Math ↠ Calculating areas with integrals ↠ Calculating areas by partitioning the interval

Calculating areas by partitioning the interval

An exception to calculating areas are functions with sign changes in the interval $[a; b]$. This means that the area is partly above and partly below the x-axis.

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Remember

If the area you are looking for is both above and below the x-axis, you must calculate the areas separately.
This is the case when the function has a zero in the interval.
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Method

  1. Determine zeros and find intervals
  2. Calculate definite integrals for each interval
  3. Determine area

Example

Calculate the area between the graph of the function $f(x)=x^2+2x$ and the x-axis over the interval $[-1; 1]$

  1. Find intervals

    First, calculate the zero(s) of the function.
    $x^2+2x=0$
    (Solve quadratic equation, e.g. remove brackets)
    $x(x+2)=0$
    $x_{N_1}=0$ and $x_{N_2}=-2$

    $x_{N_1}=0$ is in the interval $[-1; 1]$. Therefore, the area must be divided into two parts:
    $A_1$ over $[-1;0]$
    $A_2$ over $[0;1]$

  2. Find and calculate definite integrals

    For both intervals, an integral must now be calculated.
    $\int_a^b f(x) \, \mathrm{d}x$ $= [F(x) + C]_a^b$ $= F(b) - F(a)$
    $F(x)=\frac13x^3+x^2$

    $A_1$ over $[-1;0]$:
    $\int_{-1}^0 (x^2+2x) \, \mathrm{d}x$ $= [\frac13x^3+x^2]_{-1}^0$ $= \frac13\cdot0^3+0^2 -$ $\frac13\cdot(-1)^3+(-1)^2$
    $=0-\frac23$ $=-\frac23$


    $A_2$ over $[0;1]$:
    $\int_0^1 (x^2+2x) \, \mathrm{d}x$ $= [\frac13x^3+x^2]_0^1$ $= \frac13\cdot1^3+1^2 -$ $\frac13\cdot0^3+0^2$
    $=\frac43-0$ $=\frac43$

  3. Determine area

    Now the area of each partition has to be determined and then added.
    $A_1=|\int_{-1}^0 f(x)\,\mathrm{d}x|$ $=|-\frac23|$ $=\frac23$
    $A_2=\int_0^1 f(x)\,\mathrm{d}x$ $=\frac43$

    $A=A_1+A_2$ $=\frac23+\frac43=2$