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Chapter 5: Problem 4

Determine a substitution that will simplify the integral. In each problem, record your choice of \(u\) and the resulting expression for \(d u .\) Then evaluate the integral. $$ \int 24 t \sqrt{t^{2}+4} d t $$

Short Answer

Expert verified
The integral evaluates to \( 8(t^2 + 4)^{3/2} + C \).

Step by step solution

01

Choose a Substitution

To simplify the integral \( \int 24 t \sqrt{t^2 + 4} \, dt \), let's choose \( u = t^2 + 4 \) so that \( \sqrt{t^2 + 4} \) becomes \( \sqrt{u} \). The expression inside the square root can be simplified into a linear term in \( u \).
02

Differentiate and Express in terms of du

Differentiate \( u = t^2 + 4 \) to find \( du \): \( \frac{du}{dt} = 2t \). Therefore, \( du = 2t \, dt \). This allows us to express \( 24t dt \) in terms of \( du \): \( 24t dt = 12 \, du \).
03

Substitute into the Integral

Substitute \( u \) and \( du \) into the integral: \[ \int 24t \sqrt{t^2 + 4} \, dt = \int 12 \sqrt{u} \, du. \]
04

Integrate in Terms of u

Now, integrate with respect to \( u \): \[ \int 12 \sqrt{u} \, du = 12 \cdot \int u^{1/2} \, du. \] Use the power rule for integration: \( \int u^{n} \, du = \frac{u^{n+1}}{n+1} + C \), where \( n = \frac{1}{2} \). Thus, \( \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} \).
05

Simplify the Expression

The integral becomes \[ 12 \times \frac{2}{3} u^{3/2} = 8u^{3/2} + C. \] Substitute back \( u = t^2 + 4 \): \[ 8(t^2 + 4)^{3/2} + C. \]
06

Final Answer

The integral \( \int 24 t \sqrt{t^2 + 4} \, dt \) evaluates to \( 8(t^2 + 4)^{3/2} + C \).

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Key Concepts

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

Indefinite Integral
When you see the term indefinite integral, you are dealing with an integral that seeks to find an antiderivative for a function. It's different from a definite integral, which calculates the area under the curve between two points.
For example, in our exercise, we want to find the indefinite integral of the function \( 24t \sqrt{t^2 + 4} \). This means our task is to reverse the process of differentiation for this expression.
  • The solution will be a function plus a constant, often noted as \( C \). This is because there are infinitely many antiderivatives differing by a constant.
  • When you write the solution to an indefinite integral, you must always include \( + C \). This ensures that your answer covers all possible antiderivatives.
The integral in the exercise is transformed step-by-step, starting with the choice of a substitution, ultimately allowing us to find its antiderivative in a simplified manner.
U-Substitution
U-substitution is like a clever detective trick in calculus. It helps in simplifying a complex integral into something more manageable.
In our exercise, we choose \( u = t^2 + 4 \). The main reason behind this choice is that \( t^2 + 4 \) is creating a complex situation inside the integral with the square root.
  • The name "u-substitution" comes from substituting the selected part of the integral with \( u \), making it easier to integrate.
  • Derive \( du \) by differentiating \( u \) with respect to \( t \). Here, \( du = 2t \, dt \), helping convert the entire expression in terms of \( u \).
This step converts the challenging integral into a much simpler task. By rewriting our integral in terms of \( u \), we've turned the original problem into a straightforward integration problem. This trick is particularly useful when you encounter compound functions like in the given exercise.
Power Rule for Integration
The power rule is a fundamental tool in calculus that makes integrating power functions straightforward. The rule states:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]provided \( n eq -1 \).
In our case, after applying u-substitution, we end up with \( \int 12 \sqrt{u} \, du \), which simplifies to \( \int 12 u^{1/2} \, du \).
  • Applying the power rule here involves increasing the exponent by one, converting \( 1/2 \) to \( 3/2 \), and dividing by the new exponent.
  • This results in \[ 12 \times \frac{u^{3/2}}{3/2} = 8u^{3/2} + C \], where \( C \) is the constant of integration.
The transformation using the power rule allows for an elegant and simple expression, revealing how the originally complicated integral can be neatly expressed as a function of \( u \). Finally, replacing \( u \) back with \( t^2 + 4 \) leads us to the complete antiderivative.

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Most popular questions from this chapter

A function \(f\) is defined on a specified interval \(I=[a, b] .\) Calculate the area of the region that lies between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. \(f(x)=x^{3}+x+2 \quad I=[-2,1]\)

A function \(f,\) an interval \(I,\) and an even integer \(N\) are given. Approximate the integral of \(f\) over \(I\) by partitioning \(I\) into \(N\) equal length subintervals and using the Midpoint Rule, the Trapezoidal Rule, and then Simpson's Rule. $$ f(x)=1 / x \quad I=[2,6], N=4 $$

An integral \(\int_{a}^{b} f(x) d x\) and a positive integer \(N\) are given. Compute the exact value of the integral, the Simpson's Rule approximation of order \(N,\) and the absolute error \(\varepsilon\). Then find a value \(c\) in the interval \((a, b)\) such that \(\varepsilon=(b-a)^{5}\left|f^{(4)}(c)\right| /\left(180 \cdot N^{4}\right) .\) (This form of the error, which resembles the Mean Value Theorem, implies inequality \((5.8 .4) .)\) $$ \int_{1}^{e} 1 / x d x \quad N=4 $$

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The integral \(\int_{a}^{b}\left(f_{1}(x)-f_{2}(x)\right) d x\) represents the area of a region in the \(x y\) -plane that is bounded by the graphs of \(f_{1}\) and \(f_{2}\). Express the area of the region as an integral of the form \(\int_{c}^{d}\left(g_{1}(y)-g_{2}(y)\right) d y .\) For example, the integral \(\int_{0}^{1}\left(x-x^{2}\right) d x\) represents the area of the shaded region in Figure \(11 .\) This area can also be represented as \(\int_{0}^{1}(\sqrt{y}-y) d y.\) $$ \int_{0}^{4}(\sqrt{x}-x / 2) d x $$
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