Integration by Parts: Videos & Practice Problems

Integration by parts is a powerful technique for solving integrals involving the product of two functions, especially when variable substitution fails. The formula, ${\int }_u^d=u\cdot v-{\int }_v^d$, allows for systematic integration. For definite integrals, apply bounds to both parts after integration. Mastery of this method enhances problem-solving skills in calculus, particularly in complex scenarios requiring repeated application of integration by parts.

Introduction to Integration by Parts

When faced with the integral of a product of two functions, such as \( \int 6x e^{x^2} \, dx \), a common approach is to use variable substitution. However, if the problem changes to \( \int 6x e^x \, dx \), this method becomes ineffective. In such cases, integration by parts (IBP) is a valuable technique to apply.

Integration by parts is based on the product rule of differentiation, which states that for two functions \( f(x) \) and \( g(x) \), the derivative of their product is given by:

\[\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)\]

To derive the integration by parts formula, we can integrate both sides of this equation:

\[\int f'(x)g(x) \, dx = f(x)g(x) - \int f(x)g'(x) \, dx\]

Rearranging this gives us the integration by parts formula:

\[\int u \, dv = uv - \int v \, du\]

In this formula, \( u \) is a function that simplifies when differentiated, and \( dv \) is a function that can be easily integrated to find \( v \).

To apply integration by parts to the integral \( \int 6x e^x \, dx \), we first identify \( u \) and \( dv \). A suitable choice is:

• Let \( u = 6x \) (which simplifies to \( du = 6 \, dx \))
• Let \( dv = e^x \, dx \) (which integrates to \( v = e^x \))

Substituting these into the integration by parts formula yields:

\[\int 6x e^x \, dx = 6x e^x - \int e^x \cdot 6 \, dx\]

This simplifies to:

\[\int 6x e^x \, dx = 6x e^x - 6 \int e^x \, dx\]

Since the integral of \( e^x \) is simply \( e^x \), we can further simplify the expression:

\[\int 6x e^x \, dx = 6x e^x - 6e^x + C\]

Thus, the final result for the integral is:

\[\int 6x e^x \, dx = 6e^x(x - 1) + C\]

Understanding how to choose \( u \) and \( dv \) is crucial for successfully applying integration by parts. Practice with various examples will help solidify this technique, making it easier to tackle more complex integrals in the future.

Introduction to Integration by Parts Example 1

To find the integral of the natural logarithm of \( x \) with respect to \( x \), we can utilize the method of integration by parts. This technique is particularly useful when we do not have a straightforward integral for the natural logarithm and cannot apply a simple substitution.

We start by identifying parts of the integral using the formula for integration by parts, which is given by:

\[ \int u \, dv = uv - \int v \, du \]

In this case, we choose \( u = \ln(x) \) because its derivative, \( du = \frac{1}{x} \, dx \), simplifies the expression. For \( dv \), we take \( dv = dx \), which is easy to integrate, yielding \( v = x \).

Now, substituting these values into the integration by parts formula, we have:

\[ \int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx \]

Notice that the \( x \) in the numerator and denominator cancels out, simplifying our integral to:

\[ \int \ln(x) \, dx = x \ln(x) - \int 1 \, dx \]

The integral of \( 1 \) with respect to \( x \) is simply \( x \). Therefore, we can express the integral as:

\[ \int \ln(x) \, dx = x \ln(x) - x + C \]

where \( C \) is the constant of integration. This result shows how integration by parts can effectively handle the integral of the natural logarithm.

Additionally, a helpful mnemonic for choosing \( u \) and \( dv \) is the acronym "LIATE," which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions. According to this rule, logarithmic functions should be chosen as \( u \) when present, as they simplify upon differentiation.

Introduction to Integration by Parts Example 2

To find the integral of the inverse sine function, denoted as arcsin(t), with respect to t, we can utilize the technique of integration by parts. This method is particularly useful when the integral does not lend itself to straightforward substitution or application of power rules.

First, we identify the components needed for integration by parts, which follows the formula:

∫ u dv = uv - ∫ v du

In this case, we choose:

• u = arcsin(t), since its derivative is manageable.
• dv = dt, which is easy to integrate.

