Chapter 5.2, Problem 36E

To determine

To find: The value of the integral.

Answer to Problem 36E

The integral is $a^2$ .

Explanation of Solution

Given information:

The expression is: ${{\displaystyle \int }}_a^{a\sqrt{3}}xdx,a>0$

Calculation:

The given expression is: ${{\displaystyle \int }}_a^{a\sqrt{3}}xdx,a>0$

Sketch the region as shown below:

  

The area of the triangle and the rectangle (ABDC) is represented by this integral (CDE).

The rectangle width:

  $\begin{array}{c}c=d\\ =\sqrt{3}a-a\end{array}$

The rectangle height:

  $\begin{array}{c}g=j\\ =a\end{array}$

Hence, the rectangle's surface is:

  $\begin{array}{c}{A}_1=a(\sqrt{3}a-a)\\ =a^2(\sqrt{3}-1)\end{array}$

The triangle base :

  $\begin{array}{c}c=\sqrt{3}a-a\\ =a(\sqrt{3}-1)\end{array}$

The triangle height:

  $\begin{array}{c}h=\sqrt{3}a-a\\ =a(\sqrt{3}-1)\end{array}$

So the triangle area is $\frac{1}{2}$ the base times the height:

  $\begin{array}{c}{A}_2=\frac{1}{2}a(\sqrt{3}-1)a(\sqrt{3}-1)\\ =\frac{a^2}{2}(\sqrt{3}-1)(\sqrt{3}-1)\\ =\frac{a^2}{2}(3-2\sqrt{3}+1)\\ =\frac{a^2}{2}(4-2\sqrt{3})\\ =\frac{a^2}{2}\cdot 2(2-\sqrt{3})\\ =a^2(2-\sqrt{3})\end{array}$

The sum is the total area.

  $\begin{array}{c}A=A_1+A_2\\ =a^2(\sqrt{3}-1)+a^2(2-\sqrt{3})\\ =a^2[(\sqrt{3}-1)+(2-\sqrt{3})]\\ =a^2[1]\\ =a^2\end{array}$

Therefore, the required integral is $a^2$ .