Problem 15 - Calculus

Question

Evaluate \(\int_{0}^{3}(2-\sqrt{x+1})^{2} d x\)

Step-by-Step Solution

Answer

The evaluated integral is \(\frac{5}{6}\).

Step 1: Expand the Integrand

First, expand the square in the integrand: \[(2 - \sqrt{x+1})^2 = (2)^2 - 2 \cdot 2 \cdot \sqrt{x+1} + (\sqrt{x+1})^2.\] This simplifies to:\[4 - 4\sqrt{x+1} + (x+1).\]Thus, the integrand becomes:\[x + 5 - 4\sqrt{x+1}.\]

Step 2: Split the Integral

Now express the integral as a sum of three separate integrals:\[\int_{0}^{3} (x + 5 - 4\sqrt{x+1}) \, dx = \int_{0}^{3} x \, dx + \int_{0}^{3} 5 \, dx - \int_{0}^{3} 4\sqrt{x+1} \, dx.\]

Step 3: Evaluate Each Integral Separately

Calculate each integral separately:1. \(\int_{0}^{3} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{3} = \frac{9}{2} - 0 = \frac{9}{2}.\)2. \(\int_{0}^{3} 5 \, dx = 5[x]_{0}^{3} = 5 \cdot 3 = 15.\)3. To evaluate \(\int_{0}^{3} 4\sqrt{x+1} \, dx\), first simplify: \(4\int_{0}^{3} \sqrt{x+1} \, dx\). Make the substitution: let \(u = x+1\), \(du = dx\), change the limits of integration from \(x = 0\) to \(x = 3\), \(u = 1\) to \(u = 4\).So, the integral becomes: \(4 \int_{1}^{4} u^{1/2} \, du = 4\left[\frac{2}{3}u^{3/2}\right]_{1}^{4}.\)

Step 4: Simplify the Substituted Integral

Calculate the evaluated integral from Step 3:\[4 \left[\frac{2}{3} \cdot 4^{3/2} - \frac{2}{3} \cdot 1^{3/2}\right] = \frac{8}{3} (8 - 1) = \frac{8}{3} \times 7 = \frac{56}{3}.\]

Step 5: Combine Results

Combine the individual integral results:\[\frac{9}{2} + 15 - \frac{56}{3}.\] First, find a common denominator to simplify this expression:- The common denominator for 2 and 3 is 6.- Convert \(\frac{9}{2}\) to \(\frac{27}{6}\), convert 15 to \(\frac{90}{6}\), and \(\frac{56}{3}\) to \(\frac{112}{6}\).- Then, calculate:\[\frac{27}{6} + \frac{90}{6} - \frac{112}{6} = \frac{5}{6}.\]

Step 6: Final Answer

The evaluated integral is:\[\frac{5}{6}.\] This is the final result of the integral of the given function from 0 to 3.

Key Concepts

Integration Techniques

Integration is a core concept in calculus used to find areas under curves and accumulate quantities. There are various techniques to perform integration effectively:

• Basic Integration: This involves directly applying the integral rules to functions like polynomials, exponentials, and trigonometric functions.
• Substitution: A powerful technique useful for handling integrals involving composite functions. By substituting parts of the integrand, we simplify the computation.
• Integration by Parts: Useful for integrating products of functions, often employed with logarithmic, polynomial, and exponential functions.
• Partial Fraction Decomposition: Breaks down complex fractions into simpler ones that are easier to integrate.

Choosing the correct technique depends on the form of the function you're integrating. The problem in question uses a combination of expansion and substitution, splitting a complex polynomial expression into simpler integrals for effective calculation.

Substitution Method

The substitution method simplifies complex integrals by changing variables. This process transforms the derivative of one variable to another, effectively tackling integrals that appear complicated at first.

• Identify the Inner Function: Look for a composition of functions and consider the innermost one for substitution.
• Change the Variable: Assign a new variable to the inner function, differentiating it to find the corresponding differential.
• Adjust Limits of Integration: If evaluating a definite integral, adjust the limits to reflect the new variable.

In our exercise, substituting \(u = x+1\) transforms the integral \(\int 4\sqrt{x+1} \, dx\) into a simpler form \(\int 4u^{1/2} \, du\), easing the integration process. This is particularly effective when dealing with square roots or other non-linear expressions inside a polynomial.

Polynomial Functions

Polynomial functions, composed of terms with variable exponents that are non-negative integers, are among the simplest and most predictable functions to integrate.

• Structure and Simplicity: Generally expressed as \(a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\), they are straightforward to handle due to each term's predictable behavior during integration.
• Integration Rule for Polynomials: The power rule states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) (where \(C\) is the constant of integration).

In the given problem, polynomial functions are examined through expansion and integration. After simplifying \((2-\sqrt{x+1})^2\), the polynomial terms like \(x\) and constants \(5\) are integrated separately using basic rules, showcasing the straightforward aspect of polynomial integrals.