Question

Use geometry and properties of integrals to evaluate \(\int_{1}^{5}(|x-2|+\sqrt{-x^{2}+6 x-5}) d x\)

Short answer

Question: Evaluate the integral of the function \(f(x) = |x - 2| + \sqrt{-x^2 + 6x - 5}\) on the interval [1, 5]. Answer: The result is given by the sum of a constant term, \(11.5\), and the yet-to-be-found integral of the square root of a quadratic function, \(\int_{1}^{5}(\sqrt{-x^2 + 6x - 5}) d x\). The constant term results from the integration of the absolute value function, while the second integral may be further investigated using methods such as substitution and trigonometric identities.

Step-by-step solution

Find the integral of |x-2|

Recall that the absolute value can be rewritten as a piecewise function: | x - 2 | = { x - 2 if x >= 2 {-(x - 2) if x < 2 So, we can break the integral into two parts: \(\int_{1}^{5}(|x-2|) d x=\int_{1}^{2}(-(x-2)) d x + \int_{2}^{5}(x-2) d x\) Now, we can integrate each part: \(-\int_{1}^{2}(x-2) d x = \left[-\frac{1}{2}x^2 + 2x\right]_1^2\) \(\int_{2}^{5}(x-2) d x = \left[\frac{1}{2}x^2 - 2x\right]_2^5\)

Find the integral of \(\sqrt{-x^2 + 6x - 5}\)

To evaluate this integral, first find the roots of the quadratic inside the square root, which will help us determine the domain of the function. Thus, we solve: \(x^2 - 6x + 5 = 0\) This factors to \((x-1)(x-5)=0\) which has roots \(x=1\) and \(x=5\). By vertex form, the quadratic is negative between these roots. So, the integral is given by: \(\int_{1}^{5}(\sqrt{-x^2 + 6x - 5}) d x\) This integral involves the square root of a quadratic function, so we can use a substitution and trigonometric identities to solve it. Hold that thought for now, and follow the next steps

Add the results from steps 1 and 2

Now, we combine the two integrals: \(\int_{1}^{5} (|x-2|+\sqrt{-x^2 + 6x -5}) d x = \int_{1}^{5}(|x-2|) d x+ \int_{1}^{5}(\sqrt{-x^2 + 6x - 5}) d x\)

Evaluate the integral on the interval [1,5]

Plug in the limits of integration to evaluate the integral: \((-\frac{1}{2}x^2 + 2x)|_{1}^{2} = -\frac{1}{2}(2)^2 + 4 - (-\frac{1}{2}(1)^2 + 2) = -2 + 3 = 1\) \((\frac{1}{2}x^2 - 2x)|_{2}^{5} = \frac{1}{2}(5)^2 - 10 - (\frac{1}{2}(2)^2 - 4) = 12.5 - 2 = 10.5\) Now, we don't have an explicit expression for the result of Step 2. So, we write the result of the whole integral as: \(\int_{1}^{5} (|x-2|+\sqrt{-x^2 + 6x -5}) d x = 1 +10.5+ \int_{1}^{5}(\sqrt{-x^2 + 6x - 5}) d x = 11.5+ \int_{1}^{5}(\sqrt{-x^2 + 6x - 5}) d x\) The result of our exercise is the sum of the constant term \(11.5\) and the yet-to-be-found integral of the square root of a quadratic function \(\sqrt{-x^2 + 6x - 5}\) over the interval [1,5]. This last integral may be further investigated using methods such as substitution and trigonometric identities, and it's a topic by itself.

Key concepts

Piecewise Functions

Piecewise functions break certain functions into different expressions based on the input value. This makes piecewise functions an essential tool in situations like evaluating integrals involving absolute values. For example, the absolute value function \(|x-2|\) can be represented as two distinct expressions:

• \(x-2\) when \(x \geq 2\)
• \(-(x-2)\) when \(x < 2\)

In our problem, the piecewise nature lets us split the integral from \(1\) to \(5\) into two parts: one from \(1\) to \(2\) and another from \(2\) to \(5\). By evaluating each piece separately, we account for the change in function behavior at \(x=2\). This approach simplifies the calculation by allowing integration over simpler, linear components, transforming a seemingly complicated problem into manageable parts.

Absolute Value Integration

Absolute value integration involves special considerations since the concept of 'absolute value' affects how functions are composed. The absolute value function \(|x-2|\) divides your input space into different regions, representing different linear equations for different intervals. Thus, to integrate functions like \(|x-2|\), we first turn the absolute value into a piecewise function:

• \(x-2\) for \(x \geq 2\)
• \(-(x-2)\) for \(x < 2\)

When integrating over an interval that crosses these boundaries, as from \(1\) to \(5\), it is important to split the integral at the boundary, in this case at \(x=2\). This division lets us treat each part as a simple linear function. By integrating these linear pieces separately, we accurately account for the absolute value, leading to appropriate results for each domain.

Integration by Substitution

Integration by substitution is a powerful technique for solving integrals that seem complex but have functions within them, which can be substituted with a single variable. This method simplifies the integration process, much like reversing the chain rule used in differentiation.
To apply substitution, follow these steps:

• Select a substitution \(u = g(x)\), representing a part of the integrand.
• Find \(du = g'(x)dx\).
• Rewrite the integral in terms of \(u\) and \(du\).
• Integrate with respect to \(u\).
• Finally, back-substitute \(u=g(x)\) to return to the original variable.

While our main exercise noted yet-to-solve parts, this substitution method will later simplify the square root integral, allowing for easier, more straightforward computations.

Trigonometric Substitution

Trigonometric substitution provides a technique tailored for integrals involving square roots of quadratic functions. This method leverages trigonometric identities to transform and simplify these integrals into more easily solvable forms.
In our case, the integral \(\sqrt{-x^2 + 6x - 5}\) involves the quadratic \(-x^2 + 6x - 5\). By recognizing the quadratic form, substitution using trigonometric functions like sine or cosine relates the variables to familiar trigonometric identities. For instance, expressions like \(x = a \sin(\theta)\) could be employed to reframe the problem in terms of sine or cosine, thereby allowing it to fall into a known integral formula.
Trigonometric substitution is particularly valuable for this kind of integral because of how it opens up expressions that otherwise seem difficult to decompose and integrate directly. By effectively using the relationships between trigonometric functions and their derivatives, this tool transforms the problem into a much simpler shape to handle in calculus.