Problem 55 - Calculus Single Variable

Question

Calculate the integrals. $$ \int \frac{x}{\sqrt{x+3}} d x $$

Step-by-Step Solution

Answer

\( \int \frac{x}{\sqrt{x+3}} \, dx = \frac{2}{3}(x + 3)^{3/2} - 6(x + 3)^{1/2} + C. \)

Step 1: Identify Substitution

To simplify the integration, let's use substitution. Notice that the expression under the square root can be transformed into an easier form for integration. Let \( u = x + 3 \). Then \( du = dx \), and \( x = u - 3 \). Plug these into the integral.

Step 2: Substitute in the Integral

Substitute \( u \) and \( du \) into the integral:\[\int \frac{u - 3}{\sqrt{u}} \, du.\]This simplifies to:\[\int (u^{1/2} - 3u^{-1/2}) \, du.\]

Step 3: Integrate Term by Term

Now, integrate each term separately:1. Integrate \( u^{1/2} \): \[\int u^{1/2} \, du = \frac{2}{3}u^{3/2} + C_1.\]2. Integrate \( -3u^{-1/2} \): \[\int -3u^{-1/2} \, du = -3(2)u^{1/2} + C_2 = -6u^{1/2} + C_2.\]

Step 4: Combine the Results

Combine the results from each term of the integral:\[\frac{2}{3}u^{3/2} - 6u^{1/2} + C.\]Here, \(C\) is the constant of integration which combines \(C_1\) and \(C_2\).

Step 5: Back-Substitute for x

Replace \( u \) back in terms of \( x \):\[u = x + 3.\]So the integral becomes:\[\frac{2}{3}(x + 3)^{3/2} - 6(x + 3)^{1/2} + C.\]

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving integrals, especially when dealing with complicated or nested functions. It involves replacing a portion of the integrand with a new variable to simplify the expression. In this exercise, we identified that the term under the square root, \(x+3\), can be substituted. We chose \(u = x + 3\), which transforms the integral

• Recognizing the part of the integrand that complicates the equation helps in deciding what to substitute.
• Setting \(u = x+3\) helped to simplify the square root and made integration more straightforward.
• By expressing \(x\) in terms of \(u\), we carried out the substitution comprehensively.

Substitution is not always a straightforward choice, but with practice in identifying suitable expressions, it becomes an intuitive part of tackling integrals.

Definite Integrals

While this specific problem does not deal with a definite integral, understanding definite integrals can be essential in similar contexts. A definite integral involves computation between two specific boundaries, providing the area under a curve. In our exercise, the focus was on the indefinite form,

• In definite integrals, substitutions need to be considered correctly within the bounds of integration.
• Changing variables requires adjusting the limits of integration accordingly.
• Definite integrals yield a numerical value, in contrast to indefinite integrals, which include a constant of integration.

These concepts apply to the substitution technique. Although indefinite in nature, our exercise lays foundational understanding for when limits are introduced.

Indefinite Integrals

Indefinite integrals are aimed at finding the antiderivative of a function. In contrast to definite integrals, they do not have limits, hence result in a general form plus a constant \(C\) of integration. For the problem at hand,

• The integral before substitution is expressed without limits, reflective of an indefinite integral.
• The result after substitution and integration forms a function plus a constant \(C\).
• The constant accounts for the family of all possible antiderivatives.

Understanding indefinite integrals is crucial for learning how initial conditions or additional information can transform them into definite solutions.

Algebraic Manipulation

Algebraic manipulation is frequent in calculus, helping simplify integrals and derivatives, permitting easier computation. In this exercise, converting under the radical and versed as fractions was key. Actions taken included:

• Changing expression forms, such as \(\frac{u-3}{\sqrt{u}}\), into separate terms \(u^{1/2} - 3u^{-1/2}\).
• Performing multiplication or division to break down complicated terms into simpler parts.
• Integrating each resulting term independently after manipulation, following rules of integration.

Through manipulation, complex integrals were turned into more manageable expressions, revealing patterns and solutions hidden within the original form.