5.11 Definite Integral - Area Between Curves
1. The Definite Integral
The definite integral is denoted by $\int_{a}^{b}f(x)dx$ where $a$ and $b$ are real numbers within the domain of $f$.
The calculation is straightforward. For example:
Indefinite: $\int 2xdx = x^2 + c$
Definite: $$ \int_{1}^{3}2xdx = \left[ x^2 \right]_1^3 = (3^2)-(1^2) = 8 $$ Indefinite: $\int (2x+3)dx = x^2 + 3x + c$
Definite: $$ \int_{0}^{4}(2x+3)dx = \left[x^2+3x\right]_0^4 = (4^2+3(4)) - (0^2+3(0)) = 28 $$
Definite: $$ \int_{1}^{3}2xdx = \left[ x^2 \right]_1^3 = (3^2)-(1^2) = 8 $$ Indefinite: $\int (2x+3)dx = x^2 + 3x + c$
Definite: $$ \int_{0}^{4}(2x+3)dx = \left[x^2+3x\right]_0^4 = (4^2+3(4)) - (0^2+3(0)) = 28 $$
Notice: The constant $c$ is omitted in the definite integral! (Why? Because if we evaluate it, we generate $(F(b)+c) - (F(a)+c) = F(b)-F(a)$, neutralizing the term).
If $\int f(x)dx = F(x) + c$, then $\int_a^b f(x)dx = \left[F(x)\right]_a^b = F(b)-F(a)$
EXAMPLE 1
$$
\begin{aligned}
\int_{-1}^{1} (8x^3 + 12x^2 - 6x + 3)dx &= \left[ 2x^4 + 4x^3 - 3x^2 + 3x \right]_{-1}^1 \\
&= \left( 2(1)^4 + 4(1)^3 - 3(1)^2 + 3(1) \right) \\
&- \left( 2(-1)^4 + 4(-1)^3 - 3(-1)^2 + 3(-1) \right) \\
&= (2 + 4 - 3 + 3) - (2 - 4 - 3 - 3) \\
&= 6 - (-8) \\
&= \mathbf{14}
\end{aligned}
$$
EXAMPLE 2
$$
\begin{aligned}
\int_{-2}^{-3} \dfrac{8}{x^3} dx &= \int_{-2}^{-3} 8x^{-3} dx = \left[ \dfrac{8x^{-2}}{-2} \right]_{-2}^{-3} = \left[ -4x^{-2} \right]_{-2}^{-3} \\
&= \left[ \dfrac{-4}{x^2} \right]_{-2}^{-3} = \left( \dfrac{-4}{(-3)^2} \right) - \left( \dfrac{-4}{(-2)^2} \right) = -\dfrac{4}{9} - \left( -\dfrac{4}{4} \right) \\
&= -\dfrac{4}{9} + 1 = \mathbf{\dfrac{5}{9}}
\end{aligned}
$$
2. Properties of the Definite Integral
The rules defining indefinite integrals transition here:
- $\int_{a}^{b} [f(x) + g(x)]dx = \int_{a}^{b} f(x)dx + \int_{a}^{b} g(x)dx$
- $\int_{a}^{b} \alpha f(x)dx = \alpha \int_{a}^{b} f(x)dx$
Moreover:
- $\int_{a}^{b} f(x)dx + \int_{b}^{c} f(x)dx = \int_{a}^{c} f(x)dx$ (Notice that the limits $a, b, c$ must be consecutive)
- $\int_{a}^{b} f(x)dx = - \int_{b}^{a} f(x)dx$ (Accepted by definition)
- $\int_{a}^{b} f'(x)dx = \left[ f(x) \right]_a^b$ (Derivative and integral exist as inverse counterparts)
- $\int_{a}^{b} f(x)dx = \int_{a}^{b} f(y)dy = \int_{a}^{b} f(t)dt$ (The variable representation is irrelevant)
