Question

Evaluate the given integral. $$ \int_{0}^{5} x \sqrt{x+2} d x $$

Short answer

The integral evaluates to approximately \( 28.67 \).

Step-by-step solution

Substitution

To evaluate the integral \( \int_{0}^{5} x \sqrt{x+2} \, dx \), we'll begin by using substitution. Let \( u = x + 2 \). Then, differentiate: \( \frac{du}{dx} = 1 \), which implies \( du = dx \). Also, change the limits of integration: when \( x = 0 \), \( u = 2 \); when \( x = 5 \), \( u = 7 \). So, the integral becomes \( \int_{2}^{7} (u-2) \sqrt{u} \, du \).

Expand and Simplify

Next, expand the expression \( (u-2) \sqrt{u} \) as \( u^{1/2} (u - 2) = u^{3/2} - 2u^{1/2} \). The integral now is \( \int_{2}^{7} (u^{3/2} - 2u^{1/2}) \, du \).

Integrate Term by Term

Integrate each term separately: - For \( \int u^{3/2} \, du \), use the power rule: \( \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2} \).- For \( -2 \int u^{1/2} \, du \), again use the power rule: \( -2 \frac{u^{3/2}}{3/2} = -\frac{4}{3} u^{3/2} \).Thus, the integral becomes \( \left[ \frac{2}{5} u^{5/2} - \frac{4}{3} u^{3/2} \right]_{2}^{7} \).

Evaluate at Bounds

Evaluate this expression at the upper and lower bound:- At \( u = 7 \): \( \frac{2}{5} (7)^{5/2} - \frac{4}{3} (7)^{3/2} \).- At \( u = 2 \): \( \frac{2}{5} (2)^{5/2} - \frac{4}{3} (2)^{3/2} \).Compute these values to find the definite integral.

Simplify Computations

Calculate each term from Step 4. For \( 7^{5/2} \) and \( 7^{3/2} \), approximate:- \( 7^{1/2} \approx 2.64575 \)- \( 7^{3/2} = 7 \times 2.64575 \approx 18.5207 \)- \(7^{5/2} = 7^2 \times 2.64575 \approx 129.6412 \)For \( 2^{5/2} \) and \( 2^{3/2} \):- \( 2^{1/2} = \sqrt{2} \approx 1.4142 \)- \( 2^{3/2} = 2 \times 1.4142 \approx 2.8284 \)- \( 2^{5/2} = 4 \times 1.4142 \approx 5.6568 \)Finally, combine terms.

Final Computation

Substitute and calculate:- Upper bound: \( \frac{2}{5} \times 129.6412 - \frac{4}{3} \times 18.5207 \approx 51.8565 - 24.6943 \approx 27.1622 \)- Lower bound: \( \frac{2}{5} \times 5.6568 - \frac{4}{3} \times 2.8284 \approx 2.2627 - 3.7712 \approx -1.5085 \)Thus, the evaluated integral is approximately \( 27.1622 + 1.5085 \approx 28.6707 \).

Key concepts

Definite Integral

A definite integral calculates the net area under a curve, between two points on the x-axis. It gives a precise calculation of the accumulated quantity, represented by the function within specific limits.
This contrasts with an indefinite integral, which represents a family of functions. The definite integral provides a numerical value determined by evaluating the integral at the upper and lower limits.
For example, in the problem above, we need to evaluate from 0 to 5. After substituting and transforming the limits as 2 to 7, we evaluate the resulting integral to find the total area under the curve from the starting point to the endpoint.

Substitution Method

Substitution is a powerful technique in integration used to transform complex integrals into simpler ones. This method involves substituting part of the integrand with a new variable to make the integral more manageable.
In the given problem, we let \( u = x + 2 \). This substitution transforms the original limits from x to in terms of u, thus simplifying the evaluation of the integral. For substitution to work smoothly, remember to change the limits accordingly and express \( dx \) in terms of \( du \).
When you substitute, you often aim to eliminate the complicating part, such as a nested function or a composite function, making the process of integration much clearer and more straightforward.

Power Rule

The power rule is a fundamental technique in integration, incredibly useful for polynomials and expressions with powers. The rule states that to integrate a function of the form \( x^n \), we calculate \( \frac{x^{n+1}}{n+1} + C \), assuming \( n eq -1 \).
This rule applies to each term individually, which is demonstrated when integrating either \( u^{3/2} \) or \( u^{1/2} \) in our example. The power rule transforms \( u^{3/2} \) into \( \frac{2}{5} u^{5/2} \) and \( u^{1/2} \) into \( -\frac{4}{3} u^{3/2} \).
By breaking down integrals term by term using this rule, complex expressions become easier to handle, promoting better fluency with calculus operations.

Integration Techniques

Integration techniques encompass various methods designed to tackle the wide array of functions encountered in calculus. Choosing the right technique is crucial for simplifying and solving integrals efficiently.
Some foundational techniques include:

• Substitution, as seen previously, which helps in dealing with nested or composite functions.
• Integration by parts, highly beneficial for products of functions, where one part can be differentiated neatly while the other is integrated.
• Trigonometric substitution, useful when integrals involve roots and powers within trigonometric identities.

The main goal of these techniques is to transform an integral into a more standard or elementary form, where the application of the basic rules becomes straightforward, thus making a complex problem manageable and solvable.