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Chapter 5: Problem 47

Evaluate the integrals in Exercises \(17-50\) $$ \int x^{3} \sqrt{x^{2}+1} d x $$

Short Answer

Expert verified
\(\int x^3 \sqrt{x^2+1} \, dx = \frac{1}{5}(x^2+1)^{5/2} - \frac{1}{3}(x^2+1)^{3/2} + C\).

Step by step solution

01

Choose a substitution

We notice that the expression inside the square root, \(x^2+1\), could be simplified if we use a substitution. A common technique with square roots of expressions like \(x^2+1\) is trigonometric substitution, but here we use a simpler algebraic substitution: let \(u = x^2 + 1\). Then, \(du = 2x \, dx\).
02

Express \(dx\) and \(x^3\) in terms of \(u\)

From \(u = x^2 + 1\), we have \(x^2 = u - 1\). Therefore, \(x^3 = x\cdot x^2 = x(u - 1)\). Now, express \(x\) from the differential form: \(x \, dx = \frac{1}{2} du\). Thus, \(dx = \frac{du}{2x}\).
03

Reformulate the integral with substitution

Substitute back into the integral: \(\int x^3 \sqrt{x^2 + 1} \, dx = \int x(u-1)\sqrt{u} \frac{du}{2x}\). The \(x\) cancels with the numerator: we are left with \(\frac{1}{2} \int (u-1)\sqrt{u} \, du\).
04

Expand and simplify the integrand

The integrand \((u-1)\sqrt{u}\) can be expanded to \(u\sqrt{u} - \sqrt{u}\), which simplifies to \(u^{3/2} - u^{1/2}\). The integral now becomes \(\frac{1}{2} \int (u^{3/2} - u^{1/2}) \, du\).
05

Integrate each term separately

Integrate term by term: \(\int u^{3/2} \, du = \frac{2}{5} u^{5/2}\) and \(\int u^{1/2} \, du = \frac{2}{3} u^{3/2}\). Substitute these back into the integral: \(\frac{1}{2}\left( \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right) + C\).
06

Simplify and convert back to \(x\)

Simplify the expression: \(\frac{1}{2} \left( \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right) = \frac{1}{5} u^{5/2} - \frac{1}{3} u^{3/2} + C\). Replace \(u\) by \(x^2 + 1\) to get the integral in terms of \(x\): \(\frac{1}{5}(x^2 + 1)^{5/2} - \frac{1}{3}(x^2 + 1)^{3/2} + C\).

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Key Concepts

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

Trigonometric Substitution
Trigonometric substitution is a technique used in calculus to simplify integrals involving square roots of quadratic expressions. It generally applies to expressions like \( \sqrt{x^2 + a^2} \), \( \sqrt{x^2 - a^2} \), or \( \sqrt{a^2 - x^2} \). These forms resemble trigonometric identities, so by using appropriate trigonometric functions like sine, cosine, or tangent, the integrals can be converted into simpler forms.
  • Example: For \( \sqrt{x^2 + a^2} \), we can use \( x = a \tan\theta \), leading to the identity \( \sqrt{x^2 + a^2} = a \sec\theta \).
  • Benefit: This transforms the integral into one involving trigonometric functions, which are often easier to solve.
In our exercise, trigonometric substitution was an option due to the \( x^2 + 1 \) under the square root. However, the chosen technique was algebraic substitution, because it was more straightforward in this context.
Algebraic Substitution
In calculus, algebraic substitution is a powerful tool for simplifying integrals. It involves replacing a complex expression with a simpler variable. This change of variable helps render an otherwise complicated integral into a simpler one that can be solved more easily.
  • Common Usage: Often used when the integrand can be expressed as a function of another simpler variable.
  • How it Works: Substitute \( u \) for a chosen expression, find \( du \), and rewrite the integral in terms of \( u \).
In our exercise, we set \( u = x^2 + 1 \), which helped simplify the expression under the square root. Then, we used the differential \( du = 2x \, dx \) to substitute \( dx \) and \( x^3 \) in terms of \( u \). This turns the original complex integral into a simpler one in the variable \( u \). This illustrates how algebraic substitution can make it easier to solve complicated integral problems.
Definite Integral
A definite integral computes the area under a curve between two limits of integration. It features prominently in calculus due to its application in finding areas, volumes, and other physical quantities.
  • Integration Limits: A definite integral is written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits.
  • Evaluation: First, find the indefinite integral, then apply the limits of integration to find a numerical value.
In our problem, although we didn't evaluate a definite integral directly, understanding them is crucial for applying integration to real-life problems. When given a definite integral, after calculating the indefinite integral, the next step would be substituting and subtracting the antiderivative evaluated at the upper limit from that evaluated at the lower limit.
Indefinite Integral
Indefinite integrals refer to the general form of integration, which results in a family of functions or antiderivatives. Unlike definite integrals, they do not have set limits and include an arbitrary constant \( C \).
  • Form: It is represented as \( \int f(x) \, dx = F(x) + C \).
  • Importance: Finding an indefinite integral gives a function whose derivative is the original integrand \( f(x) \).
  • Example: The integral \( \int u^{n} \, du = \frac{u^{n+1}}{n+1} + C \).
In our exercise, we found the indefinite integral \( \int x^{3} \sqrt{x^{2}+1} dx \) using substitution. The process of substitution allowed us to transform it into an easier integral, which we solved to find \( \frac{1}{5}(x^2 + 1)^{5/2} - \frac{1}{3}(x^2 + 1)^{3/2} + C \). Adding \( C \) signifies that there are multiple possible functions that can derive to the original integrand.

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Most popular questions from this chapter

In Exercises \(95-98,\) use a CAS to perform the following steps: a. Plot the functions over the given interval. b. Partition the interval into \(n=100,200,\) and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation \(f(x)=(\) average value) for \(x\) using the average value calculated in part (c) for the \(n=1000\) partitioning. $$ f(x)=\sin x \quad \text { on } \quad[0, \pi] $$

Show that the value of \(\int_{0}^{1} \sqrt{x+8} d x\) lies between 2\(\sqrt{2} \approx 2.8\) and \(3 .\)

Evaluate the integrals in Exercises \(17-50\) $$ \int \frac{1}{x^{2}} \sqrt{2-\frac{1}{x}} d x $$

Find the areas of the regions enclosed by the lines and curves. $$ y=\sqrt{|x|} \text { and } 5 y=x+6 \quad \text {(How many intersection points are there?)} $$

In Exercises \(95-98,\) use a CAS to perform the following steps: a. Plot the functions over the given interval. b. Partition the interval into \(n=100,200,\) and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation \(f(x)=(\) average value) for \(x\) using the average value calculated in part (c) for the \(n=1000\) partitioning. $$ f(x)=\sin ^{2} x \quad \text { on } \quad[0, \pi] $$
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