Integrate polynomials: find indefinite integrals with a constant of integration, evaluate definite integrals, and use definite integration to find the area under a curve.

What this dot point is asking

Integration is the reverse of differentiation and the second half of the calculus strand in CCEA GCSE Further Mathematics. You must integrate polynomials, remember the constant of integration in an indefinite integral, recover a function from its derivative when given a point, evaluate definite integrals between limits, and use a definite integral to find the area between a curve and the $x$-axis. Integration pairs directly with differentiation, so the two topics are best learned together.

The reverse-power rule

Integration undoes differentiation, so its rule reverses the power rule: instead of multiplying by the power and reducing it, you increase the power by one and divide by the new power. Write each term as a power of $x$ first.

For example, $\displaystyle\int (3x^2 + 2x)\,dx = x^3 + x^2 + c$. You can always check an integration by differentiating the answer: it should return the original expression.

The constant of integration

When you differentiate, any constant disappears, so when you integrate you cannot know what constant was there. The $+c$ stands for this unknown constant, and an indefinite integral is incomplete without it.

Finding a function from its derivative

If you know a curve's gradient function and one point on the curve, you can find the curve exactly. Integrate to get the general function with its $+c$, then substitute the known point to find the value of $c$. This pins down the particular curve from the whole family.

For instance, if $\dfrac{dy}{dx} = 4x - 1$ and the curve passes through $(2, 3)$, then $y = 2x^2 - x + c$; substituting gives $3 = 8 - 2 + c$, so $c = -3$ and $y = 2x^2 - x - 3$.

Definite integrals and area

A definite integral has limits and produces a number rather than a function. You integrate as usual, then substitute the upper limit and the lower limit and subtract; the constant cancels, so it is left out.

The definite integral of a curve that lies above the $x$-axis between two values of $x$ gives the area enclosed by the curve, the axis and the two vertical boundaries. If part of the curve dips below the axis, that part gives a negative contribution, so you treat regions above and below separately when an exact area is needed.

Why this matters

Integration completes the calculus toolkit and links back to differentiation as its inverse: recovering a function from its rate of change is integration, and it appears in the Mechanics unit where integrating acceleration gives velocity and integrating velocity gives displacement. The area-under-a-curve interpretation also connects calculus to geometry. Remembering the constant of integration and handling limits carefully are the habits these questions reward.