Integration as the reverse of differentiation

An antiderivative is a function that differentiates to give the function you started with. If

, then $F$ is an antiderivative of $f$.

An indefinite integral describes a family of antiderivatives:

Don’t treat the $+C$ as decoration. Differentiation removes constants, so integration must put them back.

A definite integral is a number found by accumulating the signed values of a function over an interval:

Geometrically, the integral represents signed area. Area above the $x$-axis counts positive; area below the $x$-axis counts negative.

The first power rule for integration

When integrating powers of $x$, raise the power by one, then divide by the new power:

This is the differentiation power rule in reverse. For example, $\frac{d}{dx}\left(\frac{x^6}{6}\right)=x^5$, so $\int x^5\,dx=\frac{x^6}{6}+C$.

A constant can be carried through the integration, and each term in a sum can be integrated separately:

Roots and fractions are usually easier to integrate after rewriting them as powers. For instance, $\sqrt{x}=x^{\frac12}$ and $\frac{1}{x^4}=x^{-4}$. That quick algebra step can make the method much easier to spot.

Using a condition to find the constant

If the question gives a point on the original function, apply it after integrating. Suppose $f'(x)=6x-5$ and $f(2)=9$. Then

Now substitute $x=2$ and $f(2)=9$:

so $C=7$ and $f(x)=3x^2-5x+7$.

Definite integrals using technology

On a technology paper, a GDC can evaluate many definite integrals directly. You still need to understand what it is calculating. The result is the signed accumulation, not automatically the physical area when the curve crosses the axis.

If a curve stays above the $x$-axis on $[a,b]$, the area between the curve and the axis is

When the function goes below the axis, the same formula gives signed area. To find the true geometric area, use absolute values or split the integral at the intercepts. That fuller version comes in 5.11.

Core indefinite integrals

Here, the power rule also applies to rational powers:

There is one key exception: $n=-1$.

The absolute value is needed because $\frac{1}{x}$ is defined for both negative and positive $x$, whereas $\ln x$ is only defined for positive $x$.

The standard trigonometric and exponential integrals are

Trigonometric calculus uses radians. A calculator set to degrees may still produce convincing graphs, but it won’t give the calculus you intend.

Linear composites: the inside derivative is constant

A linear composite is a composite function with an inside function of the form $ax+b$. Here, $a$ is a non-zero constant multiplier of $x$ (unit depends on context), while $b$ is a constant shift (same unit as $x$). If $\int f(x)\,dx=F(x)+C$, then

The factor $\frac{1}{a}$ comes from reversing the chain rule. Differentiating $F(ax+b)$ introduces an extra factor of $a$, so integration requires division by $a$.

For example,

Similarly,

because the usual power rule divides by $7$, then the inside derivative gives another division by $3$.

Integration by inspection: the reverse chain rule

Integration by inspection integrates a composite function when the derivative of the inside function is already present, except perhaps for a constant factor. The syllabus form is

Ask yourself: “Do I see the derivative of the inside?” Consider

Therefore,

Now consider

Inspection can also be expressed as substitution. When the structure is clear, though, inspection is faster. Don’t turn every integral into a long substitution when the reverse chain rule is already visible.

The analytical approach to definite integrals

The key result is

So, if $F'(x)=f(x)$, then

This can be written compactly as

For example,

Some definite integrals can’t be evaluated using the elementary antiderivatives in the course. In these cases, technology is the appropriate mathematical method, not a shortcut. Write the integral with the correct limits before using the calculator.

Signed area and true area

A signed area is the value of a definite integral: regions above the $x$-axis count as positive, while regions below it count as negative. A geometric area is the non-negative size of a region.

If $f(x)\ge 0$ throughout $[a,b]$, then

When $f$ stays below the axis throughout the interval,

If $f$ crosses the axis, split the interval at the intercepts. Alternatively, write

Without technology, splitting is usually clearer since it shows exactly where the sign changes.

Write a correct expression before calculating. For example, suppose a graph lies below the axis on $[p,q]$ and above it on $[q,r]$. Its area is

where $p$, $q$, and $r$ are $x$-values marking the endpoints and intercept.

Area between curves

Suppose two continuous curves $y=f(x)$ and $y=h(x)$ enclose a region from $x=a$ to $x=b$, with $f(x)\ge h(x)$ throughout the interval. Then

I usually say “top minus bottom”. It may sound less elegant than “upper function minus lower function”, but it prevents mistakes.

