Integration

Integration reverses differentiation Paper 1

Differentiation turns a function into its gradient function. Integration goes the other way: given $\frac{\mathrm{d}y}{\mathrm{d}x}$, it finds $y$. Since $\frac{\mathrm{d}}{\mathrm{d}x}(x^3)=3x^2$, an integral of $3x^2$ is $x^3$.

But $x^3+5$ and $x^3-2$ also have derivative $3x^2$, because constants differentiate to 0. So every integral includes an arbitrary constant of integration, $c$:
$$\int 3x^2\mathrm{d}x=x^3+c$$
Reversing "multiply by the power, then reduce the power by 1" gives the rule for powers: increase the power by 1, then divide by the new power.
$$\int x^n\mathrm{d}x=\frac{x^{n+1}}{n+1}+c(n\ne -1)$$
The rule fails for $n=-1$ because it would divide by zero; $\int x^{-1}\mathrm{d}x$ is not in this paper. As with differentiation, constant multiples stay and sums are integrated term by term, and you must rewrite roots and fractions as powers first, and expand products or split fractions. You can always check an integral by differentiating your answer.

When an answer box asks for an indefinite integral on this site, type it without the "$+c$"; in the exam, always write it.

Integrating (ax + b) to a power Paper 1

The chain rule says $\frac{\mathrm{d}}{\mathrm{d}x}(ax+b{)}^{n+1}=a(n+1)(ax+b{)}^n$. Reversing it:
$$\int (ax+b{)}^n\mathrm{d}x=\frac{(ax+b{)}^{n+1}}{a(n+1)}+c(n\ne -1)$$
Increase the power, divide by the new power, and divide by $a$, the coefficient of $x$ inside the bracket. Forgetting to divide by $a$ is the classic error; differentiating your answer catches it at once.

This works only for a linear inside, $ax+b$. It does not integrate $(x^2+1{)}^3$, which you would expand.

For example, $\int \frac{1}{(2x+3{)}^2}\mathrm{d}x=\int (2x+3{)}^{-2}\mathrm{d}x=\frac{(2x+3{)}^{-1}}{2\times (-1)}+c=-\frac{1}{2(2x+3)}+c$.

Finding the constant of integration Paper 1

The constant $c$ gives a whole family of curves with the same gradient function, each a vertical translation of the others. A point on the curve picks out one member of the family: substitute its coordinates into $y=\dots +c$ and solve for $c$.

Questions often combine this with differentiation work: find the equation of the curve, then its stationary points, or the equation of a tangent.

Definite integrals Paper 1

A definite integral has limits:
$${\int }_a^b\mathrm{f}(x)\mathrm{d}x=[\mathrm{F}(x){]}_a^b=\mathrm{F}(b)-\mathrm{F}(a)$$
where $\mathrm{F}$ is any integral of $\mathrm{f}$. The constant of integration cancels in the subtraction, so it is left out. The answer is a number, and questions usually want it exact: keep fractions and surds.

Lay out the working with square brackets, substitute the upper limit, then the lower limit, and subtract the whole of the second bracket. Most lost marks here are sign errors in that subtraction.

Areas Paper 1

The definite integral ${\int }_a^{b}y\mathrm{d}x$ gives the area between a curve, the $x$-axis and the lines $x=a$ and $x=b$. Think of the region as thin vertical strips of height $y$ and width $\delta x$: adding their areas and letting the width shrink gives the integral.

• Below the axis. Where the curve is below the $x$-axis, $y$ is negative and so is the integral. The area is the size of that value. If a region crosses the axis, integrate each part separately and add the sizes; integrating straight across lets the parts cancel.
• Between a curve and the $y$-axis. Use horizontal strips: area $={\int }_c^{d}x\mathrm{d}y$, with $x$ written in terms of $y$ and limits on the $y$-axis.
• Between a curve and a line, or two curves. Area $={\int }_a^b(y_{\text{upper}}-y_{\text{lower}})\mathrm{d}x$, where $a$ and $b$ are usually the $x$-coordinates of the intersection points. Find them first by solving the equations simultaneously. This works even if part of the region is below the $x$-axis, because the difference of heights is always the strip height.

Another common approach is to find an area under the curve and subtract a triangle or trapezium found from geometry.

The strips between the curves have height $(6x-x^2)-x$.

Volumes of revolution Paper 1

When a region is rotated through ${360}^{°}$ about an axis it sweeps out a solid, a volume of revolution. Each thin vertical strip of height $y$ becomes a thin disc of radius $y$ and thickness $\delta x$, with volume $\pi y^2\delta x$. Adding the discs:
$$\text{about the}x\text{-axis:}V=\pi {\int }_a^b{y}^2\mathrm{d}x\text{about the}y\text{-axis:}V=\pi {\int }_c^d{x}^2\mathrm{d}y$$
For rotation about the $y$-axis, rearrange to get $x^2$ in terms of $y$, and use limits on the $y$-axis.

Square $y$ before integrating: $\pi \int y^2\mathrm{d}x$ is not $\pi (\int y\mathrm{d}x{)}^2$. Leave answers in terms of $\pi$ when an exact value is asked for.

A region not bounded by the axis. If the region lies between an upper boundary $y_1$ and a lower boundary $y_2$, each strip sweeps out a ring (a washer): a disc of radius $y_1$ with a disc of radius $y_2$ removed. So
$$V=\pi {\int }_a^b(y_1^2-y_2^2)\mathrm{d}x$$
Note that this is the difference of the squares, not the square of the difference.

Rotated about the $x$-axis, each strip of this region makes a ring with outer radius $10-x^2$ and inner radius 1.

Improper integrals Paper 1

Two kinds of improper integral appear in simple cases.

• An infinite limit, such as ${\int }_1^{\infty }{x}^{-2}\mathrm{d}x$. Integrate to an upper limit $X$, then let $X$ grow: $[-x^{-1}{]}_1^X=1-\frac{1}{X}$, and $\frac{1}{X}$ approaches 0, so the integral is 1. An infinitely long region can have a finite area.
• An integrand that is not defined at a limit, such as ${\int }_0^1{x}^{-\frac{1}{2}}\mathrm{d}x$, where $x^{-\frac{1}{2}}$ is not defined at $x=0$. Integrate from a small positive number and let it approach 0: $[2x^{\frac{1}{2}}{]}_0^1=2-0=2$, since $2\sqrt{x}$ approaches 0.

In each case the integral has a value only if the limit is a finite number. For example ${\int }_1^{\infty }{x}^{-\frac{1}{2}}\mathrm{d}x=[2\sqrt{x}{]}_1^{\infty }$ has no value, because $2\sqrt{X}$ grows without bound. Show the limiting step in words, such as "as $X\to \infty$, $\frac{1}{X}\to 0$".