Next, we compute the derivative of u:

du = \frac{1}{\sqrt{1 - t^2}} dt

And we find v by integrating dv:

v = t

Now we can substitute these values into the integration by parts formula:

∫ arcsin(t) dt = t * arcsin(t) - ∫ t * \frac{1}{\sqrt{1 - t^2}} dt

At this point, we need to simplify the remaining integral. To do this, we can use a substitution method. Let:

w = 1 - t^2

Then, the derivative is:

dw = -2t dt

Rearranging gives us:

dt = -\frac{dw}{2t}

Substituting these into the integral, we have:

∫ t * \frac{1}{\sqrt{1 - t^2}} dt = -\frac{1}{2} ∫ \frac{1}{\sqrt{w}} dw

Now, we can integrate:

∫ \frac{1}{\sqrt{w}} dw = 2\sqrt{w}

Thus, we have:

∫ t * \frac{1}{\sqrt{1 - t^2}} dt = -\frac{1}{2} * 2\sqrt{w} = -\sqrt{w}

Substituting back for w gives us:

∫ t * \frac{1}{\sqrt{1 - t^2}} dt = -\sqrt{1 - t^2}

Now, we can combine everything to find the integral:

∫ arcsin(t) dt = t * arcsin(t) + \sqrt{1 - t^2} + C

Here, C represents the constant of integration. This result shows how integration by parts can effectively handle integrals involving inverse trigonometric functions, emphasizing the importance of correctly identifying u and dv to simplify the process.

Find the integral.
${\int }_{}^{}s{e}^{3s}ds$

• A. $\frac{s}{9}{e}^{3s}-\frac{1}{3}{e}^{3s}+C$
• B. $3s{e}^s-\frac{1}{9}{e}^{3s}+C$
• C. $\frac{s}{3}{e}^{3s}-\frac{1}{9}{e}^{3s}+C$
• D. $3s{e}^{3s}-9e^{3s}+C$

Answer: (C)

Find the integral.
${\int }_{}^{}\theta \cos 2\theta d\theta$

• A. $\frac{\theta }{2}\sin \left(2\theta \right)+\frac{1}{4}\cos \left(2\theta \right)+C$
• B. $\frac{1}{2}\sin \left(2\theta \right)+C$
• C. $\frac{1}{2}{\theta }^2\sin \left(2\theta \right)+C$
• D. $\frac{{\theta }^2}{2}\sin \left(2\theta \right)+\frac{\theta }{4}\cos \left(2\theta \right)+C$

Answer: (A)

Find the integral.
${\int }_{}^{}{x}^2\ln xdx$

• A. $\frac{x^3}{3}\ln x+\frac{x^3}{9}+C$
• B. $\frac{x^3}{3}\ln x+\frac{1}{x}+C$
• C. $\frac{x^3}{3}\ln x-\frac{1}{x}+C$
• D. $\frac{x^3}{3}\ln x-\frac{x^3}{9}+C$

Answer: (D)

Repeated Integration by Parts

Integration by parts is a powerful technique used to evaluate integrals, particularly when dealing with the product of two functions. This method is especially useful when simpler strategies, like variable substitution, are not applicable. A key aspect of integration by parts is that it may need to be repeated multiple times to solve more complex integrals.

To apply integration by parts, we start with the formula:

$$\int u \, dv = uv - \int v \, du$$

In this formula, we need to identify two components: \(u\) and \(dv\). The choice of \(u\) should be a function that simplifies when differentiated, while \(dv\) should be a function that is easy to integrate. For example, consider the integral:

$$\int 3x^2 e^x \, dx$$

Here, we can choose \(u = 3x^2\) (since its derivative \(du = 6x \, dx\) simplifies) and \(dv = e^x \, dx\) (which integrates to \(v = e^x\)). Applying the integration by parts formula gives:

$$3x^2 e^x - \int e^x (6x) \, dx$$

At this point, we encounter another integral, \(\int 6x e^x \, dx\), which requires us to apply integration by parts again. We repeat the process by selecting \(u = 6x\) (with \(du = 6 \, dx\)) and \(dv = e^x \, dx\). This leads to:

$$6x e^x - \int e^x (6) \, dx$$

Now, we can easily integrate \(\int e^x \, dx\) to get \(e^x\). Thus, the overall expression becomes:

$$3x^2 e^x - 6x e^x + 6e^x + C$$

In summary, when faced with complex integrals, the integration by parts technique may require multiple applications of the formula. Each time, carefully select \(u\) and \(dv\) to simplify the process. This method not only helps in solving integrals but also reinforces the understanding of how functions interact through differentiation and integration.

