Calculus: free practice, theory and problems

1. Differentiation

What is it about?

The derivative $f'(x)$ measures how fast a quantity changes. Geometrically, it is the slope of the tangent line at a point. For an engineer it is everywhere: velocity is the derivative of position, current is the derivative of charge, and power is the derivative of energy. Whenever you look for an optimum – least material, highest power, lowest cost – you set the derivative equal to zero.

Formally, the derivative is the limit of the slope over a small interval:

Key concepts and formulas

• Power rule: $(x^n)' = nx^{n-1}$. Constant factors stay, constant terms vanish: $(5x^3 + 7)' = 15x^2$.
• Standard derivatives: $(e^x)' = e^x$, $(\ln x)' = \frac1x$, $(\sin x)' = \cos x$, $(\cos x)' = -\sin x$, $(\arctan x)' = \frac{1}{1+x^2}$.
• Product rule: $(fg)' = f'g + fg'$.
• Quotient rule: $\left(\frac fg\right)' = \frac{f'g - fg'}{g^2}$.
• Chain rule: $\big(f(u(x))\big)' = f'(u)\cdot u'(x)$ – outer derivative times inner derivative.
• $f' > 0$: $f$ is increasing. $f' < 0$: $f$ is decreasing. $f'' > 0$: the graph is concave up. $f'' < 0$: concave down.
• Tangent line at $x = a$: $y = f(a) + f'(a)(x - a)$.

How to solve the problems

1. Identify the outermost operation in the expression (sum, product, quotient or composition) and choose the rule accordingly.
2. Differentiate piece by piece. Write out the intermediate steps, especially the inner derivative in the chain rule.
3. Extreme points: solve $f'(x) = 0$. Check the sign of $f'$ on each side, or the sign of $f''$ at the point.
4. On a closed interval $[a, b]$ you must also check the endpoints.
5. Optimization with two unknowns: use the constraint to express everything in one variable before differentiating.

Example

Find the maximum of $f(x) = x\,e^{-2x}$ for $x \ge 0$.

1. Product rule with the chain rule: $f'(x) = 1\cdot e^{-2x} + x\cdot(-2)e^{-2x} = e^{-2x}(1 - 2x)$.
2. $e^{-2x}$ is never zero, so $f'(x) = 0$ gives $1 - 2x = 0$, i.e. $x = \frac12$.
3. Signs: $f' > 0$ for $x < \frac12$ and $f' < 0$ for $x > \frac12$. It is a maximum.
4. The value: $f\left(\frac12\right) = \frac12e^{-1} \approx 0.184$.

Answer: a maximum at $x = 0.5$, with a function value of about $0.184$.

Common mistakes

• Forgetting the inner derivative: $(\sin 3x)' = 3\cos 3x$, not $\cos 3x$.
• Differentiating a product factor by factor: $(fg)' \neq f'g'$.
• Sign errors in the quotient rule: the numerator is $f'g - fg'$, in that order.
• Believing that $f'(a) = 0$ always means a maximum or minimum. $f(x) = x^3$ has $f'(0) = 0$, but a horizontal inflection point.
• Forgetting the endpoints when looking for the largest and smallest value on an interval.

Concepts in this part

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2. Integration

What is it about?

Integration is the reverse of differentiation, and it is the tool for adding up something that changes. If you know the velocity $v(t)$, the integral gives the distance traveled. If you know a force $F(x)$ along a path, the integral gives the work. Geometrically, $\int_a^b f(x)\,dx$ is the signed area between the graph and the $x$-axis: area above the axis counts as positive, area below as negative.

Key concepts and formulas

• $F$ is an antiderivative of $f$ if $F' = f$. The indefinite integral is $\int f(x)\,dx = F(x) + C$.
• Fundamental theorem of calculus: $\int_a^b f(x)\,dx = F(b) - F(a)$.
• Table: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$, $\int \frac1x\,dx = \ln|x| + C$, $\int e^{kx}\,dx = \frac1k e^{kx} + C$, $\int \sin kx\,dx = -\frac1k\cos kx + C$ and $\int \cos kx\,dx = \frac1k\sin kx + C$.
• Substitution: with $u = g(x)$ we have $du = g'(x)\,dx$, and $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$.
• Integration by parts: $\int u\,v'\,dx = uv - \int u'\,v\,dx$.
• Area between curves: $\int_a^b \big(f(x) - g(x)\big)\,dx$ when $f \ge g$.
• Mean value: $\bar f = \frac{1}{b-a}\int_a^b f(x)\,dx$.

How to solve the problems

1. Check the table first. Rewrite the expression (expand, split fractions) if that helps.
2. If you see a composite function multiplied by (roughly) the derivative of the inner function, use substitution.
3. A product such as a polynomial times $e^x$, $\sin x$ or $\ln x$: use integration by parts. Let $u$ be the factor that gets simpler when differentiated (but for $\ln x$: let $u = \ln x$).
4. Definite integral: insert the limits into the antiderivative. With substitution, the limits must be converted too.
5. Check by differentiating your answer.

Example

Evaluate $\int_0^{\pi} x\sin x\,dx$.

1. Integration by parts with $u = x$ and $v' = \sin x$. Then $u' = 1$ and $v = -\cos x$.
2. $\int x\sin x\,dx = -x\cos x + \int \cos x\,dx = -x\cos x + \sin x + C$.
3. Insert the limits: $\big[-x\cos x + \sin x\big]_0^{\pi} = \big(-\pi\cdot(-1) + 0\big) - (0 + 0) = \pi$.

The answer is $\pi \approx 3.14$. Check: $(-x\cos x + \sin x)' = -\cos x + x\sin x + \cos x = x\sin x$.