EXAMPLE 3
Suppose that $\int_{0}^{5} f(x)dx = 10$. Given that $f(0)=15$ and $f(5)=3$, resolve the following conditions:
- $\int_{0}^{5} 2f(x)dx = 2 \cdot 10 = \mathbf{20}$
- $\int_{0}^{5} [f(x) + 2x]dx = \int_{0}^{5} f(x)dx + \int_{0}^{5} 2xdx = 10 + \left[ x^2 \right]_0^5 = 10 + 25 = \mathbf{35}$
- $\int_{5}^{0} f(x)dx = \mathbf{-10}$
- $\int_{0}^{5} f'(x)dx = \left[ f(x) \right]_0^5 = f(5) - f(0) = 3 - 15 = \mathbf{-12}$
3. Substitution and Definite Integrals
EXAMPLE 4
Find $I = \int_{0}^{2} \dfrac{x}{x^2+4} dx$
Method A: We calculate the indefinite integral first
We use $u = x^2+4$, extracting $\dfrac{du}{dx} = 2x \implies dx = \dfrac{du}{2x}$, resolving to: $$ \int \dfrac{x}{x^2+4} dx = \int \dfrac{x}{u} \dfrac{du}{2x} = \dfrac{1}{2} \int \dfrac{1}{u} du = \dfrac{1}{2} \ln|u| + c = \dfrac{1}{2} \ln(x^2+4) + c $$ Applying bounds: $$ I = \left[ \dfrac{1}{2} \ln(x^2+4) \right]_0^2 = \dfrac{1}{2}\ln(8) - \dfrac{1}{2}\ln(4) = \dfrac{1}{2}(\ln 8 - \ln 4) = \dfrac{1}{2}\ln\left(\dfrac{8}{4}\right) = \mathbf{\dfrac{1}{2}\ln 2} $$
Method B: We recalculate mapping the limits
Again we use the term $u = x^2+4$, isolating $\dfrac{du}{dx} = 2x \implies dx = \dfrac{du}{2x}$.
We map and change the limits referencing $u = x^2+4$:
For $x=0$, $u = 0^2+4 = 4$
For $x=2$, $u = 2^2+4 = 8$
Under this condition, we do not go back to $x$. Namely, $$ \begin{aligned} I &= \int_{0}^{2} \dfrac{x}{x^2+4} dx = \int_{4}^{8} \dfrac{x}{u} \dfrac{du}{2x} = \dfrac{1}{2} \int_{4}^{8} \dfrac{1}{u} du = \left[ \dfrac{1}{2}\ln|u| \right]_4^8 \\ &= \dfrac{1}{2}\ln 8 - \dfrac{1}{2}\ln 4 = \mathbf{\dfrac{1}{2}\ln 2} \end{aligned} $$
We use $u = x^2+4$, extracting $\dfrac{du}{dx} = 2x \implies dx = \dfrac{du}{2x}$, resolving to: $$ \int \dfrac{x}{x^2+4} dx = \int \dfrac{x}{u} \dfrac{du}{2x} = \dfrac{1}{2} \int \dfrac{1}{u} du = \dfrac{1}{2} \ln|u| + c = \dfrac{1}{2} \ln(x^2+4) + c $$ Applying bounds: $$ I = \left[ \dfrac{1}{2} \ln(x^2+4) \right]_0^2 = \dfrac{1}{2}\ln(8) - \dfrac{1}{2}\ln(4) = \dfrac{1}{2}(\ln 8 - \ln 4) = \dfrac{1}{2}\ln\left(\dfrac{8}{4}\right) = \mathbf{\dfrac{1}{2}\ln 2} $$
Method B: We recalculate mapping the limits
Again we use the term $u = x^2+4$, isolating $\dfrac{du}{dx} = 2x \implies dx = \dfrac{du}{2x}$.