The limits often come from the intersection points. Find them by solving $f(x)=h(x)$, or use technology when allowed. If the curves swap which one is on top, split the integral at the crossing point. One neat-looking integral in the wrong order produces a signed result, not the required area.

Technology can help build intuition: graph the curves, shade the trapped region, and check which curve is above. On a non-technology paper, though, the expression and exact working must stand on their own.

Substitution beyond inspection

Integration by substitution changes the variable in an integral, turning a difficult-looking integrand into a standard one. If the integral is not an obvious inspection integral of the form $\int k g'(x)f(g(x))\,dx$, the examination question will provide the substitution.

Let $u=g(x)$, where $u$ is the new variable and $g(x)$ is the substituted expression. Then

In a friendly case,

When a substitution is provided, it is often cleaner to write $x$ in terms of $u$ and use

where $x(u)$ is $x$ written as a function of $u$, and $\frac{dx}{du}$ is the derivative of $x$ with respect to $u$. Students most often lose the factor $\frac{dx}{du}$. Without it, the answer is usually off by a constant factor or worse.

For definite integrals, change the limits as soon as the variable changes. Suppose $x=a$ becomes $u=\alpha$ and $x=b$ becomes $u=\beta$, where $\alpha$ and $\beta$ are the new lower and upper limits in the $u$-variable. Continue entirely in $u$:

This avoids substituting back to $x$ at the end.

Integration by parts

Integration by parts integrates a product by moving one derivative from one factor to the other. The method comes from the product rule, and its formula is

where $u$ and $v$ are differentiable functions of $x$.

The main decision is which part to differentiate. One useful classroom rule is ILATE: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. The earlier type is usually chosen as $u$ because differentiating it tends to simplify the integral.

For example, in $\int x\sin x\,dx$, choose $u=x$ and $\frac{dv}{dx}=\sin x$. Then $\frac{du}{dx}=1$ and $v=-\cos x$, so

Don’t invent a rule that says the integral of a product equals the product of the integrals. That rule is false. Parts works by changing the product integral into a new integral that should be easier.

Some integrals only become products when you rewrite them. For example,

Choose $u=\ln x$ and $\frac{dv}{dx}=1$. Then $\frac{du}{dx}=\frac{1}{x}$ and $v=x$, giving

The same approach works for inverse trigonometric functions such as $\int \arcsin x\,dx$.

Repeated integration by parts

Repeated integration by parts means using the parts formula more than once in the same problem. It is needed when one application makes the integral easier but does not finish it.

For a polynomial multiplied by an exponential or trigonometric function, each application usually reduces the polynomial degree. For instance, $\int x^2e^x\,dx$ needs parts twice: first reduce $x^2$ to $2x$, then reduce $2x$ to a constant.

With products such as $\int e^x\sin x\,dx$, repeated parts can bring back the original integral. This isn’t a failure. Put the original integral on one side and solve algebraically for it. Integration has become a small simultaneous equation with itself.

Area enclosed by a curve and the y-axis

For a region bounded by a curve and the $y$-axis, integrating with respect to $y$ is often the better approach. Suppose the curve is written as $x=r(y)$ from $y=c$ to $y=d$. Here, $r(y)$ is the horizontal distance from the $y$-axis (unit depends on context), while $c$ and $d$ are the lower and upper $y$-limits. Then

When the curve lies to the left of the $y$-axis, use the positive horizontal distance, or split/sign-correct as you would for area with the $x$-axis.

The idea hasn’t changed; only the direction of the thin strips has. Vertical strips lead to $dx$, while horizontal strips lead to $dy$.

Volumes of revolution

A solid of revolution is a three-dimensional solid formed by rotating a plane region around an axis. In this course, the rotation is about either the $x$-axis or the $y$-axis.

Rotate the region under $y=f(x)$ from $x=a$ to $x=b$ about the $x$-axis. Its cross-sections are circles with radius $f(x)$, so the volume is

Don’t miss the square: the integral adds circular areas, not lengths.

For a region between $x=r(y)$ and the $y$-axis from $y=c$ to $y=d$, rotated about the $y$-axis, the volume is

The quantity being squared is the radius of each circular slice. Before integrating, identify the axis of rotation and the radius, then set the limits. These questions fit naturally into design contexts because rotating a profile curve can generate a vase, nozzle, bowl, or machine part.

Some of the resulting definite integrals have no elementary antiderivative in the course. On a technology paper, first write the correct volume expression, then evaluate the definite integral numerically.