Question

In Exercises \(23-28,\) use areas to evaluate the integral. $$\int_{a}^{\sqrt{3} a} x d x, \quad a>0$$

Short answer

The solution to the integral \(\int_{a}^{\sqrt{3} a} x d x\) is \( \frac{1}{2} (\sqrt{3} - 1) a^2\)

Step-by-step solution

Graph the function

Sketch the graph of the function \(f(x)=x\) over the interval \([a, \sqrt{3}a]\). Outlining this on a coordinate system, one can see it's a right-angled triangle with base from \(a\) to \(\sqrt{3}a\) and height of \(\sqrt{3}a\)

Calculate the Base and Height

The base of the triangle spans from \(a\) to \(\sqrt{3}a\), so the base of the triangle will be: \(Base = \sqrt{3}a - a = (\sqrt{3} - 1) a\). The height of the triangle at \(x=\sqrt{3} a\) is \(Height = f(\sqrt{3} a) = \sqrt{3} a\).

Calculate the Area

The area of a triangle is calculated as \(\frac{1}{2} \times Base \times Height\). Therefore, substitute the base and height obtained in step 2.\n Area = \( \frac{1}{2} \times (\sqrt{3} - 1) a \times \sqrt{3} a = \frac{1}{2} (\sqrt{3} - 1) a^2\).