Problem 45 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int \frac{2 x^{2}+x}{\sqrt{x}} d x $$

Step-by-Step Solution

Answer

\( \frac{4}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}} + C \)

Step 1: Simplify the Integrand

Begin by rewriting the expression \( \frac{2x^2 + x}{\sqrt{x}} \) in a form that is easier to integrate directly. We can do this by splitting the fraction: \( \frac{2x^2}{\sqrt{x}} + \frac{x}{\sqrt{x}} \). By simplifying each term, we get \( 2x^{2-\frac{1}{2}} + x^{1-\frac{1}{2}} \), or \( 2x^{\frac{3}{2}} + x^{\frac{1}{2}} \).

Step 2: Apply the Power Rule for Integration

Now apply the power rule for integration, which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for any real number \( n eq -1 \). Using this rule, integrate each term separately:- For \( 2x^{\frac{3}{2}} \), the antiderivative is \( \frac{2}{\frac{3}{2} + 1} x^{\frac{3}{2} + 1} = \frac{4}{5} x^{\frac{5}{2}} \).- For \( x^{\frac{1}{2}} \), the antiderivative is \( \frac{1}{\frac{1}{2} + 1} x^{\frac{1}{2} + 1} = \frac{2}{3} x^{\frac{3}{2}} \).

Step 3: Combine the Results

Combine the individual antiderivatives obtained from Step 2:\[ \int \frac{2 x^2 + x}{\sqrt{x}} \, dx = \frac{4}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}} + C \]The constant \( C \) represents the constant of integration.

Key Concepts

Power Rule for Integration

The power rule for integration is a valuable and frequently used tool for finding the integral of a function. When you have a term like \(x^n\) in your integrand, the power rule helps you integrate it easily.

According to the power rule, if you have an integral in the form \( \int x^n \, dx \), and \( n eq -1 \), you can find the antiderivative using the formula:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

Let's take a closer look:

• The exponent \( n \) is increased by one to become \( n + 1 \)
• The entire expression \( x^{n+1} \) is divided by the new exponent \( n+1 \)
• Finally, don't forget to add the constant of integration \( C \)

This formula simplifies the integration process significantly. However, you need to be careful with the condition \( n eq -1 \), as this would lead to a division by zero. For such cases, a special rule applies.

Integrands

The integrand is the function you aim to integrate. In the context of indefinite integrals, the choice of integrand significantly determines how complex or straightforward the integration process will be.

For example, in the given problem, the integrand is initially presented as \( \frac{2x^2 + x}{\sqrt{x}} \).
This expression can be challenging to integrate directly. Hence, simplifying the integrand is a strategic first step.

One common approach is to break down complex fractions into simpler, more manageable terms:

• Start by separating each component: \( \frac{2x^2}{\sqrt{x}} + \frac{x}{\sqrt{x}} \)
• Simplify each term: \( 2x^{\frac{3}{2}} + x^{\frac{1}{2}} \)

With these simplified forms, the integration process becomes more straightforward as it prepares each term for easy application of the power rule. A clear integrand can make the difference between a cumbersome calculation and a smooth integration process.

Constant of Integration

Whenever you compute an indefinite integral, you end up with an antiderivative plus a constant called the constant of integration, denoted by \( C \). This constant plays a crucial role in the solution of indefinite integrals.

The constant of integration arises because integration, essentially the reverse of differentiation, loses any constant term from the original function. Here’s why this is important:

• Without specifying \( C \), there could be infinitely many functions, as every constant will yield the same derivative.
• The constant \( C \) represents this family of antiderivatives.

Think of \( C \) as providing the full set of possible solutions which share a common derivative. Integrals thus specify a family of curves, all of which differ by a vertical shift.
This illustrates why the constant \( C \) is indispensable: it completes the picture when returning from derivatives to functions.