We map and change the limits referencing $u = x^2+4$:
For $x=0$, $u = 0^2+4 = 4$
For $x=2$, $u = 2^2+4 = 8$
Under this condition, we do not go back to $x$. Namely, $$ \begin{aligned} I &= \int_{0}^{2} \dfrac{x}{x^2+4} dx = \int_{4}^{8} \dfrac{x}{u} \dfrac{du}{2x} = \dfrac{1}{2} \int_{4}^{8} \dfrac{1}{u} du = \left[ \dfrac{1}{2}\ln|u| \right]_4^8 \\ &= \dfrac{1}{2}\ln 8 - \dfrac{1}{2}\ln 4 = \mathbf{\dfrac{1}{2}\ln 2} \end{aligned} $$
EXAMPLE 5
Suppose that $\int_{0}^{5} f(x)dx = 10$. Evaluating translations yields:
Let $y=x-3$. Then, $dy=dx$. The upper bound $x=3 \implies y=0$ and the lower bound $x=8 \implies y=5$, so
$$
\int_{3}^{8} f(x-3)dx = \int_{0}^{5} f(y)dy = \mathbf{10}
$$
4. Area Between Curve and x-axis
EXAMPLE 6
Calculate the area of the closed region located between the curve $y = 9 - x^2$ and the x-axis.
Solution:
First, we find the roots: $9 - x^2 = 0 \iff x = -3, x = 3$.
Hence, $$ \begin{aligned} \text{Area} &= \int_{-3}^{3} (9 - x^2)dx = \left[ 9x - \dfrac{x^3}{3} \right]_{-3}^3 = \left( 9(3) - \dfrac{3^3}{3} \right) - \left( 9(-3) - \dfrac{(-3)^3}{3} \right) \\ &= (27 - 9) - (-27 + 9) = 18 - (-18) = \mathbf{36} \end{aligned} $$ Notice that the curve has symmetry about the y-axis, allowing the area between $x=0$ and $x=3$ to be multiplied by 2: $$ \text{Area} = 2 \int_{0}^{3} (9 - x^2)dx = 2 \left[ 9x - \dfrac{x^3}{3} \right]_0^3 = 2(27 - 9) = \mathbf{36} $$
First, we find the roots: $9 - x^2 = 0 \iff x = -3, x = 3$.
Hence, $$ \begin{aligned} \text{Area} &= \int_{-3}^{3} (9 - x^2)dx = \left[ 9x - \dfrac{x^3}{3} \right]_{-3}^3 = \left( 9(3) - \dfrac{3^3}{3} \right) - \left( 9(-3) - \dfrac{(-3)^3}{3} \right) \\ &= (27 - 9) - (-27 + 9) = 18 - (-18) = \mathbf{36} \end{aligned} $$ Notice that the curve has symmetry about the y-axis, allowing the area between $x=0$ and $x=3$ to be multiplied by 2: $$ \text{Area} = 2 \int_{0}^{3} (9 - x^2)dx = 2 \left[ 9x - \dfrac{x^3}{3} \right]_0^3 = 2(27 - 9) = \mathbf{36} $$
Suppose that $f(x)$ cycles into both positive and negative domains. If $A$, $B$, and $C$ define the areas of those isolated curves, solving gives:
$$ \int_a^d f(x)dx = A - B + C $$
The integral $\int_a^d f(x)dx$ does not produce the total area. The area relies on the absolute value:
$$ \text{AREA} = \int_a^d |f(x)|dx = A + B + C $$
In practice, we evaluate the distinct boundaries, calculate definite integrals, and add the resulting absolute values.
