Integral Calculus

Look at a question JEE Main has set more than once, in one form or another:

$I={\int }_0^1\frac{8\log (1+x)}{1+x^2}dx$

There is no elementary antiderivative for this integrand. No substitution, no by-parts, no partial fractions will produce an $F(x)$. A candidate who has learnt integration as "find $F$, then compute $F(b)-F(a)$" cannot start.

It is worth $\pi \log 2$, and the work takes four lines.

That gap is the chapter.

Indefinite integration asks you to produce an antiderivative, and every technique in it transforms the integrand into something you recognise.

Definite integration hands you two limits as extra information, and the powerful methods spend that information instead — exploiting symmetry about the midpoint, evenness, periodicity. They never find $F$ at all.

Across NTA papers from 2009 onwards, at least one definite-integral question every year falls to the midpoint symmetry known as King's property, without the integral being evaluated in full — a pattern visible in the year-by-year previous-paper collections listed under Sources below. That is not one technique among seven. It is the first thing to try on any definite integral with awkward limits.

This chapter is worth about 2 questions (8 marks) in JEE Main, and integration as a whole accounts for 2–4 questions per paper.

Part I — Indefinite Integration

1. The Antiderivative and the Constant

$F$ is an antiderivative of $f$ when $F^{\prime }(x)=f(x)$. Since the derivative of a constant is zero, antiderivatives come in a family differing by a constant:

$\int f(x)dx=F(x)+C$

The $+C$ is not bookkeeping. Two functions with the same derivative differ by a constant, so $C$ is the entire ambiguity, and a definite integral is exactly the operation that cancels it.

2. Standard Integrals

Two that carry weight later:

$\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C\int \frac{dx}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C$

$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C$

Check every antiderivative by differentiating it. Integration is the only topic in the syllabus where verification is free, and in a multiple-choice paper it is often faster to differentiate the four options than to integrate the question.

3. Substitution

Substitution reverses the chain rule. Put $u=g(x)$, so $du=g^{\prime }(x)dx$:

$\int f(g(x))g^{\prime }(x)dx=\int f(u)du$

The skill is spotting that a factor of the integrand is the derivative of another part, up to a constant.

Three patterns cover most of what is set:

• $\int \frac{f^{\prime }(x)}{f(x)}dx=\log |f(x)|+C$ — the numerator is the derivative of the denominator.
• $\int [f(x){]}^n{f}^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1}+C$.
• Trigonometric substitution, driven by the surd present: $\sqrt{a^2-x^2}$ suggests $x=a\sin \theta$; $\sqrt{a^2+x^2}$ suggests $x=a\tan \theta$; $\sqrt{x^2-a^2}$ suggests $x=a\sec \theta$.

4. Integration by Parts

From the product rule $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$, integrate both sides and rearrange:

$\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx$

The whole game is choosing which factor is $u$ — the one you will differentiate. Choose it so that differentiating simplifies things. ILATE orders the candidates:

Inverse trig → Logarithmic → Algebraic → Trigonometric → Exponential.

Take $u$ from as early in that list as appears. Inverse trigonometric and logarithmic functions have no easy integral but pleasant derivatives, so they must be the part you differentiate.

Two special forms worth recognising on sight

The $e^x$ pair. If the integrand is $e^x$ times a function plus its own derivative, the answer is immediate:

$\int e^x\left[f(x)+f^{\prime }(x)\right]dx=e^{x}f(x)+C$

This is by parts done once, in general, and JEE sets it repeatedly in disguised form. The work is entirely in recognising $f$ and $f^{\prime }$ inside the bracket.

The returning integral. For $\int e^{ax}\sin bxdx$, applying by parts twice reproduces the original integral on the right. Call it $I$, solve the resulting equation for $I$ algebraically. Applying by parts a third time undoes the second and returns $0=0$, which is the standard way this is lost.

