How Do You Calculate a Definite Integral Analytically?

To calculate a definite integral analytically, find an antiderivative of the integrand and evaluate it at the upper and lower limits. If $F'(x)=f(x)$, then

$$\int_{a}^{b} f(x)\,dx=F(b)-F(a).$$

The subtraction measures the net accumulation from $a$ to $b$. Differentiation and integration are inverse processes, so $F$ converts the changing rate $f(x)$ into an accumulated quantity; evaluating at two endpoints isolates only the accumulation across the required interval.

The Method

This result is the fundamental theorem of calculus. Use these steps:

1. Find an antiderivative $F(x)$ of $f(x)$.
2. Write the evaluated integral as $[F(x)]_{a}^{b}$.

For example, calculate

$$\int_{1}^{3}(2x+1)\,dx.$$

An antiderivative of $2x+1$ is $x^2+x$. Therefore,

$$\begin{aligned} \int_{1}^{3}(2x+1)\,dx &=[x^2+x]_{1}^{3}\\ &=(3^2+3)-(1^2+1)\\ &=12-2\\ &=10. \end{aligned}$$

The constant of integration $C$ is normally omitted because it cancels in $(F(b)+C)-(F(a)+C)$.

As a quick check, the integrand $2x+1$ is positive and ranges from $3$ to $7$ on $[1,3]$. Over an interval of width $2$, an integral of is reasonable because it equals width times the function's average value, .

Integral Versus Area

A definite integral gives signed area, so regions below the $x$-axis contribute negatively.

For instance, $\int_{0}^{2}(x-1)\,dx=0$, but the total area between the graph and the $x$-axis is , because the negative and positive regions must be counted separately.

IB Exam Technique

This is IB Mathematics: Analysis and Approaches SL 5.11 and can appear on Paper 1 or Paper 2. For command terms such as calculate or find, show the antiderivative and substitution of both limits; for area questions, write the correct integral expression before evaluating it. A common misconception is that every definite integral equals geometric area: it equals signed area unless the question specifically requires total area.