Repeated Integration by Parts Example 3

To solve the indefinite integral of the product of two functions, we can utilize the integration by parts formula, which states:

Integration by Parts Formula: \[\int u \, dv = uv - \int v \, du\]

In this case, we have a polynomial and an exponential function. We start by selecting \( u \) and \( dv \). A good choice for \( u \) is the polynomial \( t^2 + 2t - 3 \), as its derivative will simplify the expression. The derivative \( du \) is calculated as follows:

\[du = (2t + 2) \, dt\]

Next, we choose \( dv \) as \( e^{4t} \, dt \). Integrating this gives us:

\[v = \frac{1}{4} e^{4t}\]

Now we can apply the integration by parts formula:

\[\int (t^2 + 2t - 3) e^{4t} \, dt = \left( t^2 + 2t - 3 \right) \left( \frac{1}{4} e^{4t} \right) - \int \left( \frac{1}{4} e^{4t} \right) (2t + 2) \, dt\]

We can simplify this to:

\[= \frac{1}{4} e^{4t} (t^2 + 2t - 3) - \frac{1}{4} \int e^{4t} (2t + 2) \, dt\]

At this point, we need to apply integration by parts again to the remaining integral \( \int e^{4t} (2t + 2) \, dt \). We set \( u = 2t + 2 \) and \( dv = e^{4t} \, dt \). The derivative \( du \) is:

\[du = 2 \, dt\]

And integrating \( dv \) gives us the same \( v \) as before:

\[v = \frac{1}{4} e^{4t}\]

Applying integration by parts again, we have:

\[\int (2t + 2) e^{4t} \, dt = (2t + 2) \left( \frac{1}{4} e^{4t} \right) - \int \left( \frac{1}{4} e^{4t} \cdot 2 \, dt \right)\]

Which simplifies to:

\[= \frac{1}{4} e^{4t} (2t + 2) - \frac{1}{2} \int e^{4t} \, dt\]

Now, we can evaluate the integral \( \int e^{4t} \, dt \), which is:

\[\int e^{4t} \, dt = \frac{1}{4} e^{4t}\]

Substituting this back into our expression gives:

\[\int (2t + 2) e^{4t} \, dt = \frac{1}{4} e^{4t} (2t + 2) - \frac{1}{2} \left( \frac{1}{4} e^{4t} \right) = \frac{1}{4} e^{4t} (2t + 2) - \frac{1}{8} e^{4t}\]

Now, substituting this result back into our original integration by parts expression, we have:

\[\int (t^2 + 2t - 3) e^{4t} \, dt = \frac{1}{4} e^{4t} (t^2 + 2t - 3) - \frac{1}{4} \left( \frac{1}{4} e^{4t} (2t + 2) - \frac{1}{8} e^{4t} \right)\]

After simplifying, we arrive at the final result:

\[\int (t^2 + 2t - 3) e^{4t} \, dt = \frac{1}{4} e^{4t} (t^2 + 2t - 3) - \frac{1}{16} e^{4t} (2t + 2) + \frac{1}{32} e^{4t} + C\]

In conclusion, the process of repeated integration by parts allows us to break down complex integrals into simpler components, ultimately leading to a solution. The key is to choose \( u \) wisely so that its derivative simplifies the integral, and to be prepared to apply the method multiple times if necessary.

Repeated Integration by Parts Example 4

To find the indefinite integral of the natural logarithm of \(x^2\), we will utilize integration by parts, a technique useful for integrating products of functions. The formula for integration by parts is given by:

\[\int u \, dv = uv - \int v \, du\]

In this case, we set \(u = \ln(x^2)\) and differentiate it to find \(du\). Using the chain rule, we have:

\[du = \frac{d}{dx}(\ln(x^2)) = \frac{2}{x} \, dx\]

Next, we choose \(dv = dx\), which integrates to \(v = x\). Now we can apply the integration by parts formula:

\[\int \ln(x^2) \, dx = x \ln(x^2) - \int x \cdot \frac{2}{x} \, dx\]

Notice that the \(x\) terms cancel, simplifying our integral to:

\[\int \ln(x^2) \, dx = x \ln(x^2) - 2 \int dx\]

Integrating \(dx\) gives us \(x\), leading to:

\[\int \ln(x^2) \, dx = x \ln(x^2) - 2x + C\]

Next, we can simplify \(x \ln(x^2)\) using the property of logarithms, \(\ln(a^b) = b \ln(a)\), which gives us:

\[x \ln(x^2) = 2x \ln(x)\]

Thus, the integral can be rewritten as:

\[\int \ln(x^2) \, dx = 2x \ln(x) - 2x + C\]

In conclusion, the indefinite integral of \(\ln(x^2)\) is:

\[\int \ln(x^2) \, dx = 2x \ln(x) - 2x + C\]

This process illustrates the effectiveness of integration by parts, especially when dealing with logarithmic functions, and highlights the importance of choosing \(u\) and \(dv\) wisely to simplify the integration process.