Illustration 1 · By parts twice

Evaluate $\int x^2\sin 2xdx$. (JEE previous-year question)

By ILATE, the algebraic $x^2$ is differentiated. Take $u=x^2$, $v^{\prime }=\sin 2x$, so $v=-\frac{\cos 2x}{2}$:

$I=-\frac{x^2\cos 2x}{2}+\int x\cos 2xdx$.

Apply by parts again to $\int x\cos 2xdx$ with $u=x$, $v=\frac{\sin 2x}{2}$:

$\int x\cos 2xdx=\frac{x\sin 2x}{2}-\frac{1}{2}\int \sin 2xdx=\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}$.

$I=-\frac{x^2\cos 2x}{2}+\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}+C$

Differentiating confirms it: the $x\cos 2x$ terms cancel, as do the $\sin 2x$ terms, leaving $x^2\sin 2x$.

Illustration 2 · The returning integral

Evaluate $\int e^x\sin xdx$. (JEE previous-year question)

Neither factor simplifies on differentiation, so expect the integral to return. Take $u=\sin x$, $v^{\prime }=e^x$:

$I=e^x\sin x-\int e^x\cos xdx$.

Apply by parts again to the new integral, keeping the same choice of which factor is exponential:

$\int e^x\cos xdx=e^x\cos x+\int e^x\sin xdx=e^x\cos x+I$.

Substituting back: $I=e^x\sin x-e^x\cos x-I$, so $2I=e^x(\sin x-\cos x)$.

$I=\frac{e^x}{2}\left(\sin x-\cos x\right)+C$

Had we swapped roles on the second application, we would have returned to $I=I$ and learnt nothing. Consistency is the whole technique.

Illustration 3 · Recognising $e^x[f+f^{\prime }]$

Evaluate $\int \frac{(x-1)e^x}{(x+1{)}^3}dx$. (AIEEE previous-year question)

Attacking this with by parts is painful. Test the $e^x[f(x)+f^{\prime }(x)]$ pattern instead by rewriting the rational factor with $x-1=(x+1)-2$:

$\frac{x-1}{(x+1{)}^3}=\frac{(x+1)-2}{(x+1{)}^3}=\frac{1}{(x+1{)}^2}-\frac{2}{(x+1{)}^3}$.

Now set $f(x)=(x+1{)}^{-2}$. Then $f^{\prime }(x)=-2(x+1{)}^{-3}$, and the bracket is exactly $f+f^{\prime }$.

$\int \frac{(x-1)e^x}{(x+1{)}^3}dx=\frac{e^x}{(x+1{)}^2}+C$

Recognition cue: an $e^x$ multiplied by a sum of two terms whose powers differ by one, with opposite signs. Split the numerator to expose it.

Illustration 4 · The same pattern in trigonometric disguise

Evaluate $\int \frac{e^x(1+\sin x)}{1+\cos x}dx$. (JEE previous-year question)

Convert to half-angles: $1+\cos x=2{\cos }^2\frac{x}{2}$ and $\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$.

$\frac{1+\sin x}{1+\cos x}=\frac{1+2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\cos }^2\frac{x}{2}}=\frac{1}{2}{\sec }^2\frac{x}{2}+\tan \frac{x}{2}$.

With $f(x)=\tan \frac{x}{2}$ we get $f^{\prime }(x)=\frac{1}{2}{\sec }^2\frac{x}{2}$, so the bracket is $f+f^{\prime }$ once more.

$\int \frac{e^x(1+\sin x)}{1+\cos x}dx=e^x\tan \frac{x}{2}+C$

5. Partial Fractions

For a rational integrand $\frac{P(x)}{Q(x)}$ with $\mathrm{deg}P<\mathrm{deg}Q$, split $Q$ into factors and decompose:

If $\mathrm{deg}P\ge \mathrm{deg}Q$, divide first. Skipping the division is the most common failure here.

Before decomposing, always check whether the numerator can be rearranged into the denominator's factors — it is frequently faster. For $\frac{(x+1{)}^2}{x(x^2+1)}$, writing $(x+1{)}^2=x^2+2x+1=(x^2+1)+2x$ splits it in one line with no unknowns to solve for.

