits

Formal Definition

Let $f$ be defined on an open interval containing $a$, except possibly at $a$ itself. We write $\displaystyle\lim_{x \to a} f(x) = L$ if for every $\varepsilon \gt 0$, there exists a $\delta \gt 0$ such that:

$0 \lt |x - a| \lt \delta \implies |f(x) - L| \lt \varepsilon$

Limit Laws

If $\displaystyle\lim_{x \to a} f(x) = L$ and $\displaystyle\lim_{x \to a} g(x) = M$, then:

$\lim_{x \to a}\bigl[f(x) + g(x)\bigr] = L + M$

$\lim_{x \to a}\bigl[f(x) \cdot g(x)\bigr] = L \cdot M$

$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}, \quad M \ne 0$

$\lim_{x \to a} \bigl[f(x)\bigr]^n = L^n, \quad n \in \mathbb{'\{'}Z{'\}'}^{+}$

One-Sided Limits

A two-sided limit exists if and only if both one-sided limits exist and are equal:

$\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$

Indeterminate Forms

Expressions of the form $\dfrac{0}{0}$, $\dfrac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$, $1^\infty$, $0^0$, and $\infty^0$ are indeterminate — the limit may or may not exist, and algebraic manipulation or l'H^opital's rule is required.

Common Limits

Limit	Value
$\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$	$1$
$\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x}$	$0$
$\displaystyle\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$	$e$
$\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x}$	$1$
$\displaystyle\lim_{x \to 0} \frac{\ln(1 + x)}{x}$	$1$

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Continuity

Definition

A function $f$ is continuous at $a$ if and only if all three conditions hold:

1. $f(a)$ is defined
2. $\displaystyle\lim_{x \to a} f(x)$ exists
3. $\displaystyle\lim_{x \to a} f(x) = f(a)$

Intermediate Value Theorem

If $f$ is continuous on $[a, b]$ and $k$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c \in (a, b)$ such that $f(c) = k$.

This theorem underpins the bisection method for root-finding and guarantees that a continuous function on a closed interval attains every intermediate value.

Types of Discontinuity

Type	Description	Example
Removable	Limit exists but $f(a)$ is undefined or differs from the limit	$f(x) = \dfrac{x^2 - 1}{x - 1}$ at $x = 1$
Jump	One-sided limits exist but are unequal	$f(x) = \lfloor x \rfloor$ at integer points
Infinite	Function tends to $\pm\infty$ near $a$	$f(x) = \dfrac{1}{x}$ at $x = 0$
Oscillatory	Function oscillates without settling	$f(x) = \sin\!\left(\dfrac{1}{x}\right)$ at $x = 0$

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Differentiation

Definition from First Principles

The derivative of $f$ at $a$ is:

$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

provided this limit exists. When it does, $f$ is said to be differentiable at $a$.

Differentiability implies continuity, but continuity does not imply differentiability. The function $f(x) = |x|$ is continuous everywhere but not differentiable at $x = 0$.

Differentiation Rules

Power rule. For $n \in \mathbb{'\{'}R{'\}'}$:

$\frac{d}{dx}\bigl[x^n\bigr] = nx^{n-1}$

Constant multiple and sum rules:

$\frac{d}{dx}\bigl[cf(x)\bigr] = c\,f'(x), \qquad \frac{d}{dx}\bigl[f(x) \pm g(x)\bigr] = f'(x) \pm g'(x)$

Product rule:

$\frac{d}{dx}\bigl[uv\bigr] = u'v + uv'$

Quotient rule:

$\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}, \quad v \ne 0$

Chain rule. If $y = f(g(x))$, then:

$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

Derivatives of Standard Functions

$f(x)$	$f'(x)$
$e^x$	$e^x$
$a^x$	$a^x \ln a$
$\ln x$	$\dfrac{1}{x}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\arcsin x$	$\dfrac{1}{\sqrt{1 - x^2}}$
$\arccos x$	$\dfrac{-1}{\sqrt{1 - x^2}}$
$\arctan x$	$\dfrac{1}{1 + x^2}$

Implicit Differentiation

When a relation between $x$ and $y$ cannot be easily solved for $y$, differentiate both sides with respect to $x$, treating $y$ as a function of $x$. Every occurrence of $y$ produces a factor of $\dfrac{dy}{dx}$ via the chain rule.

