Find antiderivatives of standard functions, apply integration by substitution and evaluate definite integrals using the Fundamental Theorem of Calculus

What this dot point is asking

NESA wants you to find antiderivatives of standard functions, apply the reverse chain rule via substitution, and evaluate definite integrals using the Fundamental Theorem of Calculus (FTC). Integration underlies areas, volumes, motion problems and growth models.

The answer

Standard antiderivatives

Memorise these. The constant $C$ is omitted in definite integrals.

Linear inside argument

If the argument is linear, divide by the coefficient.

Integration by substitution

The substitution rule reverses the chain rule. To evaluate $\int f(g(x)) g'(x) \, dx$, set $u = g(x)$, so $du = g'(x) \, dx$. The integral becomes $\int f(u) \, du$, which you evaluate, then substitute back.

For a definite integral, you can either substitute back to $x$ and use the original limits, or change the limits to values of $u$ and skip the back-substitution.

The Fundamental Theorem of Calculus

If $F$ is any antiderivative of $f$ (so $F'(x) = f(x)$), then

A second statement: the function $G(x) = \int_a^x f(t) \, dt$ satisfies $G'(x) = f(x)$. In short, differentiation and integration are inverse operations.

Worked examples

Standard antiderivative

$\int (5 x^3 + 4 x - 7) \, dx = \frac{5 x^4}{4} + 2 x^2 - 7 x + C$.

Linear inside argument

$\int e^{3 x} \, dx = \frac{e^{3 x}}{3} + C$.

$\int (2 x + 1)^{4} \, dx = \frac{(2 x + 1)^{5}}{10} + C$.

Substitution

Evaluate $\int x \sqrt{x^2 + 4} \, dx$.

Let $u = x^2 + 4$, so $du = 2 x \, dx$, giving $x \, dx = \frac{1}{2} du$.

$\int x \sqrt{x^2 + 4} \, dx = \int \sqrt{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} (x^2 + 4)^{3/2} + C$.

Definite integral with substitution and changed limits

Evaluate $\int_0^1 2 x e^{x^2} \, dx$.

Let $u = x^2$, $du = 2 x \, dx$. When $x = 0$, $u = 0$. When $x = 1$, $u = 1$.

$\int_0^1 2 x e^{x^2} \, dx = \int_0^1 e^u \, du = [e^u]_0^1 = e - 1$.

Using the FTC in reverse

If $G(x) = \int_1^x \cos(t^2) \, dt$, then $G'(x) = \cos(x^2)$. No antiderivative needed.

Common traps

Forgetting the $+ C$. An indefinite integral must include the constant of integration.

Dividing by the wrong coefficient. $\int e^{2 x} \, dx = \frac{1}{2} e^{2 x} + C$, not $2 e^{2 x} + C$.

Substituting without changing $dx$. When you substitute $u = g(x)$, the $dx$ must become $\frac{du}{g'(x)}$ (or the integrand must already contain a multiple of $g'(x) \, dx$).

Mixing limits and the variable. After a substitution, either change the limits to $u$ values or substitute back to $x$ before evaluating. Do not mix the two.

Using the FTC with $|x|$ inside a logarithm wrongly. $\int \frac{1}{x} \, dx = \ln |x| + C$. For a definite integral, the interval must not cross $x = 0$.

In one sentence

Integration finds antiderivatives, the substitution rule reverses the chain rule, and the Fundamental Theorem of Calculus evaluates a definite integral as the change of any antiderivative across the interval.