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Chapter 5: Problem 37

Evaluate the integrals by completing the square and applying appropriate formulas from geometry. $$ \int_{0}^{10} \sqrt{10 x-x^{2}} d x $$

Short Answer

Expert verified
The integral evaluates to \( \frac{25\pi}{2} \).

Step by step solution

01

Recognize the integral form

The integral given is \( \int_{0}^{10} \sqrt{10x - x^2} \, dx \). This expression under the square root suggests switching to a form that can utilize geometric formulas, particularly completing the square.
02

Completing the square

We need to complete the square inside the expression \(10x - x^2\). Rewrite it as \( -(x^2 - 10x) \). To complete the square: \( x^2 - 10x = (x - 5)^2 - 25 \). Thus, \(10x - x^2 = 25 - (x - 5)^2\).
03

Express the integral using geometric interpretation

Now the integral becomes \( \int_{0}^{10} \sqrt{25 - (x-5)^2} \, dx \). This expression is in the form \( \sqrt{R^2 - (x-a)^2} \), which represents the top half of a circle with radius \(R\) centered at \((a, 0)\). Here, \( R = 5 \) and \( a = 5 \).
04

Determine the region of integration

The integral bounds are from \( x=0 \) to \( x=10 \). This corresponds to slicing the semicircle along its diameter from the left most point at \( x=0 \) to the right most point at \( x=10 \), both points lies on the x-axis.
05

Calculate the area using geometry

The integral represents half the area of a full circle with radius \( 5 \). The area of a circle is \( \pi R^2 = \pi \times 5^2 = 25\pi \). Therefore, the area of the semicircle is \( \frac{1}{2} \times 25\pi = \frac{25\pi}{2} \).
06

Final solution

Thus, the integral \( \int_{0}^{10} \sqrt{10x - x^2} \, dx \) evaluates to the area of the semicircle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Completing the Square
Completing the square is a method used to simplify expressions, particularly quadratic expressions, and make them easier to work with. For the integral \( \int_{0}^{10} \sqrt{10x - x^2} \, dx \), we need to rewrite the expression under the square root in a form that reveals more insights. The expression \(10x - x^2 = -(x^2 - 10x)\) isn鈥檛 straightforward for geometric interpretation.
\[\]
  • We proceed by rewriting \(x^2 - 10x\) to take the form of a perfect square, \((x - 5)^2\). This involves recognizing that \(x^2 - 10x = (x - 5)^2 - 25\).
  • With these adjustments, \(10x - x^2\) can now be expressed as \(25 - (x - 5)^2\).
This transformation into \((x-a)^2\) form will help relate the expression to the geometry of a circle.
Geometric Interpretation
Once the expression \( 25 - (x-5)^2 \) is achieved, it is noted as representing a semicircle equation. The term \( \sqrt{R^2 - (x-a)^2} \) in integral calculus corresponds to a geometric shape, specifically the top half of a circle.
\[\]
  • Here, \( R \), the radius of the circle, is 5. **Why?** Because \( R^2 = 25 \) which means \( R = \sqrt{25} = 5 \).
  • The term \(a\) is the x-coordinate of our circle's center, here \(a = 5\) indicating that the center is at \((5, 0)\).
With these properties, the expression represents a circle with its center on the x-axis. Thus, recognizing this allows us to strategically calculate areas using semicircle formulas.
Area of a Circle
In the context of integrals and geometric interpretation, understanding the area of a circle is crucial. The standard formula to calculate the area of a circle is \(\pi R^2\). For our example, since we've identified the radius \(R\) as 5, this circle鈥檚 total area would be \(25\pi\).
\[\]
  • However, the integral \(\int_{0}^{10} \sqrt{25 - (x-5)^2} \, dx\) refers only to the top half of this circle, hence it is actually a semicircle.
  • The area of a semicircle can thus be found by taking half of the area of a complete circle, \(\frac{1}{2} \times 25\pi = \frac{25\pi}{2}\).
This area calculation is intuitive since the integral from \(0\) to \(10\) applies solely to the arc segment at the top, effectively cutting the circle horizontally along the diameter.

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