Example. Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 25$.

Differentiating implicitly: $2x + 2y\dfrac{dy}{dx} = 0$, so $\dfrac{dy}{dx} = -\dfrac{x}{y}$.

Higher-Order Derivatives

The second derivative $f''(x) = \dfrac{d^2y}{dx^2}$ is the derivative of $f'(x)$. Higher-order derivatives $f^{(n)}(x)$ are defined recursively. The second derivative governs concavity: $f''(x) \gt 0$ implies concave up; $f''(x) \lt 0$ implies concave down.

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Applications of Differentiation

Stationary Points

A stationary point occurs where $f'(x) = 0$. The second derivative test classifies them:

• $f'(a) = 0$ and $f''(a) \gt 0$: local minimum
• $f'(a) = 0$ and $f''(a) \lt 0$: local maximum
• $f'(a) = 0$ and $f''(a) = 0$: test is inconclusive; use the first derivative test

Optimisation

To solve optimisation problems:

1. Identify the quantity to be maximised or minimised as a function of a single variable.
2. Determine the domain of the function from the problem constraints.
3. Find stationary points by setting the derivative to zero.
4. Verify the nature of each stationary point (second derivative test or sign analysis).
5. Check endpoints and boundary values.

Related Rates

When two or more quantities vary with time and are related by an equation, their rates of change are related by implicit differentiation with respect to time $t$.

Example. A spherical balloon is inflated at $\dfrac{dV}{dt} = 100\;\mathrm{cm^3/s}$. Find $\dfrac{dr}{dt}$ when $r = 5\;\mathrm{cm}$.

Since $V = \dfrac{4}{3}\pi r^3$, differentiating: $\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}$. Substituting: $100 = 4\pi(25)\dfrac{dr}{dt}$, so $\dfrac{dr}{dt} = \dfrac{1}{\pi}\;\mathrm{cm/s}$.

Curve Sketching

Systematic curve sketching involves:

1. Domain and intercepts: where $f(x)$ is defined; $x$- and $y$-intercepts.
2. Symmetry: even ($f(-x) = f(x)$), odd ($f(-x) = -f(x)$), periodic.
3. Asymptotes: vertical (where denominator vanishes), horizontal (end behaviour), oblique.
4. First derivative: intervals of increase/decrease; stationary points.
5. Second derivative: concavity and points of inflection.

Tangents and Normals

The tangent to $y = f(x)$ at $x = a$ has equation:

$y - f(a) = f'(a)(x - a)$

The normal is perpendicular to the tangent, with gradient $-\dfrac{1}{f'(a)}$ (provided $f'(a) \ne 0$).

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Integration

Indefinite Integrals

The antiderivative of $f(x)$ is a function $F(x)$ such that $F'(x) = f(x)$. The general antiderivative includes an arbitrary constant:

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

Fundamental Theorem of Calculus

First part. If $f$ is continuous on $[a, b]$ and $F'(x) = f(x)$, then:

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

Second part. If $f$ is continuous on $[a, b]$, then:

$\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$

Definite Integrals

Properties of definite integrals (assuming $f, g$ are integrable):

$\int_a^b \bigl[cf(x)\bigr]\,dx = c\int_a^b f(x)\,dx$

$\int_a^b \bigl[f(x) \pm g(x)\bigr]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$

$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$

$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$

Area Under a Curve

If $f(x) \ge 0$ on $[a, b]$, the area between the curve and the $x$-axis is:

$A = \int_a^b f(x)\,dx$

If $f$ changes sign, compute the integral piecewise. The total enclosed area is:

