Integration — H2 Maths help

Integration is the reverse process of differentiation. Every technique you learn here — from reading off a standard result to setting up a volume of revolution — ultimately rests on one question: what function, when differentiated, gives the integrand?

What you need to know

Standard integrals you must recall: $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ $(n \neq -1)$ , $\int e^x\,dx = e^x + C$ , $\int \dfrac{1}{x}\,dx = \ln|x| + C$ , $\int \cos x\,dx = \sin x + C$ , $\int \sin x\,dx = -\cos x + C$ , $\int \sec^2 x\,dx = \tan x + C$ .
Integration by substitution replaces a complicated expression with a single variable $u$ ; always change $dx$ to $du$ using $\dfrac{du}{dx}$ .
Integration by parts follows $\displaystyle\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx$ ; choose $u$ to be the factor that simplifies on differentiation (LIATE order: Logarithm, Inverse trig, Algebraic, Trig, Exponential).
Definite integrals give a numerical value: $\displaystyle\int_a^b f(x)\,dx = [F(x)]_a^b = F(b) - F(a)$ where $F'(x) = f(x)$ .
Area between curves: if $f(x) \geq g(x)$ on $[a,b]$ , the enclosed area is $\displaystyle\int_a^b \bigl[f(x) - g(x)\bigr]\,dx$ .
Volume of revolution about the $x$ -axis: $V = \pi\displaystyle\int_a^b y^2\,dx$ .

Worked example 1 — Integration by parts

Find $\displaystyle\int x e^x\,dx$ .

Set up. Let $u = x$ and $\dfrac{dv}{dx} = e^x$ , so $\dfrac{du}{dx} = 1$ and $v = e^x$ .

Applying the formula:

\int x e^x\,dx = x e^x - \int e^x \cdot 1\,dx = x e^x - e^x + C.

Check by differentiating: $\dfrac{d}{dx}(x e^x - e^x) = e^x + x e^x - e^x = x e^x$ . Confirmed.

\boxed{\int x e^x\,dx = (x-1)e^x + C}

Worked example 2 — Area between two curves

Find the area enclosed between $y = x^2$ and $y = x + 2$ .

Step 1 — intersections. Set $x^2 = x + 2$ , so $x^2 - x - 2 = 0$ , giving $(x-2)(x+1) = 0$ , i.e. $x = -1$ and $x = 2$ .

Step 2 — which is on top? At $x = 0$ : $y = x + 2$ gives $2$ , $y = x^2$ gives $0$ , so the line is above the parabola on $[-1, 2]$ .

Step 3 — integrate.

A = \int_{-1}^{2} \bigl[(x+2) - x^2\bigr]\,dx

= \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}

At $x = 2$ : $\dfrac{4}{2} + 4 - \dfrac{8}{3} = 6 - \dfrac{8}{3} = \dfrac{10}{3}$ .

At $x = -1$ : $\dfrac{1}{2} - 2 + \dfrac{1}{3} = -\dfrac{7}{6}$ .

A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}.

\boxed{A = \tfrac{9}{2} \text{ square units}}

Worked example 3 — Integration by substitution

Find $\displaystyle\int 2x\sqrt{x^2+1}\,dx$ .

Let $u = x^2 + 1$ , so $\dfrac{du}{dx} = 2x$ , i.e. $2x\,dx = du$ .

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

Substituting back:

\boxed{\int 2x\sqrt{x^2+1}\,dx = \tfrac{2}{3}(x^2+1)^{3/2} + C}

Check: $\dfrac{d}{dx}\!\left[\tfrac{2}{3}(x^2+1)^{3/2}\right] = \tfrac{2}{3} \cdot \tfrac{3}{2}(x^2+1)^{1/2} \cdot 2x = 2x\sqrt{x^2+1}$ . Confirmed.

Common mistakes

Forgetting the constant of integration $C$ in indefinite integrals — always include it.
Wrong sign with $\int \sin x\,dx$ — the result is $-\cos x + C$ , not $+\cos x + C$ .
Not swapping limits (or the integrand) when substituting in a definite integral — either change the limits to $u$ -values, or substitute back to $x$ before evaluating.
Taking the area as negative when the curve dips below the $x$ -axis — split the integral at the $x$ -intercept and take the absolute value of each part separately.