Illustration 5 · Rearranging beats decomposing

Evaluate $\int \frac{(x+1{)}^2}{x\left(x^2+1\right)}dx$. (JEE previous-year question)

The standard route sets up $\frac{A}{x}+\frac{Bx+C}{x^2+1}$ and solves for three unknowns. Look at the numerator first:

$(x+1{)}^2=x^2+2x+1=\left(x^2+1\right)+2x$.

The decomposition is now immediate:

$\frac{(x^2+1)+2x}{x(x^2+1)}=\frac{1}{x}+\frac{2}{x^2+1}$.

$\int \frac{(x+1{)}^2}{x(x^2+1)}dx=\log |x|+2{\tan }^{-1}x+C$

Always inspect the numerator for the denominator's factors before setting up unknowns.

6. Trigonometric Identities

Powers and products of trigonometric functions are reduced by identity before integrating, never by force.

Illustration 6 · A linear combination of sine and cosine

Evaluate $\int \frac{dx}{\sin x+\cos x}$. (JEE previous-year question)

Write the denominator as a single sine. Since $\sqrt{1^2+1^2}=\sqrt{2}$,

$\sin x+\cos x=\sqrt{2}\sin \left(x+\frac{\pi }{4}\right)$.

The integral becomes $\frac{1}{\sqrt{2}}\int \mathrm{cosec}\left(x+\frac{\pi }{4}\right)dx$, and $\int \mathrm{cosec}udu=\log |\tan \frac{u}{2}|$:

$\int \frac{dx}{\sin x+\cos x}=\frac{1}{\sqrt{2}}\log \left|\tan \left(\frac{x}{2}+\frac{\pi }{8}\right)\right|+C$

The $t=\tan \frac{x}{2}$ substitution also works but produces a quadratic denominator and several more lines.

7. The Standard Forms the Syllabus Specifies

The syllabus names a specific family of integrals. Every one of them reduces to a standard result by completing the square, and all ten are two techniques wearing different clothes.

Pure quadratic denominators. Complete the square to turn $a{x}^2+bx+c$ into $a\left[(x+k{)}^2\pm m^2\right]$, then read off the matching standard result.

Linear numerators. Split the numerator into a multiple of the denominator's derivative plus a constant:

$px+q=\lambda \cdot \frac{d}{dx}\left(a{x}^2+bx+c\right)+\mu$

The $\lambda$ piece integrates to a logarithm (or a square root) by substitution; the $\mu$ piece is the pure-quadratic case above. One method, two halves, and it handles every $(px+q)$ form on the list.

The surd results come from by parts and are worth deriving once:

$\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log \left|x+\sqrt{x^2+a^2}\right|+C$

Illustration 7 · Completing the square

Evaluate $\int \frac{dx}{x^2+4x+13}$.

The denominator has no real roots ($D=16-52<0$), so partial fractions are unavailable. Complete the square:

$x^2+4x+13=(x+2{)}^2+9=(x+2{)}^2+3^2$.

With $u=x+2$ this is the standard $\int \frac{du}{u^2+a^2}$ with $a=3$:

$\int \frac{dx}{x^2+4x+13}=\frac{1}{3}{\tan }^{-1}\left(\frac{x+2}{3}\right)+C$

Illustration 8 · A linear numerator

Evaluate $\int \frac{2x+3}{x^2+4x+13}dx$.

Split the numerator into a multiple of the denominator's derivative plus a constant. Here $\frac{d}{dx}(x^2+4x+13)=2x+4$, so

$2x+3=(2x+4)-1$.

The integral separates into two pieces we already know:

$\int \frac{2x+4}{x^2+4x+13}dx-\int \frac{dx}{x^2+4x+13}$.

The first is $\int \frac{f^{\prime }}{f}=\log |f|$. The second is Example 1.