Common mistakes

• Forgetting $+ C$ in indefinite integrals.
• Applying the power rule to $\frac1x$. It gives $\frac{x^0}{0}$, which is meaningless. The correct result is $\ln|x|$.
• Forgetting the factor $\frac1k$: $\int e^{3x}\,dx = \frac13 e^{3x} + C$.
• Keeping the old limits after a substitution.
• Interpreting the integral as the total area when the function changes sign. Then the interval must be split.

Concepts in this part

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3. Limits, series and complex numbers

What is it about?

This unit collects three tools that come back throughout your studies. Limits describe what a function approaches, for example the value a signal settles at. Series and Taylor polynomials replace a difficult function with a polynomial that is easy to compute with – which is how calculators and simulation software actually work. Complex numbers make oscillations and AC circuits simple: a rotation becomes a multiplication.

Key concepts and formulas

• Standard limit: $\lim_{x\to 0}\frac{\sin x}{x} = 1$. As $x\to\infty$, the highest powers determine the limit of a ratio of polynomials.
• L'Hôpital's rule: for the forms $\frac00$ or $\frac\infty\infty$, $\lim\frac fg = \lim\frac{f'}{g'}$.
• Geometric series: $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$ when $|r| < 1$. Otherwise it diverges.
• Taylor polynomial about 0: $f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$
• Known series: $e^x = 1 + x + \frac{x^2}{2!} + \dots$, $\sin x = x - \frac{x^3}{3!} + \dots$ and $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
• Complex number: $z = a + bi$ with $i^2 = -1$. Modulus $|z| = \sqrt{a^2 + b^2}$. The argument $\theta$ is the angle from the positive real axis.
• Polar form and Euler's formula: $z = re^{i\theta} = r(\cos\theta + i\sin\theta)$, and $z^n = r^ne^{in\theta}$. In a multiplication the lengths are multiplied and the angles are added.
• Division: multiply by the conjugate of the denominator, $\frac{z}{w} = \frac{z\bar w}{|w|^2}$. In electrical engineering $j$ is often written instead of $i$.

How to solve the problems

1. Limit: substitute the value first. If you get a number, you are done. If you get $\frac00$ or $\frac\infty\infty$, use factoring, L'Hôpital or Taylor.
2. Series: identify the type and check the convergence condition before using a sum formula.
3. Taylor: often you can just substitute into a known series, for example $e^{2x} = 1 + 2x + 2x^2 + \dots$
4. Complex numbers: use $a + bi$ for addition and polar form for multiplication, division and powers. Sketch the number to get the right quadrant for the argument.

Example

a) Find $\lim_{x\to 0}\frac{1 - \cos x}{x^2}$. Substitution gives $\frac00$. Taylor: $1 - \cos x = \frac{x^2}{2} - \frac{x^4}{24} + \dots$, so the fraction is $\frac12 - \frac{x^2}{24} + \dots$, which tends to $\frac12$. Applying L'Hôpital twice gives the same result.

b) Find $z^6$ for $z = 1 + i\sqrt3$. Here $|z| = \sqrt{1 + 3} = 2$ and $\theta = \arctan\sqrt3 = 60^\circ$. Then $z^6 = 2^6e^{i\cdot 360^\circ} = 64$.

Common mistakes

• Using L'Hôpital when the expression is not of the form $\frac00$ or $\frac\infty\infty$.
• Using $\frac{1}{1-r}$ when $|r| \ge 1$.
• Forgetting the factorials in Taylor series.
• Computing the argument as $\arctan\frac ba$ without checking the quadrant: $-1 - i$ has argument $-135^\circ$, not $45^\circ$.
• Multiplying out complex numbers without using $i^2 = -1$.

Concepts in this part

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4. Applications of derivatives

What is it about?

The derivative tells you how fast something changes. Engineers use this for three things all the time: finding the largest or smallest value (optimization), estimating a value quickly (linear approximation) and solving equations numerically (Newton's method).

Concepts and formulas

• Critical point: where $f'(x) = 0$. The function may have a maximum or a minimum here.
• Second derivative test: $f''(x_0) > 0$ gives a minimum (the graph smiles), $f''(x_0) < 0$ gives a maximum (the graph frowns).
• On a closed interval $[a, b]$ you must check the critical points and the end points.
• Linear approximation near $x = a$: replace the curve by its tangent line.

• Newton's method for solving $f(x) = 0$: start with a guess $x_0$ and improve it again and again.

• Related rates: when two quantities are linked, their rates of change are linked too. Differentiate the relation with respect to time $t$ (chain rule).

How to solve the problems

1. Write the quantity you want to maximize or minimize as a function of one variable. Use the condition in the problem to eliminate the others.
2. Differentiate and set $f'(x) = 0$.
3. Check that you have a maximum or minimum (sign chart or $f''$), and check the end points.
4. Answer what the problem actually asks for, for example the area and not just $x$.

Example

A rectangular area along a river is to be fenced in. No fence is needed along the river, so you need fence on three sides. You have 100 m of fence. How large can the area be?

1. Call the sides perpendicular to the river $x$ (two of them) and the side along the river $y$. Then $2x + y = 100$, so $y = 100 - 2x$.
2. The area is $A(x) = x(100 - 2x) = 100x - 2x^2$.
3. $A'(x) = 100 - 4x = 0$ gives $x = 25$ m. $A''(x) = -4 < 0$, so it is a maximum.
4. Then $y = 50$ m and $A = 25\cdot 50 = 1250$ m².

Common mistakes

• Forgetting the end points when the variable is limited to an interval.
• Differentiating before reducing to one variable.
• Answering with the $x$-value when the problem asks for the area, or the other way round.
• In Newton's method: flipping the sign ($x_n + f/f'$), or starting at a point where $f'(x_0) = 0$.

Concepts in this part

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