Problem 1 Find the indefinite integrals of... [FREE SOLUTION]

Chapter 8: Problem 1

Find the indefinite integrals of (a) \(3 x^{2 / 3}\) (b) \(\sqrt{(2 x)}\) (c) \(2 x^{3}-2 x^{2}+\frac{1}{x}-2\) (d) \(2 \mathrm{e}^{x}+3 \cos 2 x\) (e) \(x^{2}+3 \mathrm{e}^{x}-\frac{1}{x^{2}}\) (f) \((2 x+1)^{3}\) (g) \((1-2 x)^{1 / 3}\) (h) \(\left(2 x^{2}+1\right)^{3}\) (i) \(\cos (2 x+1)\) (j) \(2^{x}\left(\right.\) Hint: \(\left.2=\mathrm{e}^{\ln 2}\right)\)

Short Answer

Expert verified

The indefinite integrals are: (a) \(\frac{9}{5}x^{5/3} + C\) (b) \(\frac{2\sqrt{2}}{3}x^{3/2} + C\) (c) \(\frac{1}{2}x^4 - \frac{2}{3}x^3 + \ln|x| - 2x + C\) (d) \(2e^x + \frac{3}{2}\sin 2x + C\) (e) \(\frac{x^3}{3} + 3e^x + \frac{1}{x} + C\) (f) \(2x^4 + 4x^3 + 3x^2 + x + C\) (g) \(-\frac{3}{8}(1-2x)^{4/3} + C\) (h) \(x^7 + \frac{12}{5}x^5 + 2x^3 + x + C\) (i) \(\frac{1}{2}\sin(2x+1) + C\) (j) \(\frac{1}{\ln 2}2^x + C\)

Step by step solution

Step 1(a): Rewrite the Expression Using Exponents

The expression \(3x^{2/3}\) is already in the form of \(ax^n\). Identify the constant \(a = 3\) and the exponent \(n = \frac{2}{3}\).

Step 2(a): Apply the Power Rule for Integration

Using the formula \(\int ax^n \, dx = \frac{a}{n+1}x^{n+1} + C\), calculate \(\int 3x^{2/3} \, dx = \frac{3}{(2/3)+1}x^{2/3+1} + C = \frac{3}{5/3} x^{5/3} + C = \frac{9}{5} x^{5/3} + C\).

Step 1(b): Simplify the Expression

Rewrite \(\sqrt{2x}\) as \((2x)^{1/2}\). Use the property \((ab)^n = a^n b^n\) to separate the terms into \(2^{1/2} x^{1/2}\).

Step 2(b): Integrate Using the Power Rule

The integral becomes \(\int (\sqrt{2} \, x^{1/2}) \, dx = \sqrt{2} \int x^{1/2} \, dx = \sqrt{2} \left(\frac{2}{3}x^{3/2} + C\right) = \frac{2\sqrt{2}}{3}x^{3/2} + C\).

Step 1(c): Break Into Separate Integrals

The expression \(2x^3 - 2x^2 + \frac{1}{x} - 2\) can be split into separate integrals: \(\int 2x^3 \, dx - \int 2x^2 \, dx + \int \frac{1}{x} \, dx - \int 2 \, dx\).

Step 2(c): Integrate Each Term

Compute each integral individually: \(\int 2x^3 \, dx = \frac{2}{4}x^4 = \frac{1}{2}x^4\), \(\int 2x^2 \, dx = \frac{2}{3}x^3\), \(\int \frac{1}{x} \, dx = \ln|x|\), and \(\int 2 \, dx = 2x\).

Step 3(c): Combine the Results

Combine the results from each term to get the final integral: \(\frac{1}{2}x^4 - \frac{2}{3}x^3 + \ln|x| - 2x + C\).

Step 1(d): Integrate Each Part Separately

The integral \(2e^x + 3\cos 2x\) consists of two terms which can be integrated separately: \(\int 2e^x \, dx\) and \(\int 3\cos 2x \, dx\).

Step 2(d): Integrate Exponential Term

Integrate the term \(2e^x\): \(\int 2e^x \, dx = 2e^x + C_1\).

Step 3(d): Integrate Trigonometric Term

For \(3 \cos 2x\), use substitution \(u = 2x\), \(du = 2 \, dx\), then \(dx = \frac{1}{2}du\). The integral becomes \(\frac{3}{2}\sin u + C_2\) which is \(\frac{3}{2}\sin 2x + C_2\).

Step 4(d): Combine the Results

Combine both parts to get the final result: \(2e^x + \frac{3}{2}\sin 2x + C\).

Step 1(e): Separate Terms

Identify separate terms in \(x^2 + 3e^x - \frac{1}{x^2}\) to integrate each part separately.

Step 2(e): Integrate Each Term

Compute: \(\int x^2 \, dx = \frac{x^3}{3} + C_1\), \(\int 3e^x \, dx = 3e^x + C_2\), \(\int -\frac{1}{x^2} \, dx = \int -x^{-2} \, dx = \frac{1}{x} + C_3\).

Step 3(e): Combine the Result

Combine the integrals: \(\frac{x^3}{3} + 3e^x + \frac{1}{x} + C\).

Step 1(f): Expand the Binomial Expression

Expand \((2x + 1)^3\) using the binomial theorem: \((2x)^3 + 3(2x)^2(1) + 3(2x)(1)^2 + 1^3 = 8x^3 + 12x^2 + 6x + 1\).

Step 2(f): Integrate Each Expanded Term

Integrate: \(\int 8x^3 \, dx = 2x^4 + C_1\), \(\int 12x^2 \, dx = 4x^3 + C_2\), \(\int 6x \, dx = 3x^2 + C_3\), \(\int 1 \, dx = x + C_4\).