$A = \int_a^b |f(x)|\,dx$

Area Between Curves

The area between $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$, where $f(x) \ge g(x)$:

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

Standard Integrals

| $f(x)$ | $\displaystyle\int f(x)\,dx$ | | :-------------------------- | :-------------------------------------- | --- | ---- | | $x^n$ | $\dfrac{x^{n+1}}{n+1} + C, \; n \ne -1$ | | | | $\dfrac{1}{x}$ | $\ln | x | + C$ | | $e^x$ | $e^x + C$ | | | | $\sin x$ | $-\cos x + C$ | | | | $\cos x$ | $\sin x + C$ | | | | $\sec^2 x$ | $\tan x + C$ | | | | $\dfrac{1}{\sqrt{1 - x^2}}$ | $\arcsin x + C$ | | | | $\dfrac{1}{1 + x^2}$ | $\arctan x + C$ | | |

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Integration Techniques

Substitution

Theorem. Let $u = g(x)$ where $g'$ is continuous and $f$ is continuous on the range of $g$. Then:

$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$

Strategy. Identify an "inner function" and its derivative as a factor. Substitute $u = g(x)$, replace $dx$ with $\dfrac{du}{g'(x)}$, and integrate with respect to $u$.

Example. Evaluate $\displaystyle\int 2x\sqrt{x^2 + 1}\,dx$.

Let $u = x^2 + 1$, so $du = 2x\,dx$. Then:

$\int 2x\sqrt{x^2 + 1}\,dx = \int \sqrt{u}\,du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(x^2 + 1)^{3/2} + C$

Integration by Parts

Theorem. If $u$ and $v$ are differentiable functions:

$\int u\,dv = uv - \int v\,du$

LIATE priority for choosing $u$: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Choose $u$ as the function that appears highest on this list.

Example. Evaluate $\displaystyle\int x e^x\,dx$.

Let $u = x$, $dv = e^x\,dx$. Then $du = dx$, $v = e^x$:

$\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = (x - 1)e^x + C$

Repeated Integration by Parts

For integrals such as $\int x^n e^x\,dx$ or $\int x^n \sin x\,dx$, apply integration by parts repeatedly. Tabular (DI) integration provides a systematic shortcut: differentiate the algebraic factor until it reaches zero, integrate the other factor, and alternate signs.

Volumes of Revolution

About the $x$-axis. Rotating $y = f(x)$ about the $x$-axis from $x = a$ to $x = b$:

$V = \pi \int_a^b \bigl[f(x)\bigr]^2\,dx$

About the $y$-axis. Rotating $x = g(y)$ about the $y$-axis from $y = c$ to $y = d$:

$V = \pi \int_c^d \bigl[g(y)\bigr]^2\,dy$

Cylindrical shells (rotating about the $y$-axis):

$V = 2\pi \int_a^b x\,|f(x)|\,dx$

Improper Integrals

An improper integral involves an infinite limit of integration or an unbounded integrand. Evaluate as a limit:

$\int_a^\infty f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx$

The integral converges if this limit exists (is finite); otherwise it diverges.

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Common Pitfalls

1. Forgetting the chain rule. When differentiating composite functions, always identify the inner function and multiply by its derivative.

2. Sign errors in the quotient rule. Remember: $u'v - uv'$, not $uv' - u'v$. A mnemonic: "low d-high minus high d-low, over the square of what's below."

3. Dropping the $+C$. Every indefinite integral requires an arbitrary constant.

4. Integration by parts choice. Choosing the wrong factor for $u$ can lead to a more complicated integral. Follow the LIATE rule.

5. Incorrect limits after substitution. When using substitution in a definite integral, either transform the limits to the new variable or back-substitute before evaluating.

6. Volume formula confusion. Ensure the correct axis of rotation is used. The variable in the integrand corresponds to the axis perpendicular to the one of rotation.

7. Treating $\dfrac{dy}{dx}$ as a fraction in implicit differentiation. While the notation suggests a ratio, it is a limit. Manipulations such as cross-multiplying are justified by the chain rule, not by treating derivatives as fractions.

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Practice Problems

Problem 1

Evaluate $\displaystyle\lim_{x \to 0} \frac{\sin 3x}{2x}$.

Problem 2

Find $\dfrac{dy}{dx}$ when $x^3 + y^3 - 3xy = 0$.

Problem 3

A rectangular box with a square base has a surface area of $150\;\mathrm{cm^2}$. Find the dimensions that maximise the volume.

Problem 4

Evaluate $\displaystyle\int_0^1 \frac{x}{\sqrt{x^2 + 1}}\,dx$.

Problem 5

Evaluate $\displaystyle\int x^2 e^{-x}\,dx$ using integration by parts.

Problem 6

Find the volume generated when the region bounded by $y = \sqrt{x}$, $x = 4$, and the $x$-axis is rotated $360^\circ$ about the $x$-axis.

Problem 7

Determine whether $\displaystyle\int_1^\infty \frac{1}{x^p}\,dx$ converges or diverges for $p \gt 0$.

Problem 8

Prove that $f(x) = x^3 - 3x + 1$ has exactly one root in the interval $(1, 2)$.

Answers to Selected Problems

Problem 1: $\displaystyle\lim_{x \to 0} \frac{\sin 3x}{2x} = \lim_{x \to 0} \frac{3}{2} \cdot \frac{\sin 3x}{3x} = \frac{3}{2} \cdot 1 = \frac{3}{2}$.

Problem 2: Differentiating implicitly: $3x^2 + 3y^2\dfrac{dy}{dx} - 3y - 3x\dfrac{dy}{dx} = 0$. Factoring out $\dfrac{dy}{dx}$: $\dfrac{dy}{dx}(3y^2 - 3x) = 3y - 3x^2$, so $\dfrac{dy}{dx} = \dfrac{y - x^2}{y^2 - x}$.

Problem 3: Let base side $= x$ and height $= h$. Surface area: $2x^2 + 4xh = 150$, so $h = \dfrac{75 - x^2}{2x}$. Volume $V = x^2 h = \dfrac{75x - x^3}{2}$. Setting $\dfrac{dV}{dx} = \dfrac{75 - 3x^2}{2} = 0$ gives $x = 5\;\mathrm{cm}$, $h = 5\;\mathrm{cm}$. Second derivative $V'' = -3x$ is negative at $x = 5$, confirming a maximum. Maximum volume is $125\;\mathrm{cm^3}$.

Problem 4: Let $u = x^2 + 1$, $du = 2x\,dx$. $\displaystyle\int_0^1 \frac{x}{\sqrt{x^2 + 1}}\,dx = \frac{1}{2}\int_1^2 u^{-1/2}\,du = \left[\sqrt{u}\right]_1^2 = \sqrt{2} - 1$.

Problem 5: Using tabular integration: differentiate $x^2 \to 2x \to 2 \to 0$, integrate $e^{-x} \to -e^{-x} \to e^{-x} \to -e^{-x}$. Result: $-x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + C = -(x^2 + 2x + 2)e^{-x} + C$.

Problem 7: $\displaystyle\int_1^\infty \frac{1}{x^p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_1^\infty$. For $p \gt 1$: $= 0 - \dfrac{1}{1-p} = \dfrac{1}{p-1}$ (converges). For $p = 1$: $\displaystyle\int_1^\infty \frac{1}{x}\,dx = \lim_{b \to \infty} \ln b = \infty$ (diverges). For $0 \lt p \lt 1$: $x^{1-p} \to \infty$ (diverges). Converges if and only if $p \gt 1$.

Problem 8: $f(1) = -1 \lt 0$ and $f(2) = 3 \gt 0$. By the Intermediate Value Theorem, a root exists in $(1, 2)$. Since $f'(x) = 3x^2 - 3 = 3(x-1)(x+1) \gt 0$ for all $x \in (1, 2)$, $f$ is strictly increasing on $[1, 2]$, so the root is unique.

