Evaluate the indefinite integral. ∫ 3 t t 2 + 7 d t \int_{}^{}3t\sqrt{t^2+7}dt ∫ 3 t t 2 + 7 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> d t

( t 2 + 7 ) 3 2 + C \left(t^2+7\right)^{\frac32}+C

( t 2 + 7 ) 2 3 + C \left(t^2+7\right)^{\frac23}+C

Evaluate the indefinite integral. ∫ 1 ( 3 x + 2 ) 5 d x \int_{}^{}\frac{1}{\left(3x+2\right)^5}dx ∫ ( 3 x + 2 ) 5 1 d x

− 1 12 ( 3 x + 2 ) 4 + C -\frac{1}{12\left(3x+2\right)^4}+C

− 1 5 ( 3 x + 2 ) 4 + C -\frac{1}{5\left(3x+2\right)^4}+C

1 12 ( 3 x + 2 ) 4 + C \frac{1}{12\left(3x+2\right)^4}+C

1 5 ( 3 x + 2 ) 4 + C \frac{1}{5\left(3x+2\right)^4}+C

Evaluate the indefinite integral. ∫ 1 2 x + 5 d x \int_{}^{}\frac{1}{2x+5}dx

1 2 ln ⁡ ( 2 x + 5 ) + C \frac12\ln\left(2x+5\right)+C

1 2 ln ⁡ x + C \frac12\ln x+C

1 2 ln ⁡ ( 2 x + 5 ) 2 + C \frac12\ln\left(2x+5\right)^2+C

2 ln ⁡ ( 2 x + 5 ) + C 2\ln\left(2x+5\right)+C

Evaluate the indefinite integral. ∫ x ( 5 + x ) 79 d x \int_{}^{}x\left(5+x\right)^{79}dx ∫ x ( 5 + x ) 79 d x

1 81 ( 5 + x ) 81 − 1 16 ( 5 + x ) 80 + C \frac{1}{81}\left(5+x\right)^{81}-\frac{1}{16}\left(5+x\right)^{80}+C

1 81 x 81 − 1 16 x 80 + C \frac{1}{81}x_{}^{81}-\frac{1}{16}x^{80}+C

1 81 ( 5 + x ) 81 − 1 80 ( 5 + x ) 80 + C \frac{1}{81}\left(5+x\right)^{81}-\frac{1}{80}\left(5+x\right)^{80}+C

1 81 x 81 − 1 80 x 80 + C \frac{1}{81}x^{81}-\frac{1}{80}x^{80}+C

Evaluate the indefinite integral. ∫ t t − 2 d t \int_{}^{}\frac{t}{\sqrt{t-2}}dt ∫ t − 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> t d t

2 3 ( t − 2 ) 3 2 + 4 ( t − 2 ) 1 2 + C \frac23\left(t-2\right)^{\frac32}+4\left(t-2\right)^{\frac12}+C

2 3 t 3 2 + 4 t 1 2 + C \frac23t^{\frac32}+4t^{\frac12}+C

2 3 ( t − 2 ) 2 3 + 2 + C \frac23\left(t-2\right)^{\frac23}+2+C

2 3 ( t + 2 ) 3 2 + 4 ( t + 2 ) 1 2 + C \frac23\left(t+2\right)^{\frac32}+4\left(t+2\right)^{\frac12}+C

Evaluate the indefinite integral. ∫ x ( x − 6 ) 5 d x \int_{}^{}\frac{x}{\left(x-6\right)^5}dx ∫ ( x − 6 ) 5 x d x

− 1 3 ( x − 6 ) 3 − 3 2 ( x − 6 ) 4 + C -\frac{1}{3\left(x-6\right)^3}-\frac{3}{2\left(x-6\right)^4}+C

− x 6 ( x − 6 ) 6 − 3 x 6 ( x − 6 ) 6 + C -\frac{x}{6\left(x-6\right)^6}-\frac{3x}{6\left(x-6\right)^6}+C

− 1 3 ( x − 6 ) 6 − 3 2 ( x − 6 ) 6 + C -\frac{1}{3\left(x-6\right)^6}-\frac{3}{2\left(x-6\right)^6}+C

1 3 ( x − 6 ) 3 + 3 2 ( x − 6 ) 4 + C \frac{1}{3\left(x-6\right)^3}+\frac{3}{2\left(x-6\right)^4}+C

8. Definite Integrals

Substitution for Indefinite Integrals

8. Definite Integrals

Substitution for Indefinite Integrals: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