Step 3(f): Combine Results

Combine all integrals: \(2x^4 + 4x^3 + 3x^2 + x + C\).

Step 1(g): Use Substitution Method

For \((1 - 2x)^{1/3}\), let \(u = 1 - 2x\) thus \(du = -2dx\) so \(dx = -\frac{1}{2}du\).

Step 2(g): Rewrite the Integral

The integral becomes \(-\frac{1}{2}\int u^{1/3} \, du\).

Step 3(g): Integrate Using Power Rule

Apply the power rule: \(-\frac{1}{2}\cdot\frac{u^{4/3}}{4/3} + C = -\frac{3}{8}u^{4/3} + C\).

Step 4(g): Substitute Back for x

Substitute \(u = 1 - 2x\) back into the expression: \(-\frac{3}{8}(1 - 2x)^{4/3} + C\).

Step 1(h): Expand the Cubic Expression

Expand \((2x^2 + 1)^3\) using the binomial theorem and integrate each term.

Step 2(h): Use Binomial Theorem

\((2x^2)^3 + 3(2x^2)^2(1) + 3(2x^2)(1)^2 + 1^3 = 8x^6 + 12x^4 + 6x^2 + 1\).

Step 3(h): Integrate Each Term

Integrate: \(\int 8x^6 \, dx = x^7 + C_1\), \(\int 12x^4 \, dx = \frac{12}{5}x^5 + C_2\), \(\int 6x^2 \, dx = 2x^3 + C_3\), \(\int 1 \, dx = x + C_4\).

Step 4(h): Combine Results

Combine the integrals: \(x^7 + \frac{12}{5}x^5 + 2x^3 + x + C\).

Step 1(i): Recognize Basic Integration Formula

The integral \(\cos(2x+1)\) requires a simple substitution for integration.

Step 2(i): Use Substitution Method

Let \(u = 2x + 1\), then \(du = 2\,dx\) hence \(dx = \frac{1}{2}du\).

Step 3(i): Integrate the Simpler Form

The integral becomes \(\frac{1}{2}\int \cos(u) \, du = \frac{1}{2}\sin(u) + C\).

Step 4(i): Substitute Back

Replace \(u\) with \(2x + 1\): \(\frac{1}{2}\sin(2x + 1) + C\).

Step 1(j): Convert the Exponent

Use the hint to express \(2^x\) as \(e^{x\ln 2}\), leading to the integration of \(e^{x\ln 2}\).

Step 2(j): Integrate the Exponential Expression

The integral \(\int e^{x\ln 2} \, dx = \frac{1}{\ln 2}e^{x \ln 2} + C\).

Step 3(j): Convert Back to Exponential

Rewrite \(e^{x \ln 2}\) as \(2^x\), so the final result is \(\frac{1}{\ln 2} 2^x + C\).

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Key Concepts

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

Power Rule for Integration

When integrating an expression with a variable raised to a power, we use the power rule for integration. This rule is essential for finding indefinite integrals of polynomial functions. The power rule for integration states:

\( \int ax^n \, dx = \frac{a}{n+1}x^{n+1} + C \)

Here, \(a\) is a constant, \(x\) is the variable, \(n\) is the exponent, and \(C\) is the constant of integration. We increment the exponent by one and divide the coefficient by this new exponent. For instance, the integral of \(3x^{2/3}\) becomes \(\frac{9}{5}x^{5/3} + C\). Using this approach can simplify the integration of terms like \(x^2\), \(-\frac{1}{x^2}\), and \(x^{1/2}\).
Make sure to rewrite any radicals or fractions with exponents to apply this rule accurately.

Substitution Method

The substitution method, often referred to as "u-substitution," is a technique used to simplify the integration process. This method is helpful when explicitly integrating complex compositions. Here is how the substitution method works:

Choose a substitution \(u\) for an inner function within the integrand.
Express \(dx\) in terms of \(du\).
Transform the entire integral into one in terms of \(u\) and then integrate.

After the integration, substitute back the original variable. This method works smoothly when dealing with integrals such as \((1 - 2x)^{1/3}\) where you let \(u = 1 - 2x\). This changes the integral into a simpler form \(-\frac{1}{2} \int u^{1/3} \, du \).
It's a common method to ease the integration of products and compositions.

Binomial Theorem

The binomial theorem is a powerful expansion tool often used in conjunction with integration. When faced with a polynomial expression raised to a power, like \((2x+1)^3\), the binomial theorem allows us to expand it before integrating:

For an expression \((a + b)^n\), the expansion is given by \(a^n + \binom{n}{1}a^{n-1}b + \ldots + b^n\).

This step results in separate, simpler polynomial terms that are easier to integrate by applying the power rule. For instance, \((2x+1)^3\) expands to \(8x^3 + 12x^2 + 6x + 1\). Each term can be integrated individually, simplifying the process significantly. Understanding this theorem helps in tackling polynomial expressions efficiently.

Exponential Integration

Exponential functions are often straightforward to integrate. The key lies in recognizing the structure of the function in terms of exponential expressions. First, let's consider the term \(e^x\):

The integral of \(e^x\) is itself: \(\int e^x \, dx = e^x + C\).

For functions like \(2^x\), it's beneficial to rewrite the function using natural logs, as in \(2^x = e^{x \ln 2}\). The integral becomes easier to handle:

\(\int e^{x \ln 2} \, dx = \frac{1}{\ln 2}e^{x \ln 2} + C = \frac{1}{\ln 2}2^x + C\).

Understanding exponential integration allows us to address exponential and logarithmic problems effectively in calculus. It's essential for functions like \(2e^x + 3\cos 2x\), where exponential terms require unique handling.

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