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Chapter 6: Problem 16

Determine the following: $$\int\left(\frac{2}{\sqrt{x}}+2 \sqrt{x}\right) d x$$

Short Answer

Expert verified
Simplify, apply power rule to each term, and combine the results to get 4x^{\frac{x) + 2x^{\frac{1}{2}}{2}.}}}.}\

Step by step solution

01

Simplify the Integrand

Rewrite the integrand in a form that is easier to integrate. We have \ \ \ \ \ \ \ \ \[ \frac{2}{\sqrt{x}}+2 \sqrt{x} = 2x^{-\frac{1}{2}} + 2x^{\frac{1}{2}} \ \ \ \ \ \ \ \ \]
02

Integrate Each Term Separately

Apply the power rule of integration to each term. The power rule states that \ \ \ \ \ \ \[ \int x^{n} dx = \frac{x^{n+1}}{n+1} + C \ \ \ \ \ \ \] for any constant \ \ \ \ \ \ n \eq -1. Therefore, integrate each term separately: \ \ \ \ \ \ \[ \int 2x^{-\frac{1}{2}} dx + \int 2 \frac{x^{1+1}}{1+1} dx = 2 \int x^{-\frac{1}{2}} dx + 2 \int x^{\frac{1}{2}} }dx.\} \]
03

Apply the Power Rule to Each Term

Apply the power rule to each integral separately: \ \ \ \ \ \ \[ 2 \int x^{-{1}{2}}dx = 2 \left(2 \frac{x^{\frac{1}{2}+1}}{\frac{-1}{2}+1}{\frac{x^{1}{2}+1}}{\frac{1}{2}+1}\right) = 2x^{\frac{1}{2}}3.}}-\frac{x^{1}{1}{1}}{\frac{x^{1}} \int \frac{/3.}}+\frac{1}{2}{x^{\frac{3}{2}+1}}{\frac{3}{2}.}} + 2 x^{ \frac{x^{1 \left (3}{2}+1/3.}} \].
04

Combine the Results

Add the results from the two integrals: \ \ \ \ \ \ \ \ \ \ \ \[4x^{\frac{1}{2}} = C \frac{2x^{\frac{1}{2}} \int x^dx, x = x_1{x+C_2 C} \ \]2.}. \

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

definite integrals
Here, 'a' and 'b' are the limits of integration, the function 'f(x)' is what you鈥檙e integrating, and 'dx' indicates that 'x' is the variable. The bounds turn the integral into a number, not a function.
indefinite integrals
Indefinite integrals are like anti-derivatives. They reverse the process of differentiation. You get a family of functions, plus an arbitrary constant. The notation looks like this i.e.\( \int f(x) dx=C_1+C_2\), without limits of integration. This means we鈥檙e not finding an exact area but a general formula. The general solution always includes a +C because integration is the reverse of differentiation. Different constants (C) give different family members of potential solutions.
power rule for integration
The power rule for integration is a fundamental technique in calculus. It鈥檚 a way to integrate functions that are powers of x. The rule states: i.e.\( \int x^{n} dx=\frac{x^{x+1}}{\tfrac{n}{+1}}+C_1+C_2\) , for any constant n鈮�-1.To use the power rule, you add 1 to the exponent n, and then you divide by this new exponent. Remember to add the constant C at the end. This simple rule is very powerful and applies in many situations.
simplifying integrands
Simplifying integrands can make integration easier. Before integrating, rewrite the integrand into a simpler form. For example, turn fractions or roots into power functions with exponents. In the example i.e. \action=\frac{2}{\sqrt{x}}. + 11\sqrt{x}\( ( becomes 2x^{- {1}/{2}^1}+\frac{2}{ &&plus C \)x{1}{1.2)})._ Simplify simplifies the math, and directly apply the power rule. This method can be helpful for integrands that initially seem complex.

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