Integration

Integration undoes differentiation

If $\frac{dy}{dx}=2x$ then $y$ must be $x^2$, because differentiating $x^2$ gives $2x$. Integration is the process of working back from the derivative to the function, so the rule that reverses $\frac{d}{dx}(x^n)=n{x}^{n-1}$ is $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$: raise the power by one, then divide by the new power.

The $+c$ is the constant of integration and it is not decoration. $x^2$, $x^2+5$ and $x^2-100$ all have derivative $2x$, so the reverse process cannot tell which one you started from. Every indefinite integral carries a $c$ until a point on the curve fixes it: substitute the coordinates of the point into your answer, solve for $c$, and write the equation of the curve.

Check every integral by differentiating it back, which takes five seconds and catches most errors. $\int 6x^{2}dx=2x^3+c$, and $\frac{d}{dx}(2x^3+c)=6x^2$, so the answer is right. If the check gives $18x^2$ you divided by the old power instead of the new one; if it gives a different power you miscounted when raising it.

The one power the rule cannot handle is $n=-1$, because dividing by $n+1$ would be dividing by zero. $\int x^{-1}dx=\ln x+c$ is a Pure 3 result, and Pure 1 integrals never contain it - so if a question puts $\frac{1}{x}$ in front of you, look again at the question rather than reaching for a logarithm.

Integrating powers, multiples, sums and differences

Three rules make almost everything in this topic mechanical. A constant multiple can be taken outside the integral sign, $\int kf(x)dx=k\int f(x)dx$. A sum or difference is integrated term by term. And a constant integrates to that constant times $x$: $\int 5dx=5x+c$, because differentiating $5x$ gives $5$.

As with differentiation, the first move is usually to rewrite. $\sqrt{x}=x^{1/2}$, $\frac{1}{x^3}=x^{-3}$, $\frac{5}{x^2}=5x^{-2}$, and a quotient with a single power of $x$ underneath is divided through term by term. $\frac{3x^2+4}{\sqrt{x}}$ becomes $3x^{3/2}+4x^{-1/2}$, which integrates to $\frac{6}{5}{x}^{5/2}+8x^{1/2}+c$.

Negative and fractional powers behave exactly as you would expect, and the answers are often tidier than they look. $\int x^{-2}dx=\frac{x^{-1}}{-1}+c=-\frac{1}{x}+c$, $\int 3x^{-4}dx=-x^{-3}+c$, and $\int x^{-1/2}dx=2x^{1/2}+c$. In each case the new power is one more than the old, and you divide by that new power - including when it is negative or a fraction.

There is no product rule and no quotient rule for integration at Pure 1. If a question hands you a product such as $x(2x+3{)}^2$ or $x^2(x-4)$, multiply it out first and then integrate term by term. That is not a trick; it is the method the syllabus expects.

Integrating $(ax+b{)}^n$ and the reverse chain rule

A bracket raised to a power is integrated with $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+c$, for any rational $n$ except $-1$. Raise the power by one, divide by the new power, and divide by $a$, the coefficient of $x$ inside the bracket. The $a$ is there because differentiating the bracket gives $a$, and the chain rule would multiply by it - integration has to undo that multiplication.

For $\int (3x+2{)}^{4}dx$ the new power is 5 and $a=3$, so the integral is $\frac{(3x+2{)}^5}{15}+c$. Differentiate it back: $\frac{5(3x+2{)}^4\times 3}{15}=(3x+2{)}^4$, so the $a$ and the new power have both been accounted for.

Fractional and negative powers work the same way, and the division by $a$ is the step that is usually dropped. $\int (2x+3{)}^{1/2}dx=\frac{(2x+3{)}^{3/2}}{2\times \frac{3}{2}}=\frac{(2x+3{)}^{3/2}}{3}+c$, and $\int \frac{1}{(4x-1{)}^2}dx=\int (4x-1{)}^{-2}dx=\frac{(4x-1{)}^{-1}}{4(-1)}+c=-\frac{1}{4(4x-1)}+c$.

Nothing else is in scope at Pure 1. There is no integration by parts and no substitution, so an integrand that is a product of two functions with $x$ in both - $x^2(2x+1{)}^3$, for instance - has to be expanded first, however long the expansion is.