Find the indefinite integral.
${\int }_{}^{}{r}^2{e}^{-r}dr$

• A. $r^2{e}^{-r}-2r{e}^{-r}+2e^{-r}+C$
• B. $-r^2{e}^{-r}-2r{e}^{-r}-2e^{-r}+C$
• C. $-\frac{1}{2}{r}^2{e}^{-r}-r{e}^{-r}-e^{-r}+C$
• D. $\frac{1}{2}{r}^2{e}^{-r}-r{e}^{-r}+e^{-r}+C$

Answer: (B)

Find the indefinite integral.
${\int }_{}^{}{\theta }^2\cos 3\theta d\theta$

• A. $\frac{{\theta }^3}{3}\sin 3\theta +\frac{2\theta }{9}\cos 3\theta -\frac{2}{27}\sin 3\theta +C$
• B. $\frac{{\theta }^2}{3}\sin 3\theta +\frac{2\theta }{9}\cos 3\theta -\frac{2}{27}\sin 3\theta +C$
• C. $\frac{{\theta }^3}{3}\sin 3\theta +3\cos \theta +C$
• D. $\frac{{\theta }^2}{3}\sin 3\theta +3\cos \theta +C$

Answer: (B)

Tabular Integration by Parts

Integration by parts is a powerful technique used to solve integrals involving the product of two functions. The fundamental formula for integration by parts is given by:

$$\int u \, dv = uv - \int v \, du$$

In this context, choosing the right functions for \( u \) and \( dv \) is crucial. A helpful mnemonic for selecting \( u \) is LIATE, which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions. Typically, you want to choose \( u \) as the algebraic function when dealing with a product of functions, as it simplifies upon differentiation.

For more complex integrals, the tabular method can streamline the process. This method organizes the derivatives and integrals into a table, making it easier to visualize the steps involved. To illustrate, consider the integral of \( 3x^2 e^{2x} \). First, identify \( u \) and \( dv \): set \( u = 3x^2 \) and \( dv = e^{2x} \, dx \). Then, compute the derivatives of \( u \) and the integrals of \( dv \).

For the derivatives:

• First derivative: \( \frac{d}{dx}(3x^2) = 6x \)
• Second derivative: \( \frac{d}{dx}(6x) = 6 \)
• Third derivative: \( \frac{d}{dx}(6) = 0 \)

For the integrals:

• First integral: \( \int e^{2x} \, dx = \frac{1}{2} e^{2x} \)
• Second integral: \( \int \frac{1}{2} e^{2x} \, dx = \frac{1}{4} e^{2x} \)
• Third integral: \( \int \frac{1}{4} e^{2x} \, dx = \frac{1}{8} e^{2x} \)

Next, arrange these results in a table with alternating signs. The first column contains the derivatives of \( u \), and the second column contains the integrals of \( dv \). The pattern of signs starts with a positive sign for the first term, followed by a negative sign for the second, and so on.

Using the table, the integral can be computed as follows:

$$\int 3x^2 e^{2x} \, dx = 3x^2 \cdot \frac{1}{2} e^{2x} - 6x \cdot \frac{1}{4} e^{2x} + 6 \cdot \frac{1}{8} e^{2x} + C$$

After simplifying, the final result is:

$$\int 3x^2 e^{2x} \, dx = \frac{3}{2} x^2 e^{2x} - \frac{3}{2} x e^{2x} + \frac{3}{4} e^{2x} + C$$

This method not only reduces the complexity of the calculations but also provides a clear visual representation of the steps involved in solving the integral. When applying the tabular method, ensure that the number of derivatives taken corresponds to the number of integrals, stopping when the product of the last row is a basic integral that can be easily integrated.

Tabular Integration by Parts Example 5

To evaluate the indefinite integral using the tabular method, we start by recognizing that integration by parts is necessary due to the product of functions involved. The tabular method simplifies this process by organizing our work into two columns: one for derivatives (denoted as the d column) and one for integrals (denoted as the I column).

First, we need to choose our u and dv. The goal is to select u such that its derivative simplifies the expression. In this case, we choose u = \ln(t) because its derivative, du = \frac{1}{t} dt, is straightforward. For dv, we select dv = t^3 dt, which integrates easily to v = \frac{t^4}{4} using the power rule.

