Integration and Area | A-Maths Study Guide

Secondary 4 Additional Mathematics · Study guide

Reverse differentiation, prepare difficult integrands and distinguish signed accumulation from geometric area.

Concept → worked examples → practice30 min read · practice linked below

Integration reverses differentiation

An antiderivative of $f(x)$ is a function $F(x)$ with $F^{\prime}(x)=f(x)$ . We write $\int f(x)\,dx=F(x)+C$ . Every constant differentiates to zero, so one derivative corresponds to a family of functions. A given point determines the constant for the required curve.

The most reliable check is to differentiate your answer. Work on an interval where the integrand is defined; do not integrate straight across a vertical asymptote using an ordinary definite-integral formula.

Standard integrals and the reverse chain factor

Integrals for a linear inner expression
Integrand	Antiderivative	Conditions
$(ax+b)^n$	$\dfrac{(ax+b)^{n+1}}{a(n+1)}+C$	$a\ne0,\ n\ne-1$ ; stay in its real domain.
$1/(ax+b)$	$\dfrac1a\ln\|ax+b\|+C$	$a\ne0,\ ax+b\ne0$ .
$e^{ax+b}$	$\dfrac1a e^{ax+b}+C$	$a\ne0$ .
$\sin(ax+b)$	$-\dfrac1a\cos(ax+b)+C$	$a\ne0$ ; radians.
$\cos(ax+b)$	$\dfrac1a\sin(ax+b)+C$	$a\ne0$ ; radians.
$\sec^2(ax+b)$	$\dfrac1a\tan(ax+b)+C$	$a\ne0$ ; radians, $\cos(ax+b)\ne0$ .

Differentiating the candidate antiderivative multiplies by $a$ , so integrating must divide by $a$ to undo it. This shortcut assumes a linear inner expression. If $a=0$ , the integrand is a constant (when defined), which integrates to that constant times $x$ . For $n=-1$ , adding 1 would produce a zero denominator, so use the logarithm rule instead.

Example 1: powers, a root and a reciprocal

Find $\int(6x^2+2\sqrt{x}-4/x^2)\,dx$ , on $x>0$ .

Write the integrand as $6x^2+2x^{1/2}-4x^{-2}$ . Increase each exponent by 1 and divide by that new exponent:

2x^3+\frac43x^{3/2}+\frac4x+C.

Differentiation gives $6x^2+2\sqrt{x}-4/x^2$ , confirming the sign of the reciprocal term.

Example 2: a linear inner expression with a negative power

Find $\int(3x-2)^{-2}\,dx$ , with $x\ne2/3$ .

\int(3x-2)^{-2}\,dx=\frac{(3x-2)^{-1}}{3(-1)}+C=-\frac1{3(3x-2)}+C.

Differentiating produces $(-1/3)(-1)(3)(3x-2)^{-2}$ , exactly the integrand. The result holds on an interval on either side of $2/3$ ; constants on the two separate intervals need not be the same.

Example 3: exponential and trigonometric standard forms

Integrate $2e^{3x}+4\sin2x-3\cos3x+5\sec^2(5x)$ , using radians.

\frac23e^{3x}-2\cos2x-\sin3x+\tan5x+C.

The sine term integrates to a negative cosine. Each inner multiplier is divided out: for instance, $d(-2\cos2x)/dx=4\sin2x$ . The result is used only on intervals where $\cos5x\ne0$ .

Example 4: a logarithm with a negative inner coefficient

Find $\int\dfrac3{1-2x}\,dx$ , where $x\ne1/2$ .

\int\frac3{1-2x}\,dx=-\frac32\ln|1-2x|+C.

Indeed the derivative is $(-3/2)(-2)/(1-2x)=3/(1-2x)$ . If a question restricts $x<1/2$ , write $-\tfrac32\ln(1-2x)+C$ ; for $x>1/2$ , write $-\tfrac32\ln(2x-1)+C$ . Absolute values allow one formula on either valid interval.

Prepare the integrand before integrating

There is no general rule that integrates a product by multiplying two antiderivatives. Expand, divide, use partial fractions or apply an identity to reveal standard forms. A question's earlier differentiation result can also supply exactly the antiderivative you need.

Example 5: expand a square first

Integrate $(x-2/x)^2$ for $x\ne0$ .

(x-2/x)^2=x^2-4+4x^{-2}.

\int(x-2/x)^2\,dx=\frac{x^3}{3}-4x-\frac4x+C.

The middle term is $-4$ , because $2(x)(-2/x)=-4$ . You cannot simply increase the outer square's exponent: its inner derivative is not a constant.

Example 6: split a rational function into partial fractions

Integrate $\dfrac{3x+5}{(x+1)(x+2)}$ , where $x\ne-1,-2$ .

\frac{3x+5}{(x+1)(x+2)}=\frac A{x+1}+\frac B{x+2}.

Multiplying through gives $3x+5=A(x+2)+B(x+1)$ , so $A+B=3$ and $2A+B=5$ . Thus $A=2,B=1$ , and

\int\frac{3x+5}{(x+1)(x+2)}\,dx=2\ln|x+1|+\ln|x+2|+C.

Differentiate and combine fractions to recover the original numerator. Use the answer on an interval avoiding both excluded inputs. If the numerator's degree is at least the denominator's, divide first.

Example 7: use an identity for a squared trigonometric function

Find $\int\sin^2(2x)\,dx$ , with radians.

From $\cos4x=1-2\sin^2(2x)$ , rewrite $\sin^2(2x)=\tfrac12(1-\cos4x)$ . Then

\int\sin^2(2x)\,dx=\frac x2-\frac{\sin4x}{8}+C.