$=\log \left(x^2+4x+13\right)-\frac{1}{3}{\tan }^{-1}\left(\frac{x+2}{3}\right)+C$

No modulus is needed on the logarithm because the quadratic is positive definite. That is worth a mark.

Illustration 9 · Deriving a surd result

Show that $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log \left|x+\sqrt{x^2+a^2}\right|+C$.

Take $u=\sqrt{x^2+a^2}$ and $v^{\prime }=1$, so $v=x$ and $u^{\prime }=\frac{x}{\sqrt{x^2+a^2}}$. Write $I$ for the integral. By parts:

$I=x\sqrt{x^2+a^2}-\int \frac{x^2}{\sqrt{x^2+a^2}}dx$.

Now the trick: write $x^2=(x^2+a^2)-a^2$ in the remaining numerator.

$I=x\sqrt{x^2+a^2}-\int \sqrt{x^2+a^2}dx+a^2\int \frac{dx}{\sqrt{x^2+a^2}}$.

The middle term is $I$ again. So $2I=x\sqrt{x^2+a^2}+a^2\log |x+\sqrt{x^2+a^2}|$, and dividing by $2$ gives the result.

This is the returning-integral idea applied to a surd, and the same manoeuvre derives $\int \sqrt{a^2-x^2}dx$ and $\int \sqrt{x^2-a^2}dx$.

Part II — Definite Integration

8. The Fundamental Theorem

If $F^{\prime }=f$ on $[a,b]$, then

${\int }_a^{b}f(x)dx=F(b)-F(a)$

The constant cancels, which is why definite integrals are unambiguous.

When substituting, change the limits. If $u=g(x)$, the new limits are $g(a)$ and $g(b)$, and you never convert back to $x$. Forgetting this is the most frequent mechanical error in the chapter.

Note that the syllabus no longer includes the integral as a limit of a sum.

9. Properties of Definite Integrals

These are the machinery. Ranked by how often JEE actually uses them:

P1 — King's property.

${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$

Proof. Substitute $t=a+b-x$, so $dt=-dx$. When $x=a$, $t=b$; when $x=b$, $t=a$. Then ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(a+b-t)dt={\int }_a^{b}f(a+b-t)dt$, and $t$ is a dummy name.

Geometrically it is a reflection about the midpoint $\frac{a+b}{2}$, which leaves the area unchanged.

Why it is so powerful. Adding the two forms gives

$2I={\int }_a^b\left[f(x)+f(a+b-x)\right]dx$

and the bracket is very often a constant, even when $f$ itself has no antiderivative. When that happens the integral collapses to a length.

Two recurring JEE shapes make the bracket constant by construction:

• $f(x)=\frac{g(x)}{g(x)+g(a+b-x)}$ gives $f(x)+f(a+b-x)=1$, so $I=\frac{b-a}{2}$.
• $f(x)=\frac{g(x)}{1+c^x}$ with $g$ even and limits $\pm a$ gives $I={\int }_0^{a}g(x)dx$.

P2 — Even and odd.

${\int }_{-a}^{a}f(x)dx=\{\begin{array}{ll}2{\int }_0^{a}f(x)dx& f\text{ even}\\ 0& f\text{ odd}\end{array}$

Check parity before integrating anything symmetric about the origin. An odd integrand makes the answer $0$ with no work.

P3 — Splitting.

${\int }_a^{b}f={\int }_a^{c}f+{\int }_c^{b}f$

Essential whenever the integrand changes formula — a modulus, a greatest-integer function, a piecewise definition. Split at every point where the behaviour changes.

P4 — Periodicity. For $f$ of period $T$:

${\int }_0^{nT}f(x)dx=n{\int }_0^{T}f(x)dx$

P5.

${\int }_0^{2a}f(x)dx={\int }_0^a\left[f(x)+f(2a-x)\right]dx$

which vanishes when $f(2a-x)=-f(x)$ and doubles when $f(2a-x)=f(x)$.