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Worked Examples

Worked Example: Evaluating a Limit Using L'Hopital's Rule

Evaluate $\displaystyle\lim_{x \to 0} \frac{e^{3x} - 1 - 3x}{x^2}$.

Solution

As $x \to 0$, both numerator and denominator approach $0$, so we have the indeterminate form $\dfrac{0}{0}$. Apply l'H^opital's rule once:

$\lim_{x \to 0} \frac{3e^{3x} - 3}{2x}$

This still gives $\dfrac{0}{0}$, so apply l'H^opital's rule a second time:

$\lim_{x \to 0} \frac{9e^{3x}}{2} = \frac{9}{2}$

Worked Example: Curve Sketching with Full Analysis

Sketch the curve $y = \dfrac{x^2 - 4}{x^2 - 1}$, identifying all asymptotes, stationary points, and intervals of increase and decrease.

Solution

Domain. All $x \in \mathbb{'\{'}R{'\}'}$ except $x = \pm 1$.

Intercepts. $y$-intercept: $x = 0 \implies y = 4$. $x$-intercepts: $x^2 - 4 = 0 \implies x = \pm 2$.

Asymptotes. Vertical: $x = 1$ and $x = -1$ (denominator zero, numerator nonzero). Horizontal: since degree of numerator equals degree of denominator, $y = \dfrac{1}{1} = 1$.

Derivative. By the quotient rule:

$\frac{dy}{dx} = \frac{2x(x^2 - 1) - (x^2 - 4)(2x)}{(x^2 - 1)^2} = \frac{6x}{(x^2 - 1)^2}$

Stationary points. $\dfrac{dy}{dx} = 0 \implies x = 0$. Substituting: $y = 4$. The second derivative at $x = 0$ is negative (since $y'' = \dfrac{6(x^2-1)^2 - 6x \cdot 2(x^2-1) \cdot 2x}{(x^2-1)^4}$, evaluated at $x = 0$ gives $y'' = \dfrac{6}{1} = 6 \gt 0$), confirming a local minimum at $(0, 4)$.

Intervals. $\dfrac{dy}{dx} \lt 0$ for $x \lt 0$ (decreasing), $\dfrac{dy}{dx} \gt 0$ for $x \gt 0$ (increasing), within each branch.

Worked Example: Integration by Parts with Definite Integral

Evaluate $\displaystyle\int_0^{\pi/2} x \cos x\,dx$.

Solution

Let $u = x$, $dv = \cos x\,dx$. Then $du = dx$, $v = \sin x$.

$\int_0^{\pi/2} x \cos x\,dx = \bigl[x \sin x\bigr]_0^{\pi/2} - \int_0^{\pi/2} \sin x\,dx$

$= \frac{\pi}{2} \cdot 1 - 0 - \bigl[-\cos x\bigr]_0^{\pi/2} = \frac{\pi}{2} - (0 - 1) = \frac{\pi}{2} + 1$

Worked Example: Volume of Revolution with Shells

Find the volume generated when the region bounded by $y = x^2$, $y = 0$, and $x = 2$ is rotated $360^\circ$ about the $y$-axis.

Solution

Using cylindrical shells about the $y$-axis:

$V = 2\pi \int_0^2 x \cdot x^2\,dx = 2\pi \int_0^2 x^3\,dx = 2\pi \left[\frac{x^4}{4}\right]_0^2 = 2\pi \cdot 4 = 8\pi$

Worked Example: Implicit Differentiation and Second Derivative

Given $x^3 + y^3 = 6xy$, find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ at the point $(3, 3)$.

Solution

Differentiating implicitly with respect to $x$:

$3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$

Rearranging: $\dfrac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$, so

$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$

At $(3, 3)$: $\dfrac{dy}{dx} = \dfrac{6 - 9}{9 - 6} = \dfrac{-3}{3} = -1$.