• $\int (5x-1{)}^{3}dx=\frac{(5x-1{)}^4}{20}+c$
• $\int (2x+7{)}^{5}dx=\frac{(2x+7{)}^6}{12}+c$
• $\int (x+4{)}^{1/2}dx=\frac{2}{3}(x+4{)}^{3/2}+c$ - here $a=1$, so the only division is by the new power
• $\int (3-2x{)}^{-2}dx=\frac{(3-2x{)}^{-1}}{-2\times (-1)}+c=\frac{1}{2(3-2x)}+c$ - here $a=-2$ and the new power is $-1$, so the two minus signs in the denominator cancel
• $\int \sqrt{4x+1}dx=\frac{(4x+1{)}^{3/2}}{6}+c$

Definite integrals and the constant of integration

A definite integral comes with limits: ${\int }_a^{b}f(x)dx=[F(x){]}_a^b=F(b)-F(a)$, where $F$ is any antiderivative of $f$. Substitute the upper limit first, then the lower limit, and subtract the second result from the first. The $c$ cancels in the subtraction, which is why a definite integral is written without one - but an indefinite integral is not.

The bracket with the limits is worth writing down even when the lower value is zero, because it shows the method and because it protects you when the lower limit is negative. For ${\int }_{-1}^2(3x^2-4x+1)dx$ the antiderivative is $x^3-2x^2+x$, the upper limit gives $8-8+2=2$, the lower limit gives $-1-2-1=-4$, and subtracting a negative adds: $2-(-4)=6$.

A definite integral can come out negative, and that is not automatically an error. ${\int }_0^1(x-2)dx=[\frac{x^2}{2}-2x{]}_0^1=\frac{1}{2}-2=-\frac{3}{2}$, because the region lies below the $x$-axis and the integral counts it as negative. If the question asks for an area, the answer is positive; if it asks for the value of the integral, keep the sign.

An integral can also come out to exactly zero without meaning anything has gone wrong. ${\int }_{-2}^2{x}^{3}dx=0$ because the two halves of the region are equal and opposite - the part below the axis cancels the part above. Sketching first tells you whether a zero is sensible.

Area under a curve, and what happens below the axis

The area of the region between the curve $y=f(x)$, the $x$-axis and the lines $x=a$ and $x=b$ is ${\int }_a^{b}f(x)dx$, provided the curve stays above the axis on that interval. The integral is a sum of thin rectangles: each strip has height $y$ and width $dx$, so it adds up area.

Where the curve crosses the axis the integral stops being the area. On a stretch where $f(x)$ is negative, every strip contributes a negative amount, and the integral returns the negative of the area. So a sketch comes first, the crossing points split the region into pieces, and each piece is integrated separately and counted as a positive number.

Here is the whole method on $y=4x-x^2$. The curve crosses the $x$-axis where $x(4-x)=0$, at $x=0$ and $x=4$, and between those values it is above the axis. Then ${\int }_0^4(4x-x^2)dx=[2x^2-\frac{x^3}{3}{]}_0^4=32-\frac{64}{3}=\frac{32}{3}$, which is about 10.67 square units.

Now turn the same shape upside down: $y=x^2-4x$ is below the $x$-axis between $x=0$ and $x=4$, so on that stretch every strip contributes a negative amount. ${\int }_0^3(x^2-4x)dx=[\frac{x^3}{3}-2x^2{]}_0^3=9-18=-9$, and the area enclosed is 9. Writing $-9$ as the area is the mistake this section exists to prevent.

• Sketch the curve and find where it crosses the $x$-axis before integrating
• Integrate each piece between consecutive crossings
• Ignore the sign of each piece when the question asks for area, but keep it when it asks for the value of the integral
• A region between a curve and the $y$-axis is found by integrating with respect to $y$, so rearrange the equation to make $x$ the subject first

Area between a curve and a line, and between two curves

When a line lies above a curve between two points, the area between them is ${\int }_a^b(y_1-y_2)dx$ - the integral of the upper function minus the lower one. The limits are the $x$-coordinates of the points where the two graphs meet, so the first step is to solve the two equations simultaneously.

Take the curve $y=x^2$ and the line $y=x+2$. Setting them equal, $x^2=x+2$, so $x^2-x-2=0$ and $(x-2)(x+1)=0$, giving $x=-1$ and $x=2$. Between those limits the line is above the curve - check by putting $x=0$ into both, giving 2 and 0 - so the area is ${\int }_{-1}^2(x+2-x^2)dx=[\frac{x^2}{2}+2x-\frac{x^3}{3}{]}_{-1}^2$. The upper limit gives $2+4-\frac{8}{3}=\frac{10}{3}$ and the lower limit gives $\frac{1}{2}-2+\frac{1}{3}=-\frac{7}{6}$, so the area is $\frac{10}{3}+\frac{7}{6}=\frac{9}{2}$ square units.

