Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007)

Chapter 2: Integration; Section 2: Definite Integrals; page 2

Applications, Exercise 1

Question

The two curves

enclose an area.

1. Make a sketch and mark the enclosed area.
2. Find the intersection points.
3. Calculate the area.

Reminder

.... one of the functions, say y₁(x), should be greater than the second y₂(x) for any point of the interval; otherwise we'll have intersection points inside the interval. From the definition of area under a curve, it follows that the enclosed area S is the difference of the integrals of the functions. Formally this can be written as

The last row of (2.2.2.1) can be interpreted as the integral, whose path is clockwise along the loop, defined by the appropriate functions.

... what happens if there are points of the loop with negative  y  values? .... it does not matter: the result of the integration gives the enclosed area with positive sign, as far as the clockwise sense is observed.

Parts 1-2

Solution of question 1

1. The functions to be plotted are
   
   $$y=x^4\text{and}y=\pm \sqrt{x}$$


2. Here is the Fig. sketch with the marked enclosed area.

Parts 3-5

Solution of question 2

3. Since y=x⁴ and y²=x one should have for the intersection
   
   $$x^8=x$$


4. One solution is $${\left(x,y\right)}_1=\left(0,0\right)$$


5. According to part 3 the non zero solution should fulfill
   
   $$x^7=1$$ yielding the second solution $${\left(x,y\right)}_2=\left(1,1\right)$$

Parts 6-7

Solution of question 3

6. The required area S should be (clockwise integration)
   
   $$S=\underset{0}{\overset{1}{\int }}\sqrt{x}\text{d}x+\underset{1}{\overset{0}{\int }}{x}^4\text{d}x$$


7. which yields
   
   $$\begin{array}{l}S=\frac{2{\sqrt{x}}^3}{3}\underset{0}{\overset{1}{|}}+\frac{x^5}{5}\underset{1}{\overset{0}{|}}=\left(\frac{2}{3}-0\right)+\left(0-\frac{1}{5}\right)=\\ =\frac{2}{3}-\frac{1}{5}=\frac{10-3}{15}=\frac{7}{15}\end{array}$$

Score

By parts: