How Do You Find the Area Under a Curve Using Integration?

You find the area between a curve $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$ using the definite integral $A=\int_{a}^{b}f(x)\,dx$, provided $f(x)\geq 0$ throughout the interval. If the curve goes below the axis, the integral gives signed area rather than total geometric area.

The Method

Integration adds infinitely many thin vertical strips. A strip of width $\Delta x$ has approximate area $f(x)\Delta x$; the limiting sum of these areas is the definite integral. The Fundamental Theorem of Calculus connects this accumulation to an antiderivative.

1. Identify $f(x)$ and the boundaries $a$ and $b$.
2. Write the definite integral before evaluating it.
3. Find an antiderivative $F(x)$, where $F'(x)=f(x)$.

For example, find the area under $y=x^2+1$ from $x=0$ to $x=2$:

$$A=\int_{0}^{2}(x^2+1)\,dx.$$

An antiderivative is $F(x)=\frac{x^3}{3}+x$. Therefore,

$$A=\left[\frac{x^3}{3}+x\right]_{0}^{2} =\left(\frac{8}{3}+2\right)-0 =\frac{14}{3}.$$

The area is $\frac{14}{3}$ square units.

A useful check is to estimate the area from the graph before calculating. Since $x^2+1$ rises from $1$ to $5$ over an interval of width $2$, an answer of $\frac{14}{3}\approx4.67$ square units is reasonable. Also remember that exchanging the limits changes the sign: . This affects signed area, not the physical size of the enclosed region.

A common misconception is that every definite integral gives geometric area. It actually gives signed area, so a region below the $x$-axis contributes negatively. For total area, split at each root and make every part positive, or evaluate $\int_{a}^{b}|f(x)|\,dx$, as developed in SL 5.11.

IB Exam Technique

For Mathematics: Analysis and Approaches, positive-curve area appears in SL 5.5, while curves crossing the axis are covered in SL 5.11. Examiners expect the correct integral expression before evaluation. On Paper 1, show the antiderivative and exact substitution. On Paper 2, still write the integral before using a GDC, and include square units when appropriate.