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Chapter 8: Problem 4

Show that $$ \int p \sqrt{1+p} \mathrm{~d} p=\frac{2}{15}(1+p)^{3 / 2}[3 p-2]+C $$

Short Answer

Expert verified
\(\int p \sqrt{1+p} \mathrm{~d} p = \frac{2}{15}(1+p)^{3/2} [3p-2]+C\)

Step by step solution

01

Substitution

Let鈥檚 use a substitution to simplify the integral. Let鈥檚 set \[u = 1 + p\]Then, the derivative of u with respect to p is \[du = dp\]
02

Rewrite the Integral

Rewrite the integral in terms of u: \[\int p \sqrt{1 + p} \mathrm{d}p\]Notice that \[p = u - 1\].Thus, we can rewrite the integral as: \[ \int (u - 1) \sqrt{u} \mathrm{d}u \]
03

Expand the Integrand

Expand the integrand: \[\int (u - 1) \sqrt{u} \mathrm{d}u = \int u^{3/2} \mathrm{d}u - \int u^{1/2} \mathrm{d}u\]
04

Integrate Term by Term

Now, integrate each term separately: \[ \int u^{3/2} \mathrm{d}u = \frac{2}{5} u^{5/2} \]\[ \int u^{1/2} \mathrm{d}u = \frac{2}{3} u^{3/2} \]
05

Combine the Results

Combine the integrated terms: \[\int (u-1)\sqrt{u} \mathrm{d}u = \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} + C\]
06

Substitute Back p

Substitute back the original variable p: \[u = 1 + p\]Thus,\[\frac{2}{5} (1 + p)^{5/2} - \frac{2}{3} (1 + p)^{3/2} + C\]
07

Rewrite in the Desired Form

Finally, factor out common terms to match the given form: \[\frac{2}{15}(1 + p)^{3/2} [3p - 2] + C\]

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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 tool in integral calculus that simplifies complex integrals by changing variables.
In this exercise, we replaced the variable \( p \) with \( u = 1 + p \). This substitution helps to transform the integral into an easier form.
When you change variables, you also need to change the differential. In this case, the differential \( \text{d}p \) becomes \( \text{d}u \).
This method works because the new variable (\( u \)) simplifies the algebraic expressions inside the integral.
Integral Expansion
Once the substitution is done, the next step is to expand the integrand.
Here, we rewrote the integral \( \text{鈭珆 p \text{d}p \) in terms of \( u \) and expanded it:
\( \text{鈭珆 (u - 1) \text{u}^{1/2} \text{d}u \).
This step allows us to handle the integral term-by-term:
  • \( \text{u}^{3/2} \)
  • \( \text{u}^{1/2} \)
Breaking down the integral in this way makes it easier to solve, as each part can be integrated separately.
Definite and Indefinite Integrals
Understanding the difference between definite and indefinite integrals is crucial.
An indefinite integral represents a family of functions and includes an arbitrary constant (\( C \)): \( \text{鈭珆 f(x) \text{d}x = F(x) + C \).
A definite integral calculates the area under the curve between two specific points, and it does not include a constant (\( C \)).
In this exercise, we solved an indefinite integral:
\( \text{鈭珆 p \text{d}p = \frac{2}{15}(1+p)^{3 / 2}[3 p-2] + C \).
Remember, always add \( C \) to indefinite integrals to represent the unknown constant.

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