Integration as the reverse of differentiation, indefinite integrals with a constant, definite integrals and the limits, and the area under a curve.

What this dot point is asking

WJEC wants you to treat integration as the reverse of differentiation, to find indefinite integrals (remembering the constant of integration), to evaluate definite integrals between limits, and to use a definite integral to find the area under a curve. Integration partners differentiation throughout the course and underpins the kinematics and differential-equation work later.

The answer

Integration as the reverse of differentiation

If differentiating $F(x)$ gives $f(x)$, then integrating $f(x)$ recovers $F(x)$ up to a constant. Because differentiating a constant gives zero, the reverse must include an unknown constant of integration $c$.

To integrate a sum, integrate each term. Rewrite roots and reciprocals as powers first, exactly as for differentiation. The rule excludes $n = -1$ because $\dfrac{x^0}{0}$ is undefined; integrating $\dfrac{1}{x}$ gives $\ln|x| + c$, a result you meet alongside the exponential and logarithm work. A quick self-check on any integration is to differentiate your answer: if it returns the original integrand, the integration is correct, which is worth doing under exam pressure.

Finding a curve from its gradient

If you know $\dfrac{dy}{dx}$ and one point on the curve, integrate to get a family of curves $y = F(x) + c$, then substitute the point to pin down $c$.

Definite integrals and area

A definite integral evaluates the antiderivative at the two limits and subtracts.

Examples in context

Example 1. Area between two curves

The area between $y = x^2$ and $y = 2x$ from $x = 0$ to $x = 2$ is $\displaystyle\int_0^2 (2x - x^2)\,dx = \left[x^2 - \tfrac{x^3}{3}\right]_0^2 = 4 - \tfrac{8}{3} = \tfrac{4}{3}$. Subtracting the lower curve from the upper before integrating gives the enclosed region directly.

Example 2. Distance from velocity

If a particle has velocity $v = 3t^2$ then its displacement from $t = 0$ to $t = 2$ is $\displaystyle\int_0^2 3t^2\,dt = [t^3]_0^2 = 8\,\text{m}$. The definite integral of velocity gives displacement, the link that the mechanics unit exploits.

Example 3. A region partly below the axis

To find the total area enclosed by $y = x^2 - 1$ and the $x$-axis between $x = 0$ and $x = 2$, split at the root $x = 1$. From $0$ to $1$ the curve is below the axis: $\displaystyle\int_0^1 (x^2 - 1)\,dx = \left[\tfrac{x^3}{3} - x\right]_0^1 = -\tfrac{2}{3}$, so this area is $\tfrac{2}{3}$. From $1$ to $2$ the curve is above: $\displaystyle\int_1^2 (x^2 - 1)\,dx = \tfrac{4}{3}$. The total area is $\tfrac{2}{3} + \tfrac{4}{3} = 2$ square units, not the single integral over $[0,2]$ which would cancel part of it.

Try this

Q1. Find $\displaystyle\int (6x^2 - 4x + 5)\,dx$. [3 marks]

• Cue. $2x^3 - 2x^2 + 5x + c$ (do not forget the constant).

Q2. Evaluate $\displaystyle\int_0^4 \sqrt{x}\,dx$. [3 marks]

• Cue. $\sqrt{x} = x^{1/2}$, integral is $\tfrac{2}{3}x^{3/2}$; at the limits $\tfrac{2}{3}(8) - 0 = \tfrac{16}{3}$.

Q3. A curve has gradient $\dfrac{dy}{dx} = 2x + 1$ and passes through $(2, 7)$. Find $y$. [3 marks]

• Cue. $y = x^2 + x + c$; substitute to get $7 = 6 + c$, so $c = 1$ and $y = x^2 + x + 1$.