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Integrals: antiderivatives, areas and methods

The integral has two faces: the indefinite one is the inverse of the derivative (finding an antiderivative) and the definite one is an area, the sum of infinitely many very thin rectangles. The fundamental theorem of calculus joins them. This topic explains both, the integration methods (substitution, by parts, partial fractions) and how to compute areas.

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Antiderivatives and the indefinite integral

$F$ is an antiderivative of $f$ if $F^{'} = f$ . If $F$ is one, so is $F + C$ for any constant, since the derivative of a constant is 0. That is why $\int f (x) d x = F (x) + C$ . An antiderivative can always be checked by differentiating.

Function	Antiderivative	Function	Antiderivative
$x^{n} (n \neq = - 1)$	$\frac{x ^{n + 1}}{n + 1}$	$cos x$	$sin x$
$\frac{1}{x}$	$ln ∣ x ∣$	$sin x$	$- cos x$
$e^{a x}$	$\frac{e ^{a x}}{a}$	$sec^{2} x$	$tan x$
$\frac{1}{1 + x ^{2}}$	$arctan x$	$\frac{1}{1 - x ^{2}}$	$arcsin x$

Definite integral and area

To find the area under $y = f (x)$ between $a$ and $b$ , divide the interval into $n$ strips of width $Δ x$ and add up rectangles: $\sum f (x_{i}) Δ x$ . As the strips get thinner, the sum tends to the definite integral $\int_{a}^{b} f (x) d x$ . Parts where $f < 0$ count with a negative sign.

Sum of rectangles under

y = x^{2}

[0, 1]

: with 5 rectangles it gives 0.44; with more, it approaches the exact value 1/3.

The fundamental theorem of calculus says there is no need to add rectangles: if $F$ is an antiderivative of a continuous $f$ ,

\int_{a}^{b} f (x) d x = F (b) - F (a) (evaluation theorem)

The area between two curves is $\int_{a}^{b} (f_{top} - f_{bottom}) d x$ , with $a$ and $b$ at the crossing points. If the interval is infinite or the function has an asymptote inside, the integral is improper: it is computed as a limit and may converge or diverge.

Integration methods

Method	When	Idea
Substitution	a function and (almost) its derivative appear	$u = g (x)$ , $d u = g^{'} (x) d x$
By parts	product of a polynomial with $e^{x}$ , $sin$ , $cos$ ; or $ln$ , $arctan$	$\int u d v = uv - \int v d u$
Partial fractions	quotient of polynomials	factor the denominator and split into simple fractions
Trigonometric	powers of $sin$ and $cos$ ; roots $a^{2} - x^{2}$ …	identities and trigonometric substitution

In «by parts», $u$ is chosen as the part that gets simpler when differentiated (a polynomial drops in degree; $ln x$ becomes $1/ x$ ) and $d v$ as the part that can be integrated.

Not every function has an antiderivative written with elementary functions:

\int e^{- x^{2}} d x

does not. In that case the calculator says so and computes the definite integral numerically (

\int_{0}^{1} e^{- x^{2}} d x \approx 0.74682413

, an example in the list). Every antiderivative it shows is first checked by differentiating it.

Common mistakes

Forgetting the constant $+ C$ in the indefinite integral.
Using $\frac{x ^{n + 1}}{n + 1}$ with $n = - 1$ : $\int \frac{d x}{x} = ln ∣ x ∣$ .
When substituting, changing $x$ to $u$ but not $d x$ to $d u$ (or not changing the limits of a definite integral).
Applying Barrow across an asymptote: $\int_{- 1}^{1} \frac{d x}{x ^{2}}$ is not $- 2$ , it diverges.

Worked examples

Example 1A direct integral

Compute

\int (3 x^{2} + 2 x) d x

Solution

Term by term with the power rule: $\int 3 x^{2} d x = 3 \frac{x ^{3}}{3} = x^{3}$ and $\int 2 x d x = x^{2}$ .
Check: $(x^{3} + x^{2})^{'} = 3 x^{2} + 2 x$ . ✓

x^{3} + x^{2} + C

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Integrate: type 3x^2+2x (leave From and To empty) and press Compute. The Answer is $x^{3} + x^{2} + C$ .