Which graph is on top can change between the limits. If a curve and a line cross three times, or two curves swap over, the integral has to be split at the crossing point and the difference taken the right way round in each piece. Substituting one convenient value of $x$ into both equations settles which is which in each interval.

The subtraction order matters for the sign, not for the size. Integrating curve minus line instead of line minus curve gives the same number with a minus sign in front, which is the classic way to lose the final mark of a five-mark question.

Volumes of revolution

Rotate the region under a curve through a full turn about the $x$-axis and it sweeps out a solid. A thin strip of height $y$ and width $dx$ becomes a disc of radius $y$, whose cross-sectional area is $\pi y^2$, so the volume is $V=\pi {\int }_a^b{y}^{2}dx$. The $y^2$ is not optional: the area of a circle of radius $y$ is $\pi y^2$.

For $y=\sqrt{x}$ between $x=0$ and $x=4$, the square root disappears when you square it: $y^2=x$, so $V=\pi {\int }_0^{4}xdx=\pi [\frac{x^2}{2}{]}_0^4=8\pi$ cubic units. Squaring first is what makes these questions short - the integration is usually easy once the square is done.

Rotation about the $y$-axis uses horizontal strips, so the variable of integration is $y$: $V=\pi \int x^{2}dy$, with the limits as $y$-values. Rearrange the equation of the curve to make $x$ the subject first. For $y=x^2$ rotated about the $y$-axis from $y=1$ to $y=4$, $x^2=y$, so $V=\pi {\int }_1^{4}ydy=\pi [\frac{y^2}{2}{]}_1^4=\frac{15\pi }{2}$.

The region does not have to reach the axis of rotation. The syllabus notes that a volume of revolution may involve a region that is not bounded by the axis of rotation, for instance the region between $y=9-x^2$ and the line $y=5$ rotated about the $x$-axis. There the solid is the big one minus the small one, so $V=\pi {\int }_{-2}^2((9-x^2{)}^2-25)dx$, with the limits from $9-x^2=5$, that is $x=\pm 2$. Integrating gives $\frac{704\pi }{5}$ cubic units.

• About the $x$-axis: $V=\pi {\int }_a^b{y}^{2}dx$, limits in $x$
• About the $y$-axis: $V=\pi {\int }_a^b{x}^{2}dy$, limits in $y$
• Square $y$ (or $x$) before integrating - the square is part of the formula, not part of the algebra
• Between a curve and a line: subtract the two squares before integrating, $\pi \int (y_1^2-y_2^2)dx$
• A full turn is ${360}^{\circ }$; the syllabus rotates through a full turn about one of the axes

Improper integrals

An improper integral is one whose interval is infinite, or whose integrand misbehaves at one end of the interval. The method is always the same: replace the awkward value with a variable, do the integral as normal, then take the limit.

For an infinite limit, replace $\infty$ by $N$ and let $N$ tend to infinity at the end. ${\int }_1^{\infty }\frac{1}{x^2}dx={lim}_{N\to \infty }{\int }_1^N{x}^{-2}dx={lim}_{N\to \infty }[-\frac{1}{x}{]}_1^N={lim}_{N\to \infty }(-\frac{1}{N}+1)=1$. The region runs for ever, yet its area is 1 - the strips get thin fast enough for the total to settle down.

For an integrand that is not defined at the end of the interval, replace the bad limit by $ϵ$ and let $ϵ$ tend to zero from above. ${\int }_0^1{x}^{-1/2}dx={lim}_{ϵ\to 0^{+}}[2x^{1/2}{]}_{ϵ}^1=2-0=2$, because $\sqrt{ϵ}$ tends to zero even though the integrand itself tends to infinity at $x=0$.

Not every improper integral settles down. ${\int }_1^{\infty }\frac{1}{x}dx={lim}_{N\to \infty }[\ln x{]}_1^N$ has no finite value, because $\ln N$ grows without bound: the integral diverges. Cambridge asks only for the cases that converge, but the limit must be written down - the mark is for showing that you took it.

• Infinite limit: replace $\infty$ by $N$, integrate, then let $N\to \infty$
• Undefined endpoint: replace the endpoint by $ϵ$, integrate, then let $ϵ\to 0^{+}$
• The answer is the value of the limit - write it as a limit before you evaluate it
• If the limit does not exist, the integral diverges