Next, we set up our table with the derivatives of u and the integrals of dv. The first row will contain u and dv, while subsequent rows will contain their derivatives and integrals, respectively. Since we only need one derivative for u, we stop there. The table will look like this:

d column:
u = \ln(t)
du = \frac{1}{t} dt

I column:
dv = t^3 dt
v = \frac{t^4}{4}

Now, we apply the integration by parts formula, which states:

∫ u \, dv = uv - ∫ v \, du

From our table, we have:

∫ \ln(t) t^3 dt = \ln(t) \cdot \frac{t^4}{4} - ∫ \frac{t^4}{4} \cdot \frac{1}{t} dt

This simplifies to:

∫ \ln(t) t^3 dt = \frac{t^4}{4} \ln(t) - \frac{1}{4} ∫ t^3 dt

We can now integrate t^3 easily:

∫ t^3 dt = \frac{t^4}{4}

Substituting this back into our equation gives:

∫ \ln(t) t^3 dt = \frac{t^4}{4} \ln(t) - \frac{1}{4} \cdot \frac{t^4}{4} + C

Combining the terms results in:

∫ \ln(t) t^3 dt = \frac{t^4}{4} \ln(t) - \frac{t^4}{16} + C

Thus, the final answer for the indefinite integral using the tabular method is:

∫ \ln(t) t^3 dt = \frac{t^4}{4} \ln(t) - \frac{t^4}{16} + C

Tabular Integration by Parts Example 6

To evaluate the indefinite integral of \( e^t \sin(3t) \) using the tabular method, we start by recognizing that we have two functions multiplied together: \( e^t \) and \( \sin(3t) \). Since both functions involve the variable \( t \) in their arguments, a simple variable substitution won't simplify the integral. Instead, we will apply integration by parts, specifically using the tabular method.

In the tabular method, we create two columns: one for derivatives and one for integrals. We need to choose which function to differentiate (let's call it \( u \)) and which to integrate (let's call it \( dv \)). Here, we set \( u = \sin(3t) \) because its derivative will yield a different function, while \( dv = e^t dt \) is straightforward to integrate.

Next, we compute the derivatives of \( u \) and the integrals of \( dv \):

• First derivative: \( u' = 3 \cos(3t) \)
• Second derivative: \( u'' = -9 \sin(3t) \)
• Third derivative: \( u''' = -27 \cos(3t) \)
• Fourth derivative: \( u^{(4)} = 81 \sin(3t) \)

For the integral column, since integrating \( e^t \) repeatedly yields the same function, we have:

• First integral: \( \int e^t dt = e^t \)
• Second integral: \( \int e^t dt = e^t \)
• Third integral: \( \int e^t dt = e^t \)
• Fourth integral: \( \int e^t dt = e^t \)

Now, we set up the tabular method by alternating the signs and multiplying diagonally:

• First term: \( e^t \sin(3t) \)
• Second term: \( -3 e^t \cos(3t) \)
• Third term: \( -(-9 e^t \sin(3t)) = 9 e^t \sin(3t) \)
• Fourth term: \( -(-27 e^t \cos(3t)) = 27 e^t \cos(3t) \)

Combining these, we have:

Integral \( = e^t \sin(3t) - 3 e^t \cos(3t) + 9 \int e^t \sin(3t) dt \)

At this point, we notice that the integral \( \int e^t \sin(3t) dt \) appears on both sides of the equation. To isolate this integral, we can add \( 9 \int e^t \sin(3t) dt \) to both sides:

Let \( I = \int e^t \sin(3t) dt \), then:

\( I = e^t \sin(3t) - 3 e^t \cos(3t) + 9I \)

Rearranging gives:

\( I - 9I = e^t \sin(3t) - 3 e^t \cos(3t) \)

\( -8I = e^t \sin(3t) - 3 e^t \cos(3t) \)

Dividing by \(-8\) yields:

\( I = -\frac{1}{8} (e^t \sin(3t) - 3 e^t \cos(3t)) \)

Thus, the final result for the integral is:

\( \int e^t \sin(3t) dt = -\frac{1}{8} (e^t \sin(3t) - 3 e^t \cos(3t)) + C \)

This method illustrates how to handle integrals that exhibit repeating patterns through the tabular integration by parts technique, allowing for a systematic approach to finding the solution.

