Evaluate definite integrals using limits and use them to find the area of a region bounded by a curve and the x-axis

What this dot point is asking

SEAB wants you to evaluate a definite integral, an integral with a lower and upper limit, and to use it to find the area enclosed between a curve and the $x$-axis. The definite integral turns the abstract antiderivative into a concrete number equal to that area.

The answer

Evaluating a definite integral

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

No constant of integration is needed, because $C$ cancels when you subtract.

Area under a curve

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:

The integral adds up infinitely many thin strips of height $y$ and width $dx$.

Regions below the axis

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.

Checking the set-up

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.

Useful properties of definite integrals

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: $\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx$. This last property is exactly what justifies splitting at an axis crossing when computing a true area. Recognising these properties lets you simplify or recombine integrals without re-evaluating from scratch.

Symmetry shortcuts for even functions

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.

Examples in context

Example 1. Distance from a velocity-time graph. The area under a velocity-time graph between two times equals the distance travelled in that interval, which is exactly the definite integral of velocity, linking area to kinematics.

Example 2. Work done by a varying force. The work done by a force that varies with position is the area under the force-distance graph, computed as a definite integral, a standard physics application of this idea.

Try this

Q1. Evaluate $\displaystyle\int_{0}^{2} 3x^2\,dx$. [2 marks]

• Cue. $[x^3]_0^2 = 8 - 0 = 8$.

Q2. Evaluate $\displaystyle\int_{1}^{2} (4x - 1)\,dx$. [3 marks]

• Cue. $[2x^2 - x]_1^2 = (8 - 2) - (2 - 1) = 6 - 1 = 5$.

Q3. Find the area under $y = x^2 + 1$ between $x = 0$ and $x = 2$. [3 marks]

• Cue. $\left[\dfrac{x^3}{3} + x\right]_0^2 = \dfrac{8}{3} + 2 = \dfrac{14}{3}$ square units.