EXAMPLE 7
Consider the function $f(x) = x^3 - 6x^2 + 8x$. Find:
- (a) $\int_{0}^{2} f(x)dx$
- (b) $\int_{2}^{4} f(x)dx$
- (c) $\int_{0}^{4} f(x)dx$
- (d) the area bounded by the curve and x-axis within the segment $[0,4]$
Solution:
(a) $$ \int_{0}^{2} (x^3 - 6x^2 + 8x)dx = \left[ \dfrac{x^4}{4} - 2x^3 + 4x^2 \right]_0^2 = 4 - 0 = \mathbf{4} $$ (b) $$ \int_{2}^{4} (x^3 - 6x^2 + 8x)dx = \left[ \dfrac{x^4}{4} - 2x^3 + 4x^2 \right]_2^4 = 0 - 4 = \mathbf{-4} $$ (c) $$ \int_{0}^{4} (x^3 - 6x^2 + 8x)dx = \left[ \dfrac{x^4}{4} - 2x^3 + 4x^2 \right]_0^4 = 0 - 0 = \mathbf{0} $$ [calculating equates to combining the sequences generated above]
(d) The total area calculates as $A = \int_{0}^{4} |x^3 - 6x^2 + 8x|dx$. Sum the absolute values: $$ A = 4 + |-4| = \mathbf{8} $$
(a) $$ \int_{0}^{2} (x^3 - 6x^2 + 8x)dx = \left[ \dfrac{x^4}{4} - 2x^3 + 4x^2 \right]_0^2 = 4 - 0 = \mathbf{4} $$ (b) $$ \int_{2}^{4} (x^3 - 6x^2 + 8x)dx = \left[ \dfrac{x^4}{4} - 2x^3 + 4x^2 \right]_2^4 = 0 - 4 = \mathbf{-4} $$ (c) $$ \int_{0}^{4} (x^3 - 6x^2 + 8x)dx = \left[ \dfrac{x^4}{4} - 2x^3 + 4x^2 \right]_0^4 = 0 - 0 = \mathbf{0} $$ [calculating equates to combining the sequences generated above]
(d) The total area calculates as $A = \int_{0}^{4} |x^3 - 6x^2 + 8x|dx$. Sum the absolute values: $$ A = 4 + |-4| = \mathbf{8} $$
5. Area Between Two Curves
In general, calculating the area between curves $f(x)$ and $g(x)$ provides:
$$ \text{Area} = \int_a^b |f(x) - g(x)| dx $$
Methodology:
- Define limits by resolving the intersection by solving $f(x) = g(x)$.
- Split limits by analyzing which interval is "above" or "below". Proceed by evaluating $(\text{top} - \text{bottom})$.
EXAMPLE 8
Calculate the area bounded by $f(x) = x^2$ and $g(x) = x + 2$.
Solution:
Intersection points:
$$ f(x) = g(x) \iff x^2 = x + 2 \iff x^2 - x - 2 = 0 \iff x = -1 \text{ or } x = 2 $$ Mapping values from $x = -1$ to $x = 2$, we identify that the line $g(x) = x+2$ is above the parabola $f(x) = x^2$.
$$ \begin{aligned} \text{Area} &= \int_{-1}^{2} [g(x) - f(x)]dx = \int_{-1}^{2} (x + 2 - x^2)dx = \left[ \dfrac{x^2}{2} + 2x - \dfrac{x^3}{3} \right]_{-1}^2 \\ &= \left( \dfrac{2^2}{2} + 2(2) - \dfrac{2^3}{3} \right) - \left( \dfrac{(-1)^2}{2} + 2(-1) - \dfrac{(-1)^3}{3} \right) \\ &= \left( 2 + 4 - \dfrac{8}{3} \right) - \left( \dfrac{1}{2} - 2 + \dfrac{1}{3} \right) \\ &= \dfrac{10}{3} - \left( -\dfrac{7}{6} \right) \\ &= \mathbf{\dfrac{9}{2}} \text{ or } \mathbf{4.5} \end{aligned} $$
Intersection points:
$$ f(x) = g(x) \iff x^2 = x + 2 \iff x^2 - x - 2 = 0 \iff x = -1 \text{ or } x = 2 $$ Mapping values from $x = -1$ to $x = 2$, we identify that the line $g(x) = x+2$ is above the parabola $f(x) = x^2$.
$$ \begin{aligned} \text{Area} &= \int_{-1}^{2} [g(x) - f(x)]dx = \int_{-1}^{2} (x + 2 - x^2)dx = \left[ \dfrac{x^2}{2} + 2x - \dfrac{x^3}{3} \right]_{-1}^2 \\ &= \left( \dfrac{2^2}{2} + 2(2) - \dfrac{2^3}{3} \right) - \left( \dfrac{(-1)^2}{2} + 2(-1) - \dfrac{(-1)^3}{3} \right) \\ &= \left( 2 + 4 - \dfrac{8}{3} \right) - \left( \dfrac{1}{2} - 2 + \dfrac{1}{3} \right) \\ &= \dfrac{10}{3} - \left( -\dfrac{7}{6} \right) \\ &= \mathbf{\dfrac{9}{2}} \text{ or } \mathbf{4.5} \end{aligned} $$