Area Between Two Functions



If two functions $f(x)$ and $g(x)$ intersect at $x=1$ and $x=3$, and $f(x) \ge g(x)$ for all $1 \le x \le 3$, then the area of the shaded region between their points of intersection is given by:
$$ \begin{align} \displaystyle
A &= \int_{1}^{3}{f(x)}dx – \int_{1}^{3}{g(x)}dx \\
&= \int_{1}^{3}{\Big[f(x)-g(x)\Big]}dx
\end{align} $$

Example 1

Find the area bounded by the $x$-axis and $y=x^2-4x+3$.

\( \begin{align} \displaystyle
x\text{-intercepts:} \\
x^2-4x+3 &= 0 \\
(x-1)(x-3) &=0 \\
x &=1 \text{ and } x=3 \\
\end{align} \)


\( \begin{align} \displaystyle
A &= \int_{1}^{3}{\Big[0-(x^2-4x+3)\Big]}dx \\
&= \int_{1}^{3}{(-x^2+4x-3)}dx \\
&= \Big[\dfrac{-x^3}{3} + 2x^2- 3x\Big]_{1}^{3} \\
&= \Big(\dfrac{-3^3}{3} + 2 \times 3^2 – 3 \times 3 \Big) – \Big(\dfrac{-1^3}{3} + 2 \times 1^2- 3 \times 1\Big) \\
&= \dfrac{4}{3} \text{ units}^2
\end{align} \)

Example 2

Find the area bounded by the $y=x+2$ and $y=x^2+x-2$.

\( \begin{align} \displaystyle
\text{intersections:} \\
x^2+x-2 &= x+2 \\
x^2-4 &= 0 \\
(x+2)(x-2) &= 0 \\
x &=-2 \text{ and } x=2 \\
\end{align} \)

\( \begin{align} \displaystyle
A &= \int_{-2}^{2}{\Big[(x+2)-(x^2+x-2)\Big]}dx \\
&= \int_{-2}^{2}{(x+2-x^2-x+2)}dx \\
&= \int_{-2}^{2}{(4-x^2)}dx \\
&= \Big[4x-\dfrac{x^3}{3}\Big]_{-2}^{2} \\
&= \Big(4 \times 2-\dfrac{2^3}{3}\Big) – \Big(4 \times (-2)-\dfrac{(-2)^3}{3}\Big) \\
&= \dfrac{32}{3} \text{ units}^2
\end{align} \)

Example 3

Find the area bounded by the $x$-axis and $y=x^3-x^2-2x$.

\( \begin{align} \displaystyle
x\text{-intercepts:} \\
x^3-x^2-2x &= 0 \\
x(x^2-x-2) &= 0 \\
x(x+1)(x-2) &= 0 \\
x &=-1, x=0 \text{ and } x=2 \\
\end{align} \)

\( \begin{align} \displaystyle
A &= A_1 + A_2 \\
&= \int_{-1}^{0}{\big[(x^3-x^2-2x)-0\big]}dx + \int_{0}^{2}{\big[0-(x^3-x^2-2x)\big]}dx\\
&= \int_{-1}^{0}{(x^3-x^2-2x)}dx – \int_{0}^{2}{(x^3-x^2-2x)}dx\\
&= \Big[\dfrac{x^4}{4}-\dfrac{x^3}{3}-x^2\Big]_{-1}^{0} – \Big[\dfrac{x^4}{4}-\dfrac{x^3}{3}-x^2\Big]_{0}^{2} \\
&= 0 – \Big[\dfrac{(-1)^4}{4}-\dfrac{(-1)^3}{3}-(-1)^2\Big] – \Big[\dfrac{2^4}{4}-\dfrac{2^3}{3}-2^2\Big]+0 \\
&= \dfrac{37}{12} \text{ units}^2 \\
\end{align} \)


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