Integral Calculus

Integral calculus was worth 14 marks on NESA's sample paper for the 2024 syllabus, and integration also turns up inside Applications of calculus (motion, rates) and Random variables (probability density functions). Expect both short "find" questions and multi-mark area problems.

• Primitives: reverse differentiation, the constant $C$, finding $f(x)$ from $f^{\prime }(x)$ and a point.
• The definite integral: a limit of sums $\sum f(x)\Delta x$, signed area, and the Fundamental Theorem of Calculus.
• Indefinite integrals of $x^n$, $(ax+b{)}^n$, $e^{ax+b}$, $a^x$, $\frac{1}{x}$, $\frac{1}{ax+b}$, $\sin$, $\cos$, ${\sec }^2$, and reverse chain rule forms such as $\int f^{\prime }(x)[f(x){]}^{n}dx$.
• Areas: between a curve and the $x$-axis or $y$-axis (including when the function changes sign) and between two curves.
• Properties: additivity, reversing limits, odd and even symmetry.
• The trapezoidal rule for approximating integrals, and using technology for integrals outside the course.

A primitive of $f(x)$ is a function $F(x)$ with $F^{\prime }(x)=f(x)$. Primitives are not unique: any two differ by a constant, so the general primitive (the indefinite integral) is written

Integration is linear: integrate term by term and pull constants out the front.

Standard integrals

The reverse chain rule

If the integrand is "a function of $f(x)$, times $f^{\prime }(x)$", integrate the outside function and leave $f(x)$ inside:

The same pattern works for $\int f^{\prime }(x)\sin f(x)dx=-\cos f(x)+C$, and similarly for $\cos$ and ${\sec }^2$. If the derivative is out by a constant factor, adjust: $\int x^3\cos (x^4)dx=\frac{1}{4}\sin (x^4)+C$.

Each result can be proved by differentiating the right-hand side. Doing that is also the best way to check an answer.

Worked example 1

Find each integral.

(a) $\int (3x-1{)}^{4}dx=\frac{(3x-1{)}^5}{15}+C$, since $a(n+1)=3\times 5$.

(b) The numerator $x^2$ is $\frac{1}{3}$ of the derivative of $x^3+4$, so

(c) $\int \sin (4x+1)dx=-\frac{1}{4}\cos (4x+1)+C$.

(d) Split the fraction first, then integrate term by term:

Finding a function from its derivative

Given $f^{\prime }(x)$ and a point, integrate to get $f(x)+C$, then substitute the point to find $C$. Never leave $C$ unknown when a condition is given.

If $f(x)$ is a rate of change, then $f(x)\Delta x$ is the change over a small interval and $\sum f(x)\Delta x$ estimates the total change from $x=a$ to $x=b$. The definite integral is the limit of this sum as $\Delta x\to 0$:

Geometrically it is a signed area: area above the $x$-axis counts as positive and area below counts as negative.

The Fundamental Theorem of Calculus

If $f$ is continuous on $[a,b]$ and $F$ is any primitive of $f$, then

The theorem connects the two halves of calculus: the area function $A(x)={\int }_a^{x}f(t)dt$ is a primitive of $f$, so $A^{\prime }(x)=f(x)$.

Properties

Symmetry. If $f$ is odd ($f(-x)=-f(x)$, such as $x^3$, $\sin x$, $\tan x$), then ${\int }_{-a}^{a}f(x)dx=0$. If $f$ is even ($f(-x)=f(x)$, such as $x^2$, $\cos x$), then ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$. Split a mixed integrand into its odd and even parts first.

The integral gives signed area, but a question asking for area wants a positive answer. So:

1. Find where the curve crosses the axis in the interval (sketch it).
2. Integrate separately over each piece.
3. Take the absolute value of each negative piece and add.

Worked example 2

Find the area between $y=x^2-4x+3$ and the $x$-axis from $x=0$ to $x=4$, and compare it with ${\int }_0^4(x^2-4x+3)dx$.

