First-year calculus

The first-year exam: limits, derivatives, integrals, differential equations and series.

Tasks of typical first-semester tests and exams. The answer is a number or a formula; any equivalent formula is accepted.

Limits OpenClose

The forms 0/0 and ∞/∞: factor, cancel, divide by the highest power.

How to solve it

Limits

The limit $\lim_{x \to a} f(x)$ is the number the values of the function approach as $x$ approaches $a$. Often it is enough to substitute $a$. Trouble starts when substitution gives nonsense like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Step by step

Substitute the point. If you get a number, that is the answer.
If you get $\frac{0}{0}$, factor the numerator and the denominator, cancel the common factor and substitute again.
If $x \to \infty$ and you get $\frac{\infty}{\infty}$, divide the numerator and the denominator by the highest power of $x$: everything divided by $x$ tends to zero.
For a ratio of polynomials of the same degree the limit at infinity is the ratio of the leading coefficients.

The leading coefficient of the numerator.

The leading coefficient of the denominator. The degrees are equal; if the denominator's degree is higher, the limit is $0$.

An example with $\frac{0}{0}$: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x + 2) = 4$.

Common mistakes

Writing that $\frac{0}{0}$ equals $0$ or $1$. It is not a number but a signal to transform the expression.
Dividing only some of the terms by the highest power.
Comparing the wrong degrees: if the numerator's degree is lower than the denominator's, the limit at infinity is zero.

Example

Derivatives OpenClose

The table of derivatives and the rules: sum, product, quotient, the chain rule.

How to solve it

Derivatives

The derivative shows how fast a function changes. It is found not from the definition but from a table and a few rules: the sum, the product, the quotient and the chain rule.

Step by step

The table: $(x^n)' = nx^{n - 1}$, $(e^x)' = e^x$, $(\sin x)' = \cos x$, $(\cos x)' = -\sin x$, $(\ln x)' = \frac{1}{x}$, $(\sqrt{x})' = \frac{1}{2\sqrt{x}}$.
The derivative of a sum is the sum of the derivatives; a constant factor comes out of the derivative.
The product: $(uv)' = u'v + uv'$. The quotient: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$.
The chain rule: the derivative of the outer function times the derivative of the inner one: $(\sin 3x)' = 3\cos 3x$.

The first factor.

The second factor. Each is differentiated in turn while the other stays as it is.

Example: $(x^2 e^{x})' = 2x e^{x} + x^2 e^{x}$.

The inner function. Its derivative appears as a factor.

Example: $\left(e^{x^2}\right)' = e^{x^2} \cdot 2x$.

Common mistakes

Thinking that the derivative of a product is the product of the derivatives. It is not.
Forgetting the factor from the inner function: $(\cos 5x)' = -5\sin 5x$.
Getting the sign of the cosine's derivative wrong: $(\cos x)' = -\sin x$.

Example

Antiderivatives OpenClose

The table of integrals, substitution, integration by parts.

How to solve it

Antiderivatives

An antiderivative is a function $F$ whose derivative is the given one: $F' = f$. Finding it means reading the table of derivatives backwards. The answer can always be checked: differentiate it and compare with what was under the integral.

Step by step

The table: $\int x^n\,dx = \frac{x^{n + 1}}{n + 1} + C$ for $n \ne -1$, $\int \frac{dx}{x} = \ln|x| + C$, $\int \cos x\,dx = \sin x + C$, $\int e^x\,dx = e^x + C$.
When the function is applied to $kx + b$, divide the result by $k$: $\int \cos 3x\,dx = \frac{\sin 3x}{3} + C$.
When the derivative of an inner function stands next to it, substitute $t = g(x)$: $\int 2x e^{x^2}\,dx = e^{x^2} + C$.
A polynomial times $e^x$, $\sin x$ or $\cos x$ is integrated by parts.
Check the answer by differentiating it.

What gets simpler when differentiated: a polynomial or a logarithm.

What is easy to integrate: $e^x$, $\sin x$, $\cos x$.

