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Chapter 7: Problem 47

Use a computer algebra system to evaluate the following indefinite integrals. Assume that a is a positive real number. $$\int \frac{x}{\sqrt{2 x+3}} d x$$

Short Answer

Expert verified
Question: Evaluate the indefinite integral of the given function, assuming that 'a' is a positive real number: $$\int \frac{x}{\sqrt{2 x+3}} d x$$ Answer: The indefinite integral of the given function is: $$\int \frac{x}{\sqrt{2 x+3}} d x = \frac{1}{2}(2x+3)^{\frac{3}{2}} - \frac{3}{2}(2x+3)^{\frac{1}{2}} + \frac{1}{4}C$$

Step by step solution

01

Identify the substitution

In this case, we'll let a new variable \(u\) equal the expression inside the square root: $$u = 2 x + 3$$
02

Differentiate \(u\) with respect to \(x\)

Differentiating \(u\) with respect to \(x\) to find the derivative, \(du/dx\) : $$\frac{d u}{d x} = 2$$
03

Solve for \(dx\)

Now, solve for \(d x\) by dividing both sides by 2: $$d x = \frac{1}{2} d u$$
04

Express the integral in terms of \(u\)

Now we need to write \(x\) in terms of \(u\). Solving the equation \(u=2x+3\) for x, we get: $$x = \frac{u-3}{2}$$ Now substitute \(x\) and \(dx\) into the original integral: $$\int \frac{x}{\sqrt{2 x+3}} d x = \int \frac{\frac{u-3}{2}}{\sqrt{u}} \cdot \frac{1}{2} d u$$
05

Simplify and integrate

Simplify the integral: $$\int \frac{\frac{u-3}{2}}{\sqrt{u}} \cdot \frac{1}{2} du = \frac{1}{4} \int \frac{u-3}{\sqrt{u}} d u$$ Now separate the fraction: $$\frac{1}{4} \int \frac{u-3}{\sqrt{u}} d u = \frac{1}{4} \int \left(\frac{u}{\sqrt{u}} - \frac{3}{\sqrt{u}}\right) d u$$ Integrate each term in the parentheses: $$\frac{1}{4} \int \left(u^{\frac{1}{2}} - 3 u^{-\frac{1}{2}}\right) d u = \frac{1}{4}\left(\int u^{\frac{1}{2}} d u - 3 \int u^{-\frac{1}{2}} d u\right)$$
06

Integrate

Now integrate the terms inside the brackets: $$\frac{1}{4}\left(\int u^{\frac{1}{2}} d u - 3 \int u^{-\frac{1}{2}} d u\right) = \frac{1}{4}\left(2u^{\frac{3}{2}} - 6 u^{\frac{1}{2}} + C\right)$$
07

Substitute \(x\) back

Now, we need to substitute our original variable, \(x\), back in place of \(u\). Recall that \(u=2x+3\): $$\frac{1}{4}\left(2u^{\frac{3}{2}} - 6 u^{\frac{1}{2}} + C\right) = \frac{1}{2}(2x+3)^{\frac{3}{2}} - \frac{3}{2}(2x+3)^{\frac{1}{2}} + \frac{1}{4}C$$ So, the indefinite integral is: $$\int \frac{x}{\sqrt{2 x+3}} d x = \frac{1}{2}(2x+3)^{\frac{3}{2}} - \frac{3}{2}(2x+3)^{\frac{1}{2}} + \frac{1}{4}C$$

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Key Concepts

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

U-Substitution
Understanding u-substitution is vital for tackling complex integrals in calculus. It's a technique that simplifies the integration process by transforming an integral into a simpler form. The essential idea is to choose a new variable, usually denoted by u, which is a function of x, and then express the given integral in terms of u instead of x.

This substitution helps in converting the given integral into an easier form that can usually be integrated directly. The process involves finding the derivative of u with respect to x, denoted as du/dx, and then solving for dx. By substituting both x and dx with equivalent expressions in terms of u, we obtain an integral that is more straightforward to evaluate.
Integral Calculus
Integral calculus is one of the two principal branches of calculus, with the other being differential calculus. It is concerned with the concept of an integral, which represents the area under a curve, or more technically, the accumulation of quantities. Integral calculus is typically used for finding lengths, areas, and volumes.

Essentially, when you are integrating a function, you are finding the antiderivative or the original function whose rate of change (derivative) gives you the function you started with. Indefinite integrals, which are characterized by not having specified limits, result in a general form that includes an arbitrary constant, denoted by C, because there are infinitely many antiderivatives for any given function.
Antiderivatives
In calculus, the term antiderivative refers to the inverse operation of differentiating a function. If you have a function f(x), its antiderivative is another function F(x) such that the derivative of F(x) is equal to f(x). It's important to note that antiderivatives are not unique; adding a constant to an antiderivative results in another valid antiderivative.

When solving for the antiderivative, we look for a function whose rate of change corresponds to the integrand. This is because integration, in a broad sense, is the process of summing infinitesimal parts to find the whole, and antiderivatives provide a way to reverse the action of derivatives.
Computer Algebra System
A Computer Algebra System (CAS) is a powerful software tool that is particularly useful in the field of mathematics and engineering. CAS enables users to perform symbolic calculations such as algebra, calculus, and other mathematical manipulations. It can simplify complex algebraic expressions, solve equations, and compute integrals and derivatives.

These systems play a significant role in aiding students and professionals alike by executing calculations that might be prone to human error or would otherwise be incredibly time-consuming to solve manually. When confronted with an indefinite integral like the one in our exercise, a CAS can automate the process, handling u-substitution, simplification, and the integration steps reliably, providing users with a solution that includes the arbitrary constant representing the family of antiderivatives.

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Most popular questions from this chapter

Use the indicated methods to solve the following problems with nonuniform grids. A hot-air balloon is launched from an elevation of 5400 ft above sea level. As it rises, its vertical velocity is computed using a device (called a variometer) that measures the change in atmospheric pressure. The vertical velocities at selected times are shown in the table (with units of \(\mathrm{ft} / \mathrm{min}\) ). $$\begin{array}{|l|c|c|c|c|c|c|c|} \hline t \text { (min) } & 0 & 1 & 1.5 & 3 & 3.5 & 4 & 5 \\ \hline \begin{array}{l} \text { Velocity } \\ \text { (ft/min) } \end{array} & 0 & 100 & 120 & 150 & 110 & 90 & 80 \\ \hline \end{array}$$ a. Use the Trapezoid Rule to estimate the elevation of the balloon after five minutes. Remember that the balloon starts at an elevation of \(5400 \mathrm{ft}\) b. Use a right Riemann sum to estimate the elevation of the balloon after five minutes. c. A polynomial that fits the data reasonably well is $$g(t)=3.49 t^{3}-43.21 t^{2}+142.43 t-1.75$$ Estimate the elevation of the balloon after five minutes using this polynomial.

Use the reduction formulas in a table of integrals to evaluate the following integrals. $$\int x^{3} e^{2 x} d x$$

Evaluate the following integrals. Assume a and b are real numbers and \(n\) is an integer. $$\int \frac{x}{a x+b} d x \text { (Use } u=a x+b$$

For what values of \(p\) does the integral \(\int_{2}^{\infty} \frac{d x}{x \ln ^{p} x}\) exist and what is its value (in terms of \(p\) )?

Let \(a>0\) and let \(R\) be the region bounded by the graph of \(y=e^{-a x}\) and the \(x\) -axis on the interval \([b, \infty)\) a. Find \(A(a, b),\) the area of \(R\) as a function of \(a\) and \(b\) b. Find the relationship \(b=g(a)\) such that \(A(a, b)=2\) c. What is the minimum value of \(b\) (call it \(b^{*}\) ) such that when \(b>b^{*}, A(a, b)=2\) for some value of \(a>0 ?\)
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