Quick questions on Definite integrals and area under a curve explained: O-Level A-Maths

A definite integral is found by integrating, then substituting the upper limit minus the lower limit:

For a curve lying above the $x$-axis between $x = a$ and $x = b$, the area of the region bounded by the curve, the axis and the two vertical lines is:

Where the curve dips below the $x$-axis, the integral gives a negative value, because the heights $y$ are negative. The actual area is the modulus of that value. If a region is partly above and partly below the axis, split the integral at the crossing point and add the absolute values, or the positive and negative parts will cancel and understate the true area.

Always confirm the limits match the region and check whether the curve crosses the axis between them, since that decides whether you need to split the calculation.

A few properties make definite integrals quicker to handle and are worth knowing. Swapping the limits flips the sign: $\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$. An integral over a zero-width interval is zero: $\int_a^a f(x)\,dx = 0$. And an integral can be split at any interior point: $\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx$.

When the integrand is an even function (only even powers of $x$, so $f(-x) = f(x)$) and the limits are symmetric about zero, the area on each side of the $y$-axis is equal, so $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$. This halves the work, as in the worked example where $\int_{-2}^{2}(4 - x^2)\,dx = 2\int_0^2 (4 - x^2)\,dx$. By contrast, an odd function integrated over symmetric limits gives zero, because the negative part exactly cancels the positive part. Spotting symmetry before integrating can save time and provides a useful check on the answer.

If the curve crosses the $x$-axis between the limits, integrate each piece separately and add the absolute values.

When the region is bounded by the curve meeting the axis, find those intersection points to use as limits.

Evaluate $\displaystyle\int_{0}^{2} 3x^2\,dx$. [2 marks]

Evaluate $\displaystyle\int_{1}^{2} (4x - 1)\,dx$. [3 marks]

Find the area under $y = x^2 + 1$ between $x = 0$ and $x = 2$. [3 marks]