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The area between a graph and the x-axis, and between two graphs

The area between a graph and the x-axis, and between two graphs

An area above the x\displaystyle x-axis

Where the graph of f\displaystyle f is above the x\displaystyle x-axis for a≤x≤b\displaystyle a\le x\le b, the area between the graph and the axis is the definite integral ∫abf(x) dx\displaystyle \int_a^b f(x)\,dx. The example below calculates the area bounded by the graph of f(x)=ex\displaystyle {f(x)}={e^x}, the x\displaystyle x-axis and the lines x=0\displaystyle {x}={0} and x=1\displaystyle {x}={1}:

f(x)=ex\displaystyle f(x)=e^{x}
S=∫01ex dx=[ex]01=e−1≈1.72S=\int_0^1e^x\,dx=\left[e^x\right]_0^1=e-1\approx1.72
An area below the x\displaystyle x-axis

Where the graph is below the x\displaystyle x-axis, the integral is negative, and the area is its opposite – or the integral of −f(x)\displaystyle -f(x). The graph of f(x)=ex−3\displaystyle {f(x)}={e^x-3} crosses the x\displaystyle x-axis at x=ln⁡3\displaystyle {x}={\ln3}, and to the left of that point it is below the axis. The example below calculates the area bounded by the graph, the x\displaystyle x-axis and the y\displaystyle y-axis:

f(x)=ex−3\displaystyle f(x)=e^{x}-3
S=∫0ln⁡3(3−ex)dx=[3x−ex]0ln⁡3=(3ln⁡3−3)−(0−1)=3ln⁡3−2≈1.30S=\int_0^{\ln3}\left(3-e^x\right)dx=\left[3x-e^x\right]_0^{\ln3}=(3\ln3-3)-(0-1)=3\ln3-2\approx1.30
A function that changes sign

Where the function changes sign in the interval, we split the interval at its zero and calculate each part on its own. The function f(x)=2x−1\displaystyle {f(x)}={\frac{2}{x}-1} is zero at x=2\displaystyle {x}={2}: between 1 and 2 it is positive, between 2 and 4 negative.

f(x)=2x−1\displaystyle f(x)=\frac{2}{x}-1
  • The part above the axis: ∫12(2x−1)dx=[2ln⁡x−x]12=2ln⁡2−1\displaystyle {\int_1^2\left(\frac{2}{x}-1\right)dx}={\left[2\ln x-x\right]_1^2}={2\ln2-1}.
  • The part below the axis: ∫24(1−2x)dx=[x−2ln⁡x]24=2−2ln⁡2\displaystyle {\int_2^4\left(1-\frac{2}{x}\right)dx}={\left[x-2\ln x\right]_2^4}={2-2\ln2}.
  • The whole area: (2ln⁡2−1)+(2−2ln⁡2)=1\displaystyle {(2\ln2-1)+(2-2\ln2)}={1}.
  • The integral from 1 to 4 without splitting gives 4ln⁡2−3≈−0.23\displaystyle 4\ln2-3\approx-0.23: the parts cancel each other out, and that is not the area.
The area between two graphs

The area between two graphs, in an interval where f(x)≥g(x)\displaystyle f(x)\ge g(x):

S=∫ab(f(x)−g(x))dx\displaystyle S=\int_{a}^{b}\left(f(x)-g(x)\right)dx

The example below calculates the area bounded by the graphs of f(x)=e3x−12e2x+36ex\displaystyle {f(x)}={e^{3x}-12e^{2x}+36e^x} and of g(x)=e3x\displaystyle {g(x)}={e^{3x}} and the y\displaystyle y-axis:

f(x)=e3x−12e2x+36ex\displaystyle f(x)=e^{3x}-12e^{2x}+36e^{x}
g(x)=e3x\displaystyle g(x)=e^{3x}
  • The meeting point: f(x)=g(x)\displaystyle {f(x)}={g(x)} gives 36ex−12e2x=0\displaystyle {36e^x-12e^{2x}}={0}, that is 12ex(3−ex)=0\displaystyle {12e^x(3-e^x)}={0}. ex>0\displaystyle e^x>0, so ex=3\displaystyle {e^x}={3} and x=ln⁡3\displaystyle {x}={\ln3}.
  • For 0≤x≤ln⁡3\displaystyle 0\le x\le\ln3: f(x)−g(x)=36ex−12e2x≥0\displaystyle {f(x)-g(x)}={36e^x-12e^{2x}\ge0}.
  • The area:
S=∫0ln⁡3(36ex−12e2x)dx=[36ex−6e2x]0ln⁡3=(108−54)−(36−6)=24S=\int_0^{\ln3}\left(36e^x-12e^{2x}\right)dx=\left[36e^x-6e^{2x}\right]_0^{\ln3}=(108-54)-(36-6)=24
A parameter from a given area

When the area is given, a limit of the integral can be the unknown. The example below finds a>1\displaystyle a>1 such that the area bounded by the graph of f(x)=1x\displaystyle {f(x)}={\frac{1}{x}}, the x\displaystyle x-axis and the lines x=1\displaystyle {x}={1} and x=a\displaystyle {x}={a} is 2:

  • ∫1a1x dx=[ln⁡x]1a=ln⁡a−ln⁡1=ln⁡a\displaystyle {\int_1^a\frac{1}{x}\,dx}={\left[\ln x\right]_1^a}={\ln a-\ln1}={\ln a}.
  • ln⁡a=2\displaystyle {\ln a}={2}, so a=e2≈7.39\displaystyle {a}={e^2\approx7.39}.
Exercise 1

Calculate the area bounded by the graph of f(x)=e2x\displaystyle {f(x)}={e^{2x}}, the x\displaystyle x-axis and the lines x=0\displaystyle {x}={0} and x=1\displaystyle {x}={1}.

f(x)=e2x\displaystyle f(x)=e^{2x}

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

    Exercise 2

    The function f(x)=ex−e\displaystyle {f(x)}={e^x-e} is given.

    a

    Mark the points where the graph meets the axes.

    f(x)=ex−e\displaystyle f(x)=e^{x}-e

    b

    Calculate the area bounded by the graph of the function, the x\displaystyle x-axis and the y\displaystyle y-axis.

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

      Exercise 3

      The functions f(x)=3x\displaystyle {f(x)}={\frac{3}{x}} and g(x)=4−x\displaystyle {g(x)}={4-x} are given, for x>0\displaystyle x>0.

      a

      Find the x\displaystyle x-coordinates of the points where the two graphs meet.

        b

        Calculate the area enclosed between the two graphs.

        f(x)=3x\displaystyle f(x)=\frac{3}{x}
        g(x)=4−x\displaystyle g(x)=4-x

          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

          Exercise 4

          The function f(x)=1−1x\displaystyle {f(x)}={1-\frac{1}{x}} is given.

          a

          Find the zero of the function.

            b

            Calculate the area bounded by the graph of the function, the x\displaystyle x-axis and the lines x=12\displaystyle {x}={\frac{1}{2}} and x=2\displaystyle {x}={2}.

            f(x)=1−1x\displaystyle f(x)=1-\frac{1}{x}

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

              Exercise 5

              k\displaystyle k is a positive parameter.
              The area bounded by the graph of f(x)=ex\displaystyle {f(x)}={e^x}, the axes and the line x=k\displaystyle {x}={k} is 3.
              Find k\displaystyle k: write an equation in k\displaystyle k and solve it.