For the second derivative, differentiate $\dfrac{dy}{dx}$ using the quotient rule, then substitute $x = 3$, $y = 3$, and $\dfrac{dy}{dx} = -1$. Alternatively, start from

$(3y^2 - 6x)\frac{dy}{dx} = 6y - 3x^2$

Differentiating both sides:

$6y\left(\frac{dy}{dx}\right)^2 + (3y^2 - 6x)\frac{d^2y}{dx^2} - 6 = 6\frac{dy}{dx} - 6x$

At $(3, 3)$ with $\dfrac{dy}{dx} = -1$:

$6(3)(1) + 3 \cdot \frac{d^2y}{dx^2} - 6 = -6 - 18$

$18 + 3\frac{d^2y}{dx^2} - 6 = -24$

$3\frac{d^2y}{dx^2} = -36 \implies \frac{d^2y}{dx^2} = -12$

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Additional Common Pitfalls

• L'H^opital's rule overuse. Only apply when the limit is genuinely indeterminate ($\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$). Applying it to determinate forms produces incorrect results.

• Second derivative test inconclusiveness. When $f''(a) = 0$, the test fails. Use a sign chart of $f'$ around $a$ (the first derivative test) to classify the stationary point.

• Area vs. signed integral. $\displaystyle\int_a^b f(x)\,dx$ gives the signed area. When the curve crosses the $x$-axis, split the integral at each root and take absolute values to find total enclosed area.

• Forgetting substitution limits. In $\displaystyle\int_a^b f(g(x))g'(x)\,dx$, if you substitute $u = g(x)$, change the limits to $g(a)$ and $g(b)$. Failing to do so and using the original limits gives the wrong answer.

• Chain rule on trigonometric functions. $\dfrac{d}{dx}[\sin(f(x))] = f'(x)\cos(f(x))$, not $\cos(f(x))$. Students frequently omit the inner derivative factor.

• Power rule for negative exponents. $\dfrac{d}{dx}\bigl[x^{-3}\bigr] = -3x^{-4}$, not $-3x^{-2}$. Subtract one from the exponent, then multiply by the original exponent.

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Exam-Style Problems

Problem 9

Evaluate $\displaystyle\lim_{x \to \infty} \left(\frac{3x + 1}{2x - 5}\right)^x$.

Problem 10

Find the coordinates of the point on the curve $y = \dfrac{1}{x^2 + 1}$ where the tangent is parallel to the line $x + 8y = 1$.

Problem 11

Evaluate $\displaystyle\int_0^{\pi/4} \sec^2 x\,e^{\tan x}\,dx$.

Problem 12

A cylindrical can with radius $r$ and height $h$ has volume $V = \pi r^2 h = 500\;\mathrm{cm^3}$. Find the values of $r$ and $h$ that minimise the surface area $A = 2\pi r^2 + 2\pi r h$.

Problem 13

Use the substitution $x = 3\sin\theta$ to evaluate $\displaystyle\int_0^{3} \frac{dx}{\sqrt{9 - x^2}}$.

Problem 14

Find the equation of the normal to the curve $y = x^2 e^{-x}$ at the point where $x = 1$.

Problem 15

Show that $\displaystyle\int_0^1 \frac{x}{x^2 + 4}\,dx = \dfrac{1}{2}\ln\!\left(\dfrac{5}{4}\right)$.

Answers to Additional Problems

Problem 9: Write $\left(\dfrac{3x+1}{2x-5}\right)^x = \exp\!\left(x \ln\dfrac{3x+1}{2x-5}\right)$. As $x \to \infty$, $\dfrac{3x+1}{2x-5} \to \dfrac{3}{2}$, and $x\ln\!\left(\dfrac{3}{2} + \dfrac{13}{2(2x-5)}\right) \approx x \cdot \dfrac{13}{3(2x)} = \dfrac{13}{6}$. The limit is $e^{13/6}$.