The decision procedure

On any definite integral, in this order:

1. Is the integrand odd with symmetric limits? Answer is $0$.
2. Does $f(x)+f(a+b-x)$ simplify? Use King's property.
3. Does the integrand contain $\lfloor \cdot \rfloor$, $|\cdot |$ or a piecewise rule? Split.
4. Only then look for an antiderivative.

Reaching step 4 first is what makes an unsolvable-looking problem unsolvable.

Illustration 10 · The integral with no antiderivative

Evaluate ${\int }_0^1\frac{8\log (1+x)}{1+x^2}dx$. (AIEEE 2011)

The integrand has no elementary antiderivative, so the limits must do the work.

The $1+x^2$ invites $x=\tan \theta$, giving $dx={\sec }^2\theta d\theta$ and limits $0$ to $\frac{\pi }{4}$:

$I=8{\int }_0^{\pi /4}\log \left(1+\tan \theta \right)d\theta$.

Now apply King's property on $[0,\frac{\pi }{4}]$, where $a+b=\frac{\pi }{4}$. Write $J$ for the integral without the $8$:

$J={\int }_0^{\pi /4}\log \left(1+\tan \left(\frac{\pi }{4}-\theta \right)\right)d\theta$.

Evaluate that tangent using the subtraction formula:

$1+\tan \left(\frac{\pi }{4}-\theta \right)=1+\frac{1-\tan \theta }{1+\tan \theta }=\frac{2}{1+\tan \theta }$.

So the new integrand is $\log 2-\log (1+\tan \theta )$. Adding the two expressions for $J$, the awkward term cancels:

$2J={\int }_0^{\pi /4}\log 2d\theta =\frac{\pi }{4}\log 2$.

Hence $J=\frac{\pi }{8}\log 2$ and

$I=8J=\pi \log 2$

Nothing was ever integrated. The limits supplied the answer.

Illustration 11 · The $\frac{g}{g+h}$ shape

Evaluate ${\int }_2^4\frac{\log x^2}{\log x^2+\log \left(36-12x+x^2\right)}dx$. (JEE Main 2015)

First notice $36-12x+x^2=(x-6{)}^2$. Here $a+b=2+4=6$, so King's reflection sends $x\mapsto 6-x$.

Under that reflection $x^2\mapsto (6-x{)}^2$ and $(x-6{)}^2\mapsto x^2$ — the two logarithms swap.

So if $f(x)=\frac{\log x^2}{\log x^2+\log (x-6{)}^2}$, then $f(6-x)$ is the same fraction with numerator and the other log exchanged, and

$f(x)+f(6-x)=1$.

Therefore $2I={\int }_2^{4}1dx=2$, giving

$I=1$

Cue: whenever the integrand is one term over itself plus its reflection, the answer is half the interval length. Check for it before anything else.

Illustration 12 · The $1+a^x$ trick

Evaluate ${\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}x}{1+2^x}dx$. (JEE Main 2018)

The limits are symmetric, but the integrand is neither even nor odd because of $2^x$.

Apply King's property with $a+b=0$, so $x\mapsto -x$:

$I={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}x}{1+2^{-x}}dx={\int }_{-\pi /2}^{\pi /2}\frac{2^x{\sin }^{2}x}{1+2^x}dx$,

using ${\sin }^2(-x)={\sin }^{2}x$ and multiplying numerator and denominator by $2^x$.

Adding the two expressions, the denominators combine:

$2I={\int }_{-\pi /2}^{\pi /2}\frac{\left(1+2^x\right){\sin }^{2}x}{1+2^x}dx={\int }_{-\pi /2}^{\pi /2}{\sin }^{2}xdx$.

Now ${\sin }^{2}x$ is even, so this is $2{\int }_0^{\pi /2}{\sin }^{2}xdx=2\cdot \frac{\pi }{4}=\frac{\pi }{2}$.

$I=\frac{\pi }{4}$

This works for any even $g$ and any base $c>0$: ${\int }_{-a}^a\frac{g(x)}{1+c^x}dx={\int }_0^{a}g(x)dx$. The exponential always cancels.