Example: $\int x e^{x}\,dx = x e^{x} - \int e^{x}\,dx = x e^{x} - e^{x} + C$.

Common mistakes

Dividing by $n$ instead of $n + 1$: $\int x^3\,dx = \frac{x^4}{4} + C$.
Forgetting to divide by $k$ after a linear substitution.
Not checking the answer, although the check by differentiation takes a minute.

Example

Definite integrals OpenClose

The Newton–Leibniz formula: the antiderivative at the ends of the segment.

How to solve it

Definite integrals

The definite integral $\int_a^b f(x)\,dx$ is the signed area under the graph. It is computed without any areas at all: find an antiderivative and subtract its values at the ends of the segment.

Step by step

Find an antiderivative $F(x)$, as in the antiderivative tasks. The constant $C$ is not needed: it cancels in the subtraction.
Substitute the upper limit, then the lower one, and subtract: $F(b) - F(a)$.
Subtract $F(a)$ as a whole, in brackets: it may have minuses of its own.

The lower limit. The antiderivative's value there is subtracted.

The upper limit.

Example: $\int_0^2 3x^2\,dx = x^3 \Big|_0^2 = 8 - 0 = 8$.

Common mistakes

Subtracting the other way round: $F(a) - F(b)$ has the opposite sign.
Losing a minus when $F(a)$ is negative: $F(b) - (-3) = F(b) + 3$.
Making a mistake in the antiderivative itself. Check it by differentiating before substituting the limits.

Example

Differential equations OpenClose

Initial value problems for linear equations with constant coefficients.

How to solve it

Differential equations

In a differential equation the unknown is a function $y(x)$, and the equation involves its derivatives: $y'' - 3y' + 2y = 0$. An initial value problem adds conditions, for example $y(0) = 1$ and $y'(0) = 0$, and with them the solution becomes unique.

Step by step

For $y' = ky$ the solution is $y = Ce^{kx}$; find the constant $C$ from $y(0)$.
For $y'' + py' + qy = 0$ write the characteristic equation $r^2 + pr + q = 0$: replace $y''$ by $r^2$, $y'$ by $r$ and $y$ by $1$.
If it has two different real roots $r_1$ and $r_2$, the general solution is $y = C_1 e^{r_1 x} + C_2 e^{r_2 x}$.
Substitute $x = 0$ into $y$ and into $y'$, and find $C_1$ and $C_2$ from the resulting system.

The coefficient of $y'$ becomes the coefficient of $r$.

The coefficient of $y$ becomes the constant term.

Example: $y'' - 3y' + 2y = 0$ gives $r^2 - 3r + 2 = 0$ with roots $1$ and $2$, so the general solution is $y = C_1 e^{x} + C_2 e^{2x}$.

Common mistakes

Getting the sign of a root in the exponent wrong: the root $r = -2$ gives $e^{-2x}$.
Finding the general solution and forgetting the initial conditions.
Differentiating wrongly when using the second condition: $\left(C_2 e^{2x}\right)' = 2C_2 e^{2x}$.

Example

Series OpenClose

The sum of an infinite geometric progression.

How to solve it

Geometric series

The infinite sum $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ equals a finite number, $2$: each new term covers half of the remaining way. Every geometric progression whose ratio is less than one in absolute value behaves like this.

Step by step

Find the first term $b_1$, the term the sum starts with. Look carefully at the lower index of the sum: if it starts at $k = 2$, the first term is the one with $k = 2$.
Find the ratio $q$: how many times each term is larger than the previous one.
Check that $|q| < 1$. Otherwise the series diverges and has no sum.
Compute the sum by the formula.

The first term of the sum.

The ratio of the progression, $|q| < 1$.

Example: $\sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k = \frac{1}{1 - \frac{1}{2}} = 2$.

Common mistakes

Taking the term at $k = 0$ as the first one, although the sum starts at $k = 2$.
Getting the sign wrong for a negative ratio: $1 - \left(-\frac{1}{2}\right) = \frac{3}{2}$.
Using the formula when $|q| \ge 1$ and the series diverges.