Problem 10: The line $x + 8y = 1$ has gradient $-\dfrac{1}{8}$. Differentiating: $\dfrac{dy}{dx} = \dfrac{-2x}{(x^2+1)^2}$. Setting $= -\dfrac{1}{8}$: $\dfrac{16x}{(x^2+1)^2} = 1 \implies (x^2+1)^2 = 16x$. Solving: $x^4 + 2x^2 - 16x + 1 = 0$. By inspection $x = 1$ is a root: $1 + 2 - 16 + 1 = -12 \ne 0$. Trying $x = \dfrac{1}{2}$: $\dfrac{1}{16} + \dfrac{1}{2} - 8 + 1 \ne 0$. Since $\dfrac{dy}{dx} = -\dfrac{1}{8}$ at $x = 2$: $\dfrac{16}{25} \ne 1$. At $x = 1$: $\dfrac{2}{4} = \dfrac{1}{2} \ne \dfrac{1}{8}$. At $x = \sqrt[3]{2} - 1$ check numerically: $x \approx 0.26$, $\dfrac{dy}{dx} \approx -0.48$. Solving numerically gives $x \approx 0.065$ and $x \approx 1.48$. At $x = 1.48$: $\dfrac{dy}{dx} \approx -0.124$ close to $-\dfrac{1}{8} = -0.125$.

Problem 11: Let $u = \tan x$, so $du = \sec^2 x\,dx$. Limits: $x = 0 \implies u = 0$; $x = \pi/4 \implies u = 1$. $\displaystyle\int_0^1 e^u\,du = \bigl[e^u\bigr]_0^1 = e - 1$.

Problem 12: From $V = 500$: $h = \dfrac{500}{\pi r^2}$. Then $A = 2\pi r^2 + \dfrac{1000}{r}$. $\dfrac{dA}{dr} = 4\pi r - \dfrac{1000}{r^2} = 0 \implies r^3 = \dfrac{250}{\pi} \implies r = \left(\dfrac{250}{\pi}\right)^{1/3} \approx 4.30\;\mathrm{cm}$. Then $h = \dfrac{500}{\pi r^2} = 2r \approx 8.60\;\mathrm{cm}$. Checking $A'' = 4\pi + \dfrac{2000}{r^3} \gt 0$, confirming a minimum.

Problem 13: Let $x = 3\sin\theta$, $dx = 3\cos\theta\,d\theta$. When $x = 0$, $\theta = 0$; when $x = 3$, $\theta = \pi/2$. $\displaystyle\int_0^{\pi/2} \frac{3\cos\theta}{3\cos\theta}\,d\theta = \int_0^{\pi/2} d\theta = \frac{\pi}{2}$.

Problem 14: $y = x^2 e^{-x}$, $\dfrac{dy}{dx} = 2xe^{-x} - x^2 e^{-x} = e^{-x}(2x - x^2)$. At $x = 1$: $y = e^{-1} = \dfrac{1}{e}$, $\dfrac{dy}{dx} = e^{-1}(2 - 1) = \dfrac{1}{e}$. Normal gradient $= -e$. Normal equation: $y - \dfrac{1}{e} = -e(x - 1)$, i.e. $y = -ex + e + \dfrac{1}{e}$.

Problem 15: Let $u = x^2 + 4$, $du = 2x\,dx$. $\displaystyle\int_0^1 \frac{x\,dx}{x^2 + 4} = \frac{1}{2}\int_4^5 \frac{du}{u} = \frac{1}{2}\bigl[\ln u\bigr]_4^5 = \frac{1}{2}\ln\!\left(\frac{5}{4}\right)$.

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If You Get These Wrong, Revise:

• Differentiation rules and chain rule → Review ./calculus (sections on Differentiation Rules and Implicit Differentiation)
• Integration techniques → Review ./calculus (sections on Substitution and Integration by Parts)
• Trigonometric identities → Review the geometry and trigonometry topics in this subject
• Exponential and logarithmic functions → Review algebra and functions material
• Curve sketching fundamentals → Review ./calculus (section on Curve Sketching)

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Related Content at Other Levels

• A-Level Differentiation and Integration: Mathematics
• University Calculus: Multivariable Calculus