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Chapter 5: Problem 44

Use a change of variables to evaluate the following definite integrals. $$\int_{0}^{4} \frac{p}{\sqrt{9+p^{2}}} d p$$

Short Answer

Expert verified
Answer: The integral does not exist, as the substitution leads to a limit that is not valid.

Step by step solution

01

Change of variables

Let's change the variable \(p\) to a new variable \(u\) by using the following substitution: $$p = 3 \sin u$$ Now, we need to find the differential \(dp\) in terms of \(du\), for that we take the derivative of \(p\) with respect to \(u\): $$\frac{dp}{du} = 3 \cos u$$ Hence, the differential \(dp\) is: $$dp = 3 \cos u \, du$$
02

Change the limits of integration

We need to convert the limits of the integration from \(p\) to \(u\). The given limits in terms of \(p\) are 0 to 4. We use the substitution \(p = 3 \sin u\) to find the corresponding limits in terms of \(u\): When \(p=0\), \(u=\arcsin\left(\frac{0}{3}\right)=0\) When \(p=4\), \(u=\arcsin\left(\frac{4}{3}\right)\). Since \(\frac{4}{3}\) is greater than 1, the integral does not exist; therefore, we cannot evaluate this integral with this substitution.

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