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Chapter 5: Problem 48

Use a change of variables to evaluate the following definite integrals. $$\int_{0}^{3} \frac{v^{2}+1}{\sqrt{v^{3}+3 v+4}} d v$$

Short Answer

Expert verified
The final result for the integral \(\int_{0}^{3} \frac{v^{2}+1}{\sqrt{v^{3}+3 v+4}} d v\) can be obtained by calculating the simplified integral \[3\int_{4}^{31} \frac{(u-1)}{\sqrt{u}(v^2+1)}du.\] The key in this problem was to find a suitable substitution \(u = v^3 + 3v + 4\), and then rewrite the integral in terms of this new variable, simplifying it in the process.

Step by step solution

01

Identify the substitution

In this case, we have a function of the form \(\frac{v^2+1}{\sqrt{v^3+3v+4}}\) under the integral. We can notice that the derivative of the denominator involves \(v\) in the numerator, therefore, it's a good idea to use a substitution that will simplify the denominator. Let us choose the substitution \(u = v^3 + 3v + 4\).
02

Calculate the differential

Now we compute the differential \(du\): $$du = \frac{d}{dv}(v^3 + 3v + 4) dv = (3v^2 + 3) dv$$
03

Solve for \(v^2+1\) and \(dv\)

We need to write \(v^2+1\) in terms of \(u\) and also find \(dv\) in terms of \(du\). Notice that we can rewrite \(v^2+1\) as the coefficient of \(dv\) in \(du\) and then we have$$v^2 + 1 = 3v^2 + 3 - 2v^2 = 3(u - 1) - 2v^2$$Now we just need to find \(dv\) in terms of \(du\), so we can substitute into the original integral. Solve the previous equation for \(dv\): $$dv = \frac{du}{3v^2 + 3}$$
04

Change the limits of integration

Since we are changing the variable of integration, we also need to change the limits accordingly. When \(v = 0\), $$u = 0^3 + 3(0) + 4 = 4$$When \(v = 3\), $$u = 3^3 + 3(3) + 4 = 31$$So the new limits of integration are \(u = 4\) to \(u = 31\).
05

Substitute into the integral

Now we substitute our expressions for \(v^2+1\), \(dv\), and the limits of integration into the original integral using \(u\): $$\int_{0}^{3} \frac{v^{2}+1}{\sqrt{v^{3}+3 v+4}} d v = \int_{4}^{31} \frac{3(u-1) - 2v^2}{\sqrt{u}} \cdot \frac{du}{3v^2 + 3}$$
06

Simplify and evaluate the integral

Notice that the "- 2v^2" term cancels with the \(3v^2\) term on the denominator, and we are left with: $$\int_{4}^{31} \frac{3(u-1)}{\sqrt{u}(3v^2+3)}du$$Since there is a constant multiple of \(3\) outside of the integral, we can simplify this further: $$3\int_{4}^{31} \frac{(u-1)}{\sqrt{u}(v^2+1)}du$$The integral is now simpler, and can be evaluated through multiple methods such as partial fraction decomposition or integration by parts. Once the simplified integral has been calculated, multiply it by the constant factor \(3\) to obtain the final result.

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Most popular questions from this chapter

Average value of sine functions Use a graphing utility to verify that the functions \(f(x)=\sin k x\) have a period of \(2 \pi / k,\) where \(k=1,2,3, \ldots . .\) Equivalently, the first "hump" of \(f(x)=\sin k x\) occurs on the interval \([0, \pi / k] .\) Verify that the average value of the first hump of \(f(x)=\sin k x\) is independent of \(k .\) What is the average value?

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