The curve is $y=(x-1)(x-3)$, crossing at $x=1$ and $x=3$; it is below the axis between them. With $F(x)=\frac{x^3}{3}-2x^2+3x$:

But ${\int }_0^{4}f(x)dx=\frac{4}{3}$, because the middle piece cancels one of the others. The single integral is the signed area, not the area.

Between two curves

If $f(x)\ge g(x)$ on $[a,b]$, the area between them is

It is always "top minus bottom", and it does not matter whether the curves are above or below the $x$-axis.

Worked example 3

Find the area enclosed by $y=6-x^2$ and $y=x$.

Intersections: $6-x^2=x$, so $x^2+x-6=0$ and $(x+3)(x-2)=0$, giving $x=-3$ and $x=2$. Test $x=0$: the parabola ($y=6$) is above the line ($y=0$).

Areas with the y-axis

For a region bounded by a curve, the $y$-axis and horizontal lines $y=a$ and $y=b$, either rewrite the curve as $x=g(y)$ and use ${\int }_a^{b}g(y)dy$, or work with $x$ and subtract from a rectangle. For exponential and log curves, the syllabus expects you to use the fact that $y=a^x$ and $y={\log }_{a}x$ are reflections of each other in $y=x$, so an area against one axis for one curve equals an area against the other axis for the other.

Example. The region between $y=e^x$, the $y$-axis and the line $y=4$ lies between $x=0$ and $x=\ln 4$, under the line and above the curve:

By the reflection property, this equals ${\int }_1^4\ln ydy$.

When a function can't be integrated exactly, or only a table of values is given, approximate the area with trapezia of width $h=\frac{b-a}{n}$:

The end values are counted once and every middle value twice. Note that $n$ subintervals use $n+1$ function values.

Worked example 4

Use the trapezoidal rule with 4 subintervals to estimate ${\int }_0^1{e}^{-x^2}dx$.

Here $h=0.25$. The function values (4 d.p.) at $x=0,0.25,0.5,0.75,1$ are $1$, $0.9394$, $0.7788$, $0.5698$, $0.3679$.

The true value (from technology) is about 0.7468. The rule overestimates where the curve is concave up and underestimates where it is concave down. More subintervals give a better estimate.

• Forgetting $+C$ on an indefinite integral (often a lost mark).
• Dividing by the wrong thing: $\int e^{3x}dx=\frac{1}{3}{e}^{3x}+C$, not $3e^{3x}+C$.
• Writing $\int \frac{1}{x^2}dx=\ln (x^2)$. Only $\frac{1}{x}$ (or $\frac{f^{\prime }}{f}$) integrates to a logarithm; $x^{-2}$ uses the power rule.
• Finding "area" with a single integral across a root, so positive and negative parts cancel.
• Using "bottom minus top" between curves, giving a negative area.
• Using $n$ instead of $n+1$ function values in the trapezoidal rule, or forgetting to double the middle values.
• Using degrees with trig integrals.

• Find / evaluate: show the primitive in square brackets with the limits, then the substitution, then the value. Exact answers (fractions, $\ln$, $\pi$) unless told otherwise.
• Area questions: draw a quick sketch even if not asked. Marks are commonly given for finding intersection points, setting up the correct integral(s), and evaluating.
• Show that an integral equals a value: differentiate a given expression first if the question provides one ("hence" means use it).
• If a question gives $\frac{d}{dx}$ of something, it is almost always a hint for a later integral.
• Give units² for areas when the context has units.
• For the trapezoidal rule, set out a table of $x$ and $f(x)$ values; it is easy to mark and easy to check.

• MA12-DC Differential calculus: every integral here is a derivative read backwards.
• MA12-AC Applications of calculus: integrating velocity gives displacement, and integrating a rate gives a total change.
• MA12-RV Random variables: probabilities, the CDF, the mean and the variance of continuous random variables are all definite integrals of a PDF.
• MA11-FN and MA11-EL: graph sketching, factorising and log laws are needed to set up and simplify area problems.