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Example 2Substitution

Compute

\int 2 x cos (x^{2}) d x

Solution

Inside the cosine there is $x^{2}$ and outside its derivative $2 x$ : $u = x^{2}$ , $d u = 2 x d x$ .
$\int cos u d u = sin u + C = sin (x^{2}) + C$ .

sin (x^{2}) + C

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Integrate: type 2x cos(x^2) (leave From and To empty) and press Compute. The Answer is $sin (x^{2}) + C$ .
The steps say why $u$ is chosen, solve for $d x$ and undo the change at the end.

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Example 3By parts

Compute

\int x e^{x} d x

Solution

$u = x$ (it becomes 1 when differentiated) and $d v = e^{x} d x$ : $d u = d x$ , $v = e^{x}$ .
$\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x} + C$ .
Check: $(x e^{x} - e^{x})^{'} = e^{x} + x e^{x} - e^{x} = x e^{x}$ . ✓

x e^{x} - e^{x} + C

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Integrate: type x e^x (leave From and To empty) and press Compute. The Answer is $x e^{x} - e^{x} + C$ .

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Example 4Partial fractions

Compute

\int \frac{d x}{x ^{2} - 1}

Solution

$x^{2} - 1 = (x - 1) (x + 1)$ , so $\frac{1}{x ^{2} - 1} = \frac{A}{x - 1} + \frac{B}{x + 1}$ , that is, $1 = A (x + 1) + B (x - 1)$ .
With $x = 1$ : $A = \frac{1}{2}$ ; with $x = - 1$ : $B = - \frac{1}{2}$ .
$\int = \frac{1}{2} ln ∣ x - 1∣ - \frac{1}{2} ln ∣ x + 1∣ + C$ .

\frac{ln ∣ x - 1∣}{2} - \frac{ln ∣ x + 1∣}{2} + C

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Integrate: type 1/(x^2-1) (leave From and To empty) and press Compute. The Answer is $\frac{l n ∣ x - 1∣}{2} - \frac{l n ∣ x + 1∣}{2} + C$ .
The steps set up $A$ and $B$ as letters and match coefficients.

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Example 5Definite integral (Barrow)

Compute

\int_{0}^{1} x^{2} d x

, the area in figure 1.

Solution

Antiderivative: $F (x) = \frac{x ^{3}}{3}$ .
$F (1) - F (0) = \frac{1}{3} - 0 = \frac{1}{3}$ . The 5 rectangles gave 0.44, too much, because they take the height at the right end of each strip and the function increases.

\frac{1}{3}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Integrate: type x^2, From = 0 and To = 1 and press Compute. The Answer is $\frac{1}{3}$ .
The small graph shades the area under the curve.

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Example 6Area between two curves

Compute the area enclosed between

y = x

and

y = x^{2}

Solution

Crossings: $x = x^{2}$ → $x (x - 1) = 0$ → $x = 0$ and $x = 1$ . At $x = 0.5$ : $0.5 > 0.25$ , so $y = x$ is on top.
$A = \int_{0}^{1} (x - x^{2}) d x = [\frac{x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ .

A = \frac{1}{6}

Solve it in Simuveliq

In the Compute operation type area(x, x^2) and press Compute: the Answer is $A = \frac{1}{6}$ ; the steps find the crossings $x = 0$ and $x = 1$ and subtract the lower curve from the upper one.

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Example 7An improper integral

Compute

\int_{0}^{\infty} \frac{d x}{1 + x ^{2}}

Solution

Antiderivative: $arctan x$ . The upper limit is infinite: $lim_{b \to \infty} (arctan b - arctan 0)$ .
$arctan b \to \frac{π}{2}$ , so the integral converges to $\frac{π}{2} \approx 1.5707963$ : an infinitely long but finite area.

\frac{π}{2} \approx 1.5707963

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Integrate: type 1/(1+x^2), From = 0 and To = inf and press Compute. The Answer is $\frac{π}{2} \approx 1.5707963$ .

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Integrals: antiderivatives, areas and methods

Antiderivatives and the indefinite integral

Definite integral and area

Integration methods

Common mistakes

Worked examples

Example 1A direct integral

Solution

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Example 2Substitution

Solution

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Example 3By parts

Solution

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Example 4Partial fractions

Solution

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Example 5Definite integral (Barrow)

Solution

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Example 6Area between two curves

Solution

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Example 7An improper integral

Solution

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