Area Under a Curve and Average Value of a Function – Area & Average

Key Formulas

Area under the curve

$$A = \int_a^b f(x)\,dx$$

Average value of a function

$$f_{\text{avg}} = \dfrac{1}{b-a}\int_a^b f(x)\,dx$$

The integral gives the area under the curve. Dividing by the length of the interval $b-a$ converts that area into the average height of the function.

Tip: Think of it as the height of a rectangle that has the same base and the same area as the region under the curve.

Worked Example

Find the area of the region bounded by the graphs of $ y=\dfrac{4}{x} $, $ x=1 $, $ x=e $, and $ y=0 $. Sketch the region. Then find the average value of the function over the same region.

1. Sketch the region

The graph of $ y=\dfrac{4}{x} $ is a rational function with a vertical stretch of 4 from $ x=1 $ to $ x=e $.

2. Find the area

$$A = \int_1^e \dfrac{4}{x}\,dx = 4\ln|x|\Big|_1^e = 4(\ln e – \ln 1) = 4(1-0) = 4$$

3. Find the average value

$$f_{\text{avg}} = \dfrac{1}{e-1}\int_1^e \dfrac{4}{x}\,dx = \dfrac{4}{e-1}$$

Final Answers

Area = 4

Average value = $\dfrac{4}{e-1}$

Level 1

Easy

Straightforward area + average value on a closed interval

Find the area of the region bounded by the graphs of $ y=\dfrac{5}{x} $, $ x=1 $, $ x=e $, and $ y=0 $. Then find the average value of the function on that interval.

Show Answer

Area

$$\int_1^e \dfrac{5}{x}\,dx = 5\ln|x|\Big|_1^e = 5(\ln e – \ln 1) = 5$$

Average value

$$\dfrac{1}{e-1}\cdot 5 = \dfrac{5}{e-1}$$

Area = 5 Average value = $\dfrac{5}{e-1}$

Find the area of the region bounded by the graphs of $ y=\dfrac{3}{x} $, $ x=1 $, $ x=e^{2} $, and $ y=0 $. Then find the average value of the function on that interval.

Show Answer

Area

$$\int_1^{e^{2}} \dfrac{3}{x}\,dx = 3\ln|x|\Big|_1^{e^{2}} = 3(2-0) = 6$$

Average value

$$\dfrac{1}{e^{2}-1}\cdot 6 = \dfrac{6}{e^{2}-1}$$

Area = 6 Average value = $\dfrac{6}{e^{2}-1}$

Find the area of the region bounded by the graphs of $ y=\dfrac{2}{x} $, $ x=1 $, $ x=4 $, and $ y=0 $. Then find the average value of the function on that interval.

Show Answer

Area

$$\int_1^4 \dfrac{2}{x}\,dx = 2\ln|x|\Big|_1^4 = 2(\ln 4 – \ln 1) = 2\ln 4 = 4\ln 2$$

Note: $2\ln 4$ and $4\ln 2$ are equivalent. In calculus it is standard to simplify logarithms this way so the final answer is written in terms of the simpler constant $\ln 2$, but $2\ln 4$ is fine.

Average value

$$\dfrac{1}{4-1}\cdot 4\ln 2 = \dfrac{4\ln 2}{3}$$

Area = $4\ln 2$

Average value = $\dfrac{4\ln 2}{3}$

Ready to master integration techniques and calculus?

Book a Session with Amy