Evaluate the indefinite integral.
${\int }_{}^{}(s^2+4)e^{3s}ds$

• A. $\frac{e^{3s}}{3}(s^2+4)-\frac{2}{9}s{e}^{3s}+\frac{2}{27}{e}^{3s}+C$
• B. $\frac{e^{3s}}{3}(s^2+4)-\frac{2}{9}s{e}^{3s}+C$
• C. $(s^3+\frac{4}{3}s)e^{3s}+C$
• D. $(\frac{s^3}{3}+\frac{4}{3}s)e^{3s}+C$

Answer: (A)

Evaluate the indefinite integral.
${\int }^{}\frac{\ln x}{x^4}dx$

• A. $-\frac{\ln x}{3x^3}+\frac{1}{x}+C$
• B. $\frac{\ln x}{3x^3}+\frac{1}{x}+C$
• C. $\frac{\ln x}{3x^3}-\frac{1}{9x^3}+C$
• D. $-\frac{\ln x}{3x^3}-\frac{1}{9x^3}+C$

Answer: (D)

Evaluate the indefinite integral.
${\int }_{}^{}{x}^2\cos 4xdx$

• A. $\frac{x^2}{4}\sin \left(4x\right)+\frac{1}{4}\cos \left(4x\right)-\frac{1}{16}\cos \left(4x\right)+C$
• B. $\frac{x^3}{3}\cos \left(4x\right)+4x^2\sin \left(4x\right)+C$
• C. $\frac{x^2}{4}\sin \left(4x\right)+\frac{1}{4}\cos \left(4x\right)-\frac{1}{16}\sin \left(4x\right)+C$
• D. $\frac{4x^3}{3}\cos \left(4x\right)+4x^2\sin \left(4x\right)+C$

Answer: (C)

Integration by Parts for Definite Integrals

Integration by parts is a powerful technique used to evaluate integrals, and it can be applied to both indefinite and definite integrals. The fundamental formula for integration by parts states that the integral of \( u \, dv \) is equal to \( u \cdot v - \int v \, du \). When dealing with definite integrals, the process remains largely the same, but it is crucial to apply the bounds to both parts of the equation at the end of the calculation.

To illustrate this, consider a definite integral where we choose \( u \) and \( dv \). For example, if we let \( u = 6x \) and \( dv = e^x \, dx \), then the derivative \( du = 6 \, dx \) and the integral \( v = e^x \). Applying the integration by parts formula, we have:

\[\int_0^2 6x \cdot e^x \, dx = \left[ 6x \cdot e^x \right]_0^2 - \int_0^2 e^x \cdot 6 \, dx\]

After calculating the first term, \( 6x \cdot e^x \), we evaluate it at the bounds of 0 and 2. This means substituting 2 into the expression to get \( 12e^2 \) and substituting 0, which results in 0. Thus, the first term simplifies to \( 12e^2 - 0 = 12e^2 \).

Next, we evaluate the integral \( \int_0^2 6e^x \, dx \). Since 6 is a constant, it can be factored out, leading to:

\[6 \int_0^2 e^x \, dx = 6 \left[ e^x \right]_0^2 = 6(e^2 - 1)\]

Combining these results, we have:

\[12e^2 - 6(e^2 - 1) = 12e^2 - 6e^2 + 6 = 6e^2 + 6\]

This can be factored to yield:

\[6(e^2 + 1)\]

For those interested in a numerical approximation, substituting \( e \approx 2.718 \) gives an approximate value of 50.33. This demonstrates how to effectively apply integration by parts to definite integrals, ensuring that bounds are applied correctly to both components of the result.

Evaluate the definite integral.
${\int }_1^{2}x\ln xdx$

• A. $-0.11$
• B. $0.64$
• C. $0.89$
• D. $1.89$

Answer: (B)

Evaluate the definite integral.
${\int }_0^{\pi }{t}^2\sin tdt$

• A. $7.87$
• B. $5.87$
• C. $-9.51$
• D. $-9.87$

Answer: (B)

Integration by Parts for Definite Integrals Example 7

To find the area of the region bounded by the function \( f(x) = (x - 1)e^{3x} \) and the x-axis over the interval from \( x = 1 \) to \( x = 4 \), we can set up a definite integral. The area \( A \) can be expressed as:

\[A = \int_{1}^{4} f(x) \, dx = \int_{1}^{4} (x - 1)e^{3x} \, dx\]

Since the area is calculated between the curve and the x-axis, we recognize that the lower function \( g(x) \) is the x-axis itself, which is represented by \( g(x) = 0 \). Thus, the integral simplifies to just the function \( f(x) \).

