The area between two graphs

When the graph of $\displaystyle f$ lies above the graph of $\displaystyle g$ between $\displaystyle {x}={a}$ and $\displaystyle {x}={b}$, the area between them is the integral of the difference:

• $\displaystyle f(x)-g(x)$ is the height of the strip between the graphs at each $\displaystyle x$, so it is always positive – even when the graphs are below the $\displaystyle x$-axis;
• the area between a graph and the $\displaystyle x$-axis is a special case: $\displaystyle {g(x)}={0}$;
• when the graphs enclose an area, $\displaystyle a$ and $\displaystyle b$ are their meeting points: solve $\displaystyle {f(x)}={g(x)}$.

$\displaystyle {f(x)}={4x-x^2}$, $\displaystyle {g(x)}={x}$.
In the example below we work out the area the graphs enclose:

• The meeting points: $\displaystyle {4x-x^2}={x}$, that is $\displaystyle {3x-x^2}={0}$, $\displaystyle {x(3-x)}={0}$: $\displaystyle {x}={0}$, $\displaystyle {x}={3}$;
• between 0 and 3 the parabola is above the line (for example, at $\displaystyle {x}={1}$: $\displaystyle 3>1$);
• $\displaystyle {S}={\int_0^3\left((4x-x^2)-x\right)dx}={\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3}={\frac{27}{2}-9}={\frac{9}{2}}$.

$\displaystyle {f(x)}={x^2}$. The tangent to the graph at the point $\displaystyle (2,4)$ is $\displaystyle {y}={4x-4}$ (the slope $\displaystyle {f'(2)}={4}$), and it crosses the $\displaystyle x$-axis at $\displaystyle {x}={1}$.
In the solution below we work out the area bounded by the graph, the tangent and the $\displaystyle x$-axis – it is made of two parts:

• From 0 to 1 – between the graph and the $\displaystyle x$-axis: $\displaystyle {\int_0^1x^2\,dx}={\frac{1}{3}}$;
• from 1 to 2 – between the graph and the tangent: $\displaystyle {\int_1^2\left(x^2-(4x-4)\right)dx}={\left[\frac{x^3}{3}-2x^2+4x\right]_1^2}={\frac{8}{3}-\frac{7}{3}}={\frac{1}{3}}$;
• the whole area: $\displaystyle {\frac{1}{3}+\frac{1}{3}}={\frac{2}{3}}$.

Another way: the area under the graph from 0 to 2, $\displaystyle \frac{8}{3}$, minus the triangle under the tangent, $\displaystyle {\frac{1\cdot4}{2}}={2}$: $\displaystyle \frac{2}{3}$ again.

$\displaystyle k>0$, and the area bounded by the line $\displaystyle {y}={2x}$, the $\displaystyle x$-axis and the line $\displaystyle {x}={k}$ is 9.
In the solution below we write the area in terms of $\displaystyle k$, and solve an equation:

• $\displaystyle {S}={\int_0^k2x\,dx}={\left[x^2\right]_0^k}={k^2}$;
• $\displaystyle {k^2}={9}$, so $\displaystyle {k}={3}$ ($\displaystyle {k}={-3}$ is ruled out: $\displaystyle k>0$).

On the board – the graphs of $\displaystyle {f(x)}={x+2}$ and $\displaystyle {g(x)}={x^2}$, and the area they enclose.

Find the meeting points of the graphs, and mark them on the board.

Work out the area: write the integral in the first line.

Work out the area bounded by the graphs of $\displaystyle {f(x)}={8-x^2}$ and $\displaystyle {g(x)}={x^2}$: find the meeting points, and then the area.

On the board – the area bounded by the graph of $\displaystyle {f(x)}={\frac{4}{x^2}}$, the line $\displaystyle {y}={1}$ and the line $\displaystyle {x}={1}$.

Find where the graph of $\displaystyle f$ meets the line $\displaystyle {y}={1}$, to the right of the $\displaystyle y$-axis.

Work out the area.

$\displaystyle a>0$, and the area bounded by the graph of $\displaystyle {y}={x^2}$ and the line $\displaystyle {y}={a^2}$ is 36.

The graphs meet at $\displaystyle {x}={-a}$ and at $\displaystyle {x}={a}$. Work out the area in terms of $\displaystyle a$.

Find $\displaystyle a$.

$\displaystyle {f(x)}={x^2}$.
A tangent is drawn to the graph of $\displaystyle f$ at the point $\displaystyle (1,1)$.

Write the equation of the tangent on the board.

The tangent crosses the $\displaystyle x$-axis at $\displaystyle {x}={\frac{1}{2}}$. Work out the area bounded by the graph of $\displaystyle f$, the tangent and the $\displaystyle x$-axis.