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Magazine Chapter 1: Problem 69
$$ \text { Evaluate } \int \frac{\left(x+\sqrt{1+x^{2}}\right)^{15}}{\sqrt{1+x^{2}}} d x $$
Short Answer
Expert verified
The integral evaluates to \(\frac{(x+\sqrt{1+x^2})^{15}}{15} + C\)."
Step by step solution
01
Identify a Suitable Substitution
Notice that the expression involves \(x + \sqrt{1+x^2}\) and its derivative appears in the denominator. This suggests using a substitution. Let's set \(u = x + \sqrt{1+x^2}\).
02
Find the Derivative of the Substitution
Differentiating \(u\), we get \(\frac{du}{dx} = 1 + \frac{x}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}}\). This simplifies to \(\frac{du}{dx} = \frac{x + \sqrt{1+x^2}}{\sqrt{1+x^2}}\), making \(dx = \frac{\sqrt{1+x^2}}{u} du\).
03
Express the Integrand in Terms of u
The original integrand is \(\frac{(x+\sqrt{1+x^2})^{15}}{\sqrt{1+x^2}}\). Substitute \(u\) for \(x+\sqrt{1+x^2}\) to get \(\frac{u^{15}}{u} = u^{14}\).
04
Transform the Integral to a Simpler Form
Using the substitution and expression derived, the integral becomes \(\int u^{14} du\), since \(dx\) terms cancel out.
05
Integrate the Simplified Expression
Compute the integral \(\int u^{14} du = \frac{u^{15}}{15} + C\), where \(C\) is the constant of integration.
06
Substitute Back to Original Variable
Replace \(u\) with the original expression \(x + \sqrt{1+x^2}\) to obtain the final result: \(\frac{(x+\sqrt{1+x^2})^{15}}{15} + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique used in integral calculus to make complex integrals easier to solve. It involves replacing a part of the integral with a new variable, typically called \( u \), which simplifies the expression. Here are the steps to effectively use the substitution method:
- Identify a part of the integrand that can be substituted. Look for expressions where a direct derivative is present, as in our exercise with \( x + \sqrt{1+x^2} \).
- Define the substitution, \( u = x + \sqrt{1+x^2} \), and compute its derivative \( \frac{du}{dx} \).
- Replace all instances of the original variable in the integrand and \( dx \) with terms of \( u \) and \( du \).
- Simplify the integral to a standard form that is easier to integrate.
Definite Integration
Definite integration involves evaluating the integral of a function over a specified interval. Unlike indefinite integrals that result in a general function plus a constant \( C \), definite integrals yield a numerical value representing the area under the curve between two limits. Here's how definite integration works:
- Define the integral bounds \( [a, b] \), which specify the interval on the x-axis where the area is calculated.
- Perform the integration as usual, possibly using techniques such as substitution as in our problem.
- Evaluate the antiderivative at the upper and lower bounds. Subtract the lower bound value from the upper bound value.
- The result gives the area under the curve of the function from \( x = a \) to \( x = b \).
Integral Calculus
Integral calculus is a branch of mathematics that focuses on the process of integration, which involves finding the integral of a function. Integrals are essential for:
- Solving areas between curves, where they provide the total area accumulated between the different functions over a certain interval.
- Determining physical quantities such as displacement when given a velocity-time graph.
- Calculating volumes of solids of revolution, like those generated by rotating a curve around an axis.