To solve the integral, we will use integration by parts, which is based on the formula:

\[\int u \, dv = uv - \int v \, du\]

We choose \( u = x - 1 \) and \( dv = e^{3x} \, dx \). The derivative \( du \) is \( dx \), and to find \( v \), we integrate \( dv \):

\[v = \int e^{3x} \, dx = \frac{1}{3} e^{3x}\]

Now we can apply the integration by parts formula:

\[\int (x - 1)e^{3x} \, dx = (x - 1) \cdot \frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} \, dx\]

Next, we simplify the integral on the right:

\[\int \frac{1}{3} e^{3x} \, dx = \frac{1}{9} e^{3x}\]

Substituting back, we have:

\[\int (x - 1)e^{3x} \, dx = \frac{(x - 1)}{3} e^{3x} - \frac{1}{9} e^{3x}\]

Now we evaluate this expression from \( x = 1 \) to \( x = 4 \):

\[A = \left[ \frac{(x - 1)}{3} e^{3x} - \frac{1}{9} e^{3x} \right]_{1}^{4}\]

Calculating the upper limit at \( x = 4 \):

\[= \left( \frac{(4 - 1)}{3} e^{12} - \frac{1}{9} e^{12} \right) = \left( \frac{3}{3} e^{12} - \frac{1}{9} e^{12} \right) = e^{12} - \frac{1}{9} e^{12} = \frac{8}{9} e^{12}\]

Now calculating the lower limit at \( x = 1 \):

\[= \left( \frac{(1 - 1)}{3} e^{3} - \frac{1}{9} e^{3} \right) = 0 - \frac{1}{9} e^{3} = -\frac{1}{9} e^{3}\]

Putting it all together, we have:

\[A = \left( \frac{8}{9} e^{12} \right) - \left( -\frac{1}{9} e^{3} \right) = \frac{8}{9} e^{12} + \frac{1}{9} e^{3}\]

Thus, the area under the curve from \( x = 1 \) to \( x = 4 \) is given by:

\[A = \frac{8}{9} e^{12} + \frac{1}{9} e^{3}\]

For numerical approximation, substituting the values of \( e^{12} \) and \( e^{3} \) into a calculator will yield an approximate area of 44,673.

Integration by Parts for Definite Integrals Example 8

To find the volume of the solid generated by revolving the region bounded by the curve \( y = \sqrt{x} \cos(x) \), the x-axis, and the vertical lines \( x = 0 \) and \( x = \frac{\pi}{2} \) around the x-axis, we can use the disk method. The volume \( V \) can be expressed using the integral:

\[V = \pi \int_{0}^{\frac{\pi}{2}} \left( \sqrt{x} \cos(x) \right)^2 \, dx \]

First, we simplify the integrand:

\[V = \pi \int_{0}^{\frac{\pi}{2}} x \cos^2(x) \, dx \]

Next, we can factor out the constant \( \pi \) from the integral:

\[V = \pi \int_{0}^{\frac{\pi}{2}} x \cos^2(x) \, dx \]

To evaluate this integral, we will use integration by parts, which is given by the formula:

\[\int u \, dv = uv - \int v \, du \]

We choose \( u = x \) (which simplifies upon differentiation) and \( dv = \cos^2(x) \, dx \). To find \( v \), we need to integrate \( \cos^2(x) \). Using the identity \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \), we can integrate:

\[v = \int \cos^2(x) \, dx = \frac{1}{2} \left( x + \frac{\sin(2x)}{2} \right) \]

Now, applying integration by parts:

\[\int_{0}^{\frac{\pi}{2}} x \cos^2(x) \, dx = \left[ x \cdot v \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} v \, du \]

Calculating \( \left[ x \cdot v \right]_{0}^{\frac{\pi}{2}} \):

At \( x = \frac{\pi}{2} \):

\[\frac{\pi}{2} \cdot \left( \frac{1}{2} \left( \frac{\pi}{2} + 0 \right) \right) = \frac{\pi^2}{8} \]

At \( x = 0 \):

\[0 \cdot v = 0 \]

Thus, we have:

\[\left[ x \cdot v \right]_{0}^{\frac{\pi}{2}} = \frac{\pi^2}{8} \]

Next, we need to evaluate the integral \( \int_{0}^{\frac{\pi}{2}} v \, du \). Since \( du = dx \), we have:

\[\int_{0}^{\frac{\pi}{2}} v \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \left( x + \frac{\sin(2x)}{2} \right) \, dx \]

Evaluating this integral gives us two parts. The first part is straightforward:

\[\int_{0}^{\frac{\pi}{2}} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{\pi^2}{8} \]