To evaluate integrals involving composite functions, the substitution method, or u-substitution, is essential. By selecting an appropriate u, typically the inner function, and finding du, the integral can be simplified. For example, in the integral of $x^{2} + 1$ cubed times $2 x$ dx, choosing $u = x^{2} + 1$ simplifies the integral to $u^3$ du. This method streamlines the integration process, allowing for easier evaluation and application of the power rule.

concept

Indefinite Integrals

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Indefinite Integrals Video Summary

In calculus, evaluating integrals involving composite functions can be simplified using a technique known as substitution, or u-substitution. This method is particularly useful when dealing with integrals that contain a function within another function, as well as a derivative of that inner function. The process mirrors the chain rule used in differentiation, allowing us to rewrite complex integrals into simpler forms.

Consider the integral of $ (x^2 + 1)^3 \cdot 2x \, dx $. Here, we identify the inner function as $ u = x^2 + 1 $. To find $ du $, we differentiate $ u $ with respect to $ x $, yielding $ du = 2x \, dx $. This means that the integral can be rewritten in terms of $ u $ as $ \int u^3 \, du $, which is significantly easier to evaluate.

Applying the power rule for integration, we find the antiderivative of $ u^3 $ to be $ \frac{1}{4} u^4 + C $, where $ C $ is the constant of integration. Substituting back for $ u $, we arrive at the final result: $ \frac{1}{4} (x^2 + 1)^4 + C $.

In another example, we evaluate the integral of $ 4\sqrt{4x - 1} \, dx $. Here, we again choose $ u = 4x - 1 $. Differentiating gives $ du = 4 \, dx $, which means $ dx = \frac{du}{4} $. To adjust for the constant, we multiply the integral by $ \frac{1}{4} $ to maintain its value, leading to $ \frac{1}{4} \int \sqrt{u} \, du $.

Recognizing that $ \sqrt{u} = u^{1/2} $, we apply the power rule to find the antiderivative: $ \frac{1}{4} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{6} u^{3/2} + C $. Finally, substituting back for $ u $ gives us the result $ \frac{1}{6} (4x - 1)^{3/2} + C $.

In summary, the substitution method streamlines the process of evaluating integrals involving composite functions. By carefully selecting $ u $ and determining $ du $, we can transform complex integrals into manageable forms, apply integration rules, and revert to the original variable to present our final answer.

example

Indefinite Integrals Example 1

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Indefinite Integrals Example 1 Video Summary

To evaluate the integral of the expression $ (2x^3 + x + 7)^5 (6x^2 + 1) \, dx $ using substitution, we start by identifying a suitable substitution. We choose $ u = 2x^3 + x + 7 $, which is the inner function raised to a power. Next, we compute the derivative $ du $ by differentiating $ u $ with respect to $ x $. The derivative is given by:

\[du = (6x^2 + 1) \, dx\]

This allows us to rewrite the integral in terms of $ u $ and $ du $. Substituting, we have:

\[\int (2x^3 + x + 7)^5 (6x^2 + 1) \, dx = \int u^5 \, du\]

Now, we can integrate $ u^5 $ using the power rule. The integral becomes:

\[\int u^5 \, du = \frac{1}{6} u^6 + C\]

Substituting back for $ u $, we find:

\[\frac{1}{6} (2x^3 + x + 7)^6 + C\]

This is our final result for the integral. To verify our solution, we differentiate the result:

\[\frac{d}{dx} \left( \frac{1}{6} (2x^3 + x + 7)^6 + C \right)\]

Using the chain rule, we differentiate the outer function first, which gives:

\[\frac{1}{6} \cdot 6 (2x^3 + x + 7)^5 \cdot \frac{d}{dx}(2x^3 + x + 7)\]

The derivative of $ 2x^3 + x + 7 $ is $ 6x^2 + 1 $. Thus, we have:

\[(2x^3 + x + 7)^5 (6x^2 + 1)\]

This matches the original integrand, confirming that our integration was performed correctly. Therefore, the evaluated integral is:

\[\frac{1}{6} (2x^3 + x + 7)^6 + C\]

example

Indefinite Integrals Example 2

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Indefinite Integrals Example 2 Video Summary

To evaluate the indefinite integral of $ x^2 e^{x^3} \, dx $, we can use the method of substitution. The first step is to choose a suitable substitution for $ u $. In this case, since $ e $ is raised to the power of $ x^3 $, we can let $ u = x^3 $. Consequently, the differential $ du $ is derived from the derivative of $ u $, which gives us:

$ du = 3x^2 \, