Mathematics Admissions Test (MAT) Calculus Syllabus

For what values of $x$ is a function $f$ increasing?

Where $f'(x) > 0$ — the curve rises as $x$ increases.

For what values of $x$ is a function $f$ decreasing?

Where $f'(x) < 0$ — the curve falls as $x$ increases.

What does it mean for a function to be increasing on an interval in terms of values?

For any $x_{1} < x_{2}$ in the interval, $f(x_{1}) \leq f(x_{2})$ (strictly increasing if $f(x_{1}) < f(x_{2})$).

How do you find the intervals where $f(x)=x^{3}-3x$ is increasing?

$f'(x)=3x^{2}-3 = 3(x-1)(x+1)$. This is positive for $x<-1$ and $x>1$, so $f$ is increasing on $(-\infty,-1)$ and $(1,\infty)$.

What is the indefinite integral (antiderivative) power rule $\displaystyle\int x^{n}\,dx$?

$\displaystyle\int x^{n}\,dx = \dfrac{x^{n+1}}{n+1} + C$, valid for $n \neq -1$.

Why must you always add a constant $C$ to an indefinite integral?

Differentiation removes constants, so any antiderivative is determined only up to an additive constant; $+C$ represents the whole family of antiderivatives.

Evaluate $\displaystyle\int (6x^{2} - 4x + 5)\,dx$.

$2x^{3} - 2x^{2} + 5x + C$ (integrate term by term).

Evaluate $\displaystyle\int \dfrac{1}{x^{2}}\,dx$.

Write as $\int x^{-2}\,dx = -x^{-1} + C = -\dfrac{1}{x} + C$.

If $\dfrac{dy}{dx} = 3x^{2}$ and the curve passes through $(1,4)$, find $y$.

$y = x^{3} + C$; using $(1,4)$: $4 = 1 + C$, so $C = 3$ and $y = x^{3} + 3$.

State the Fundamental Theorem of Calculus linking integration and differentiation.

If $F'(x) = f(x)$, then $\displaystyle\int_{a}^{b} f(x)\,dx = F(b) - F(a)$. Differentiation and definite integration are inverse processes.

How do you evaluate a definite integral $\displaystyle\int_{a}^{b} f(x)\,dx$?

Find an antiderivative $F$, then compute $\big[F(x)\big]_{a}^{b} = F(b) - F(a)$. No constant of integration is needed.

Evaluate $\displaystyle\int_{1}^{2} 3x^{2}\,dx$.

$\big[x^{3}\big]_{1}^{2} = 8 - 1 = 7$.

What is the value of $\displaystyle\int_{a}^{a} f(x)\,dx$?

$0$ — integrating over a zero-width interval gives zero.

How does swapping the limits affect a definite integral, i.e. $\displaystyle\int_{b}^{a} f(x)\,dx$?

It negates the value: $\displaystyle\int_{b}^{a} f(x)\,dx = -\int_{a}^{b} f(x)\,dx$.

State the linearity / additivity property $\displaystyle\int_{a}^{c} f\,dx + \int_{c}^{b} f\,dx = \,?$

$\displaystyle\int_{a}^{b} f(x)\,dx$ — adjacent intervals combine.

What does the definite integral $\displaystyle\int_{a}^{b} f(x)\,dx$ represent when $f(x)\geq 0$ on $[a,b]$?

The area between the curve $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$.

How do you find the area enclosed between a curve and the $x$-axis when the curve dips below the axis?

The integral counts area below the axis as negative. Split at the roots and integrate each region separately, taking the absolute value of negative parts, then sum the magnitudes.

What is the formula for the area between two curves $y=f(x)$ (upper) and $y=g(x)$ (lower) from $x=a$ to $x=b$?

$\displaystyle\int_{a}^{b} \big(f(x) - g(x)\big)\,dx$, where $f(x) \geq g(x)$ on $[a,b]$.

What is the first step in finding the area enclosed between two intersecting curves?

Find the intersection points by solving $f(x) = g(x)$; these give the limits of integration $a$ and $b$.

Compute the area between $y = x$ and $y = x^{2}$ for $0 \le x \le 1$.

$\displaystyle\int_{0}^{1}(x - x^{2})\,dx = \Big[\tfrac{x^{2}}{2} - \tfrac{x^{3}}{3}\Big]_{0}^{1} = \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6}$.

On a graph of $y=f(x)$, what feature corresponds to a root of $f'(x)=0$?

A stationary point (turning point or stationary inflection) of $y=f(x)$ — a peak, trough, or flat inflection.

If the gradient graph $y=f'(x)$ is positive on an interval, what does the original graph $y=f(x)$ do there?

It increases (rises) on that interval.

How can you spot a point of inflection of $y=f(x)$ from the second derivative?

It occurs where $f''(x)=0$ and $f''$ changes sign (concavity changes from concave up to concave down or vice versa).

What does $f''(x) > 0$ tell you about the shape (concavity) of the curve $y=f(x)$?

The curve is concave up (convex), shaped like a valley; the gradient is increasing.

What does $f''(x) < 0$ tell you about the concavity of $y=f(x)$?

The curve is concave down, shaped like a hill; the gradient is decreasing.

If $y=f'(x)$ crosses the $x$-axis from positive to negative, what does the original curve $y=f(x)$ have there?

A local maximum (gradient goes from $+$ to $0$ to $-$).

How can a definite integral be estimated geometrically without exact antiderivatives?

Approximate the area with simple shapes — e.g. rectangles (left/right/midpoint Riemann sums) or trapezia (trapezium rule) — and sum their areas.

For a decreasing positive function on $[a,b]$, how do left- and right-endpoint rectangle sums bound the true integral?

The left-endpoint sum overestimates (upper bound) and the right-endpoint sum underestimates (lower bound) the integral; for an increasing function the roles reverse.

How can the comparison $f(x) \leq g(x)$ on $[a,b]$ give bounds on an integral?

If $f(x) \leq g(x)$ for all $x\in[a,b]$, then $\displaystyle\int_{a}^{b} f(x)\,dx \leq \int_{a}^{b} g(x)\,dx$ — integrals preserve inequalities, giving usable bounds.

In a multi-step optimisation problem (e.g. minimise the surface area of an open box of fixed volume), how is the constraint used?

Use the constraint (e.g. fixed volume $V = x^{2}h$) to express one variable in terms of another, substitute to write the target quantity (e.g. surface area) as a single-variable function, then differentiate, set to zero, solve, and classify the stationary point as the optimum.