Area between functions

This looks at signed and unsigned area between functions.

Contents

Fundamental theorem of calculus

Previously, we touched upon derivatives as the instantaneous change at one point, and integrals as the accumulation of change on an interval.

Loosely speaking, the fundamental theorem of calculus says that this accumulation manifests as the area under the curve and above the xx-axis. If the curve is below the xx-axis, then the enclosed area has a negative sign.

Unsigned vs signed area

By default, a definite integral gives the signed area mentioned above.

signed area between f and y=0,x=x1,x=x2:∫x1x2f(x) dx\text{signed area between }f \text{ and }y = 0, x= x_1, x=x_2: \int_{x_1}^{x_2} f(x)\d x

The unsigned area, ie taking all areas as positive, is

unsigned area between f and y=0,x=x1,x=x2:∫x1x2∣f(x)∣ dx\text{unsigned area between }f \text{ and }y = 0, x= x_1, x=x_2: \int_{x_1}^{x_2} \lvert f(x)\rvert \d x

Signed area is useful when you want to find the net effect of increases and decreases; unsigned area is useful to find the literal area.

When question says “find the area between …” they refer to the unsigned area.

When finding unsigned area by hand, it involves

  1. Solve f(x)=0f(x) = 0
  2. Split f(x)f(x) into the regions separated by its zeros
  3. Integrate for each region. Typically this means finding the antiderivative once, then evaluate using different limits of integration.
  4. Add up the absolute values of the areas from step 3.

In most circumstances, there is only one contiguous bounded area. Only for rare cases like manually finding “distance traveled” in kinematics is this procedure necessary.

Area between curves

For a contiguous enclosed region bounded by functions f(x)f(x) and g(x)g(x), such that

  • x1<x2x_1 < x_2, AND
  • g(x)≥f(x)g(x) \geq f(x) for all x∈[x1,x2]x\in\left[x_1, x_2\right]

Equivalently, each vertical line passing through the region must intersect g(x)g(x) above, and f(x)f(x) below.

Then the bounded region has area

area=∫x1x2g(x)−f(x) dx\text{area} = \int_{x_1}^{x_2} g(x) - f(x)\d x

In cases where the region has more than one function for the top or bottom portions, then the area must be subdivided by vertical lines such that each region is bounded by a single function on top, and a single function below.

Example: Find the area enclosed between x=y2x = y^2 and y=x−6y = x - 6.

The area is bounded by g(x)=xg(x) = \sqrt x on top, but by both f1(x)=−xf_1(x) = -\sqrt x and f2(x)=x−6f_2(x) = x - 6 below.

Solving x=y2x = y^2 and y=x−6y = x - 6 yield (4,−2)(4, -2), (9,3)(9, 3) as the two intersection points. The area can be calculated by splitting into two regions down x=4x = 4.

left =∫04x−(−x) dx=2∫04x dx=43x32∣04=43(4)32−0=43(8)=323=10+23\begin{align*}\text{left } &= \int_{0}^{4} \sqrt{x} - (-\sqrt{x})\d x \\ &= 2\int_{0}^{4} \sqrt{x} \d x \\ &= \frac43 x^\frac32 \Big\vert_0^4 \\ &= \frac43 (4)^\frac32 - 0 \\ &= \frac43(8) \\ &= \frac{32}{3} = 10 + \frac23 \end{align*}
right =∫49x−(x−6) dx=23x32−12x2+6x∣49=23((9)32−(4)32)−12((9)2−(4)2)+6(9−4)=23(27−8)−12(81−16)+30=383−652+30=12+23−32−12+30=10+16\begin{align*}\text{right } &= \int_{4}^{9} \sqrt{x} - (x - 6)\d x \\ &= \frac23 x^\frac32 - \frac12x^2 + 6x \Big\vert_4^9 \\ &= \frac23 ((9)^\frac32 - (4)^\frac32) - \frac12((9)^2 - (4)^2) + 6(9 - 4) \\ &= \frac23(27 - 8) - \frac12(81 - 16) + 30 \\ &= \frac{38}{3} - \frac{65}{2} + 30 \\ &= 12 + \frac23 - 32 - \frac 12 + 30 \\ &= 10 + \frac 16 \end{align*}
left +right =10+23+10+16=20+56■\begin{align*}\text{left } + \text{right } = 10 + \frac23+10+\frac16 = 20 + \frac56 \qed\end{align*}

See sketch of the curves .

Always quickly sketch the functions or curves using function transformations to visualize the area.