Illustration 13 · Greatest integer, killed by King

Evaluate ${\int }_0^{\pi }\lfloor \cot x\rfloor dx$. (AIEEE 2009)

The integrand jumps infinitely often, so splitting is hopeless. Use the symmetry of $\cot$ instead.

Here $a+b=\pi$, and $\cot (\pi -x)=-\cot x$. So King's property gives

$I={\int }_0^{\pi }\lfloor -\cot x\rfloor dx$.

For any non-integer $y$, $\lfloor y\rfloor +\lfloor -y\rfloor =-1$. Since $\cot x$ is an integer only at isolated points, which contribute nothing to an integral, adding the two forms gives

$2I={\int }_0^{\pi }(-1)dx=-\pi$.

$I=-\frac{\pi }{2}$

Illustration 14 · Greatest integer, killed by splitting

Evaluate ${\int }_{\pi /2}^{3\pi /2}\lfloor 2\sin x\rfloor dx$.

Here no symmetry helps, so split wherever $2\sin x$ crosses an integer. On $\left[\frac{\pi }{2},\frac{3\pi }{2}\right]$, $\sin x$ falls from $1$ to $-1$, crossing $\frac{1}{2},0,-\frac{1}{2}$ at $\frac{5\pi }{6},\pi ,\frac{7\pi }{6}$.

Summing: $\frac{\pi }{3}+0-\frac{\pi }{6}-\frac{2\pi }{3}=\frac{2\pi -\pi -4\pi }{6}$.

$I=-\frac{\pi }{2}$

Note $\lfloor y\rfloor =-2$ on $(-2,-1)$, not $-1$: the floor of a negative number moves away from zero. That single sign is what most attempts get wrong.

Illustration 15 · A functional equation

A polynomial satisfies $p^{\prime }(x)=p^{\prime }(1-x)$ for all $x$, with $p(0)=1$ and $p(1)=41$. Find ${\int }_0^{1}p(x)dx$. (AIEEE 2010)

You are not given $p$, so no antiderivative is available even in principle.

By King's property on $[0,1]$, $I={\int }_0^{1}p(1-x)dx$. Adding:

$2I={\int }_0^1\left[p(x)+p(1-x)\right]dx$.

Now use the given condition. Let $h(x)=p(x)+p(1-x)$. Differentiating, $h^{\prime }(x)=p^{\prime }(x)-p^{\prime }(1-x)=0$ by hypothesis, so $h$ is constant.

Evaluate it anywhere, say at $x=0$: $h(0)=p(0)+p(1)=1+41=42$.

$2I={\int }_0^{1}42dx=42$, so

$I=21$

The chain — King's property, then differentiate the sum to show it is constant — is worth learning as a unit. It appears whenever a question gives a symmetry condition on $f^{\prime }$ rather than on $f$.

Part III — Area Under Curves

The area between a curve and the $x$-axis from $a$ to $b$ is ${\int }_a^{b}ydx$, provided $y\ge 0$ throughout. Where the curve dips below the axis the integral counts that stretch as negative, so:

$\text{Area}={\int }_a^b|y|dx$

Split at every crossing point and add the magnitudes. An answer of zero for a genuine region means the crossings were not split.

For the area between two curves, integrate the difference, upper minus lower:

$\text{Area}={\int }_a^b\left[y_{\text{upper}}-y_{\text{lower}}\right]dx$

The limits are the $x$-coordinates of the intersections, so solve the curves simultaneously first.

Integrate along $y$ when the region is easier that way. For a region bounded by a rightward parabola and a line, horizontal strips need one integral where vertical strips need two. Choosing the wrong variable is the difference between three lines and a page.

Method

1. Sketch both curves, however roughly.
2. Solve simultaneously for the intersections.
3. Decide strip direction; identify which curve is upper (or right).
4. Integrate the difference between the intersection values.
5. Report a positive number, with units of area.

Illustration 16 · Parabola and line, integrating along $y$

Find the area bounded by $y^2=2x$ and $x-y=4$.