For the second part, we integrate \( \frac{\sin(2x)}{2} \):

\[\int \frac{\sin(2x)}{2} \, dx = -\frac{1}{4} \cos(2x) \]

Evaluating from \( 0 \) to \( \frac{\pi}{2} \):

At \( x = \frac{\pi}{2} \): \( -\frac{1}{4} \cos(\pi) = \frac{1}{4} \)

At \( x = 0 \): \( -\frac{1}{4} \cos(0) = -\frac{1}{4} \)

Thus, the total contribution from this part is:

\[\frac{1}{4} - \left( -\frac{1}{4} \right) = \frac{1}{2} \]

Combining these results, we find:

\[\int_{0}^{\frac{\pi}{2}} v \, dx = \frac{\pi^2}{8} + \frac{1}{2} \]

Finally, substituting back into our volume formula:

\[V = \pi \left( \frac{\pi^2}{8} - \left( \frac{\pi^2}{8} + \frac{1}{2} \right) \right) = -\frac{\pi}{2} \]

However, since volume cannot be negative, we take the absolute value, leading to the final volume of the solid:

\[V \approx 1.79 \]

This process illustrates how to find the volume of a solid of revolution using integration by parts and the disk method, emphasizing the importance of careful integration and evaluation of bounds.

Here’s what students ask on this topic:

What is the formula for integration by parts and how is it derived?

The formula for integration by parts is:

$\int udv=uv-\int vdu$

It is derived from the product rule for differentiation, which states:

$(f\cdot g)\text{'}=f\text{'}g+fg\text{'}$

Integrating both sides with respect to $x$ gives:

$\int (f\text{'}g)dx=fg-\int fg\text{'}dx$

By substituting $u$ for $f$ and $v$ for $g$, the integration by parts formula is obtained.

How do you choose $u$ and $\mathrm{dv}$ in integration by parts?

Choosing $u$ and $\mathrm{dv}$ is crucial for simplifying the integral. A common guideline is the LIATE rule, which prioritizes functions in this order:

• L: Logarithmic functions (e.g., $\ln \left(x\right)$)
• I: Inverse trigonometric functions (e.g., $\arctan \left(x\right)$)
• A: Algebraic functions (e.g., $x,x^2$)
• T: Trigonometric functions (e.g., $\sin \left(x\right)$)
• E: Exponential functions (e.g., $e^x$)

Choose $u$ as the function that simplifies when differentiated, and $\mathrm{dv}$ as the function that is easy to integrate. For example, in $\int x{e}^{x}dx$, set $u=x$ and $\mathrm{dv}=e^{x}dx$.

What is the tabular method for integration by parts?

The tabular method is a systematic way to apply integration by parts multiple times, especially for polynomials multiplied by exponential or trigonometric functions. Here's how it works:

1. Create two columns: one for derivatives (D) and one for integrals (I).
2. In the D column, write the function to differentiate (e.g., a polynomial) and repeatedly take its derivative until it becomes zero.
3. In the I column, write the function to integrate (e.g., exponential or trigonometric) and repeatedly integrate it.
4. Alternate signs starting with positive (+, -, +, ...).
5. Multiply diagonally (from D to I) and sum the terms, stopping when the derivative reaches zero.

For example, to integrate $\int x^2{e}^{x}dx$, the tabular method simplifies the process by organizing derivatives and integrals into a table, avoiding repeated manual application of the formula.

How do you apply integration by parts to definite integrals?

To apply integration by parts to definite integrals, follow these steps:

1. Use the formula $\int a_{\mathrm{b}}udv=\left[uv\right]a_{\mathrm{b}}-{\int }_{a}vdu$.
2. Find $u$, $\mathrm{dv}$, $\mathrm{du}$, and $v$ as usual.
3. Evaluate $\left[uv\right]$ at the bounds and subtract the integral of $vdu$ over the same bounds.

For example, to evaluate $\int 0_{2}x{e}^{x}dx$, set $u=x$ and $\mathrm{dv}=e^{x}dx$. After applying the formula, evaluate the resulting terms at $x=2$ and $x=0$.

When should you use repeated integration by parts?

Repeated integration by parts is necessary when the integral does not simplify after a single application of the formula. This often occurs when the product involves higher-degree polynomials or functions that require multiple reductions.

For example, consider $\int x^2{e}^{x}dx$. After the first application of integration by parts, the resulting integral still contains a product of $x$ and $e^x$, requiring a second application. Repeat the process until the integral becomes straightforward to solve.

Using the tabular method can streamline repeated integration by parts, as it organizes derivatives and integrals into a table, reducing manual repetition and errors.