Its derivative is $\tfrac12-\tfrac12\cos4x$ , the original squared sine. Similarly, $\cos^2u=(1+\cos2u)/2$ and $\tan^2u=\sec^2u-1$ prepare other integrals.

Example 8: use a given derivative to integrate a product

Differentiate $x\sin x$ , and hence find $\int x\cos x\,dx$ .

\frac{d}{dx}(x\sin x)=\sin x+x\cos x.

Rearrange the identity before integrating:

x\cos x=\frac{d}{dx}(x\sin x)-\sin x.

\int x\cos x\,dx=x\sin x+\cos x+C.

Differentiating the answer makes the two $\sin x$ terms cancel. This “hence” method uses the supplied derivative; it does not require a new general integration rule.

Recover the function from its derivatives

Example 9: one point fixes one constant

A curve has $dy/dx=3x^2-4$ and passes through $(2,5)$ . Find its equation.

y=x^3-4x+C.

Substitute the point into the function, not into its derivative: $5=8-8+C$ , so $C=5$ . The equation is $y=x^3-4x+5$ . Its derivative and its value at 2 both match the given information.

Example 10: integrate twice and use two independent conditions

Given $y^{\prime\prime}=6x$ , $y^{\prime}(1)=5$ and $y(1)=4$ , find $y$ .

y^{\prime}=3x^2+C_1,\quad 5=3+C_1\Rightarrow C_1=2.

y=x^3+2x+C_2,\quad 4=1+2+C_2\Rightarrow C_2=1.

Therefore $y=x^3+2x+1$ . Each integration introduces its own constant. If instead two points on the curve are given, integrate to $y=x^3+C_1x+C_2$ and solve the two resulting equations together.

Definite integrals give signed accumulation

If $F^{\prime}=f$ throughout an interval where $f$ is continuous, then $\int_a^b f(x)\,dx=F(b)-F(a)$ . As the upper endpoint moves, the accumulated signed area changes at rate $f$ ; this is why an antiderivative measures the accumulation.

With $a<b$ , portions above the x-axis contribute positively and portions below contribute negatively. Geometric area is non-negative: split at roots and add the magnitudes. Reversing the limits changes an integral's sign, and equal limits give zero. There is no $+C$ in the final definite value because it cancels in $F(b)-F(a)$ .

Example 11: signed integral versus total area

For $y=x-1$ , find the signed integral and the total area between the graph and the x-axis from $x=0$ to $x=3$ .

\int_0^3(x-1)\,dx=\left[\frac{x^2}2-x\right]_0^3=\frac32.

The graph crosses the axis at 1. It is negative on $[0,1)$ and positive on $(1,3]$ . Therefore

\text{Area}=-\int_0^1(x-1)\,dx+\int_1^3(x-1)\,dx=\frac12+2=\frac52.

Taking the absolute value of the unsplit integral would give $3/2$ , which is still wrong for the total area. Two triangles of areas $1/2$ and 2 confirm the answer.

Regions bounded by a curve and straight lines

Sketch the boundaries and find their intersections. For vertical strips, area is $\int(\text{upper }y-\text{lower }y)\,dx$ . If the upper boundary changes, split the integral there. For horizontal strips, use $\int(\text{right }x-\text{left }x)\,dy$ with limits in $y$ . Choosing the variable can turn an awkward region into a standard integral.

Example 12: a line and a parabola

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

At the intersections, $x=x^2$ , so $x(x-1)=0$ and $x=0,1$ . For $0<x<1$ , the line is above the parabola. Hence

A=\int_0^1(x-x^2)\,dx=\left[\frac{x^2}2-\frac{x^3}3\right]_0^1=\frac16\text{ square units}.

The integrand is the vertical distance between the two boundaries. Subtracting in the opposite order would produce a negative signed value.

The right boundary minus the left boundary is (1 + eʸ) − 1 = eʸ. Integrate this width with respect to y.

Example 13: horizontal strips avoid integrating a logarithm

Find the area bounded by $y=\ln(x-1)$ , $x=1$ , $y=0$ and $y=\ln3$ .

Rearrange the curve as $x=1+e^y$ . For each $0\le y\le\ln3$ , the left boundary is $x=1$ and the right boundary is $x=1+e^y$ . Therefore

A=\int_0^{\ln3}[(1+e^y)-1]\,dy=[e^y]_0^{\ln3}=3-1=2.

The units are square units. The curve itself requires $x>1$ ; the separate boundary line $x=1$ is not being substituted into the logarithm. The limits are y-values because the strips are horizontal.

Explore cancellation across the x-axis

Keep the lower limit at 0 and move the upper limit $b$ for $y=x-1$ . Before moving it, predict the signed integral when $b=2$ : equal triangles cancel even though their total area is positive.

Upper limit b:3

Shading below the axis contributes negatively to the integral. Both shaded parts count positively towards geometric area.

From 0 to 3: signed integral = 1.5; total area = 2.5 square units.

Common mistakes and useful checks

Using the power rule for $1/x$ : its antiderivative is $\ln|x|+C$ , not $x^0/0$ .
Multiplying by the inner coefficient: a linear inner expression requires division when integrating.
Forgetting a constant: include one per indefinite integration and use the correct point or gradient condition.
Dropping absolute values: $\ln x$ alone does not cover a negative-x interval for $1/x$ .
Taking one absolute value at the end: split the area at every sign change first.
Crossing an undefined input: the ordinary definite-integral rule cannot be applied across a vertical asymptote.
Mixing strip directions: use x-limits with $dx$ and y-limits with $dy$ .

Continue with topic practice

Use the Secondary 4 integration practice bank for exam questions and their worked solutions. Attempt a question on paper before comparing methods. This guide supports Secondary 4 learning and revision of the full Secondary Additional Mathematics course.