Intersections: substitute $x=y+4$ into $y^2=2x$:

$y^2=2y+8\Rightarrow y^2-2y-8=0\Rightarrow (y-4)(y+2)=0$,

so $y=-2$ and $y=4$, giving the points $(2,-2)$ and $(8,4)$.

      y
      4 |            o (8,4)
        |         / |
      0 +------o-----+---- x
        |    /(2,-2)
     -2 |  o
        parabola opens right; line cuts it twice

Vertical strips would need two integrals, because the parabola supplies both the upper and lower boundary to the left of $x=2$. Horizontal strips need one.

For each $y$ between $-2$ and $4$, the strip runs from the parabola on the left, $x=\frac{y^2}{2}$, to the line on the right, $x=y+4$:

$A={\int }_{-2}^4\left[(y+4)-\frac{y^2}{2}\right]dy={\left[\frac{y^2}{2}+4y-\frac{y^3}{6}\right]}_{-2}^4$.

At $y=4$: $8+16-\frac{64}{6}=\frac{72-32}{3}=\frac{40}{3}$.

At $y=-2$: $2-8+\frac{8}{6}=-6+\frac{4}{3}=-\frac{14}{3}$.

$A=\frac{40}{3}+\frac{14}{3}=18\text{ square units}$

Illustration 17 · Curve, tangent and axis

Find the area bounded by $y=x^2-1$, the tangent to it at $(2,3)$, and the $y$-axis.

The tangent first: $\frac{dy}{dx}=2x=4$ at $x=2$, so

$y-3=4(x-2)\Rightarrow y=4x-5$.

Between $x=0$ and $x=2$ the parabola lies above the tangent — a tangent touches without crossing, and the parabola is concave up. So the height of a vertical strip is

$\left(x^2-1\right)-\left(4x-5\right)=x^2-4x+4=(x-2{)}^2$.

That it factors as a perfect square is the check that the tangency was computed correctly: the difference must have a double root at the point of contact.

$A={\int }_0^2(x-2{)}^{2}dx={\left[\frac{(x-2{)}^3}{3}\right]}_0^2=0-\left(\frac{-8}{3}\right)$

$A=\frac{8}{3}\text{ square units}$

Summary

Indefinite integration produces an antiderivative; definite integration often should not.

Check any antiderivative by differentiating it — verification is free here and nowhere else.

Substitution reverses the chain rule; look for a factor that is the derivative of another part. $\int \frac{f^{\prime }}{f}=\log |f|$ is the most-used case.

By parts differentiates the factor chosen by ILATE. Two forms recur: $\int e^x[f+f^{\prime }]=e^{x}f$, and the returning integral, which requires a consistent choice on the second application.

Divide before decomposing when the numerator's degree is not smaller, and inspect the numerator for the denominator's factors before setting up unknowns.

The syllabus's ten standard forms are two techniques: complete the square, and split a linear numerator into a multiple of the denominator's derivative plus a constant.

Change the limits when substituting in a definite integral, and never convert back.

On any definite integral, in order: check odd symmetry, then King's property, then split at breakpoints, and only then hunt for an antiderivative.

King's property reflects about the midpoint. Adding $f(x)$ and $f(a+b-x)$ often gives a constant, which collapses the integral to a length — and it works on integrands that have no antiderivative at all.

Two shapes recur: $\frac{g}{g+g(\text{reflected})}$ integrates to $\frac{b-a}{2}$, and $\frac{\text{even}}{1+c^x}$ over $[-a,a]$ integrates to ${\int }_0^a$ of the even part.

For $\lfloor \cdot \rfloor$ and $|\cdot |$, either exploit a symmetry or split at every breakpoint; remember $\lfloor y\rfloor +\lfloor -y\rfloor =-1$ for non-integer $y$.

Area needs $\int |y|$, split at every axis crossing; between curves, integrate upper minus lower after solving for the intersections.

Choose horizontal strips when the region is bounded left and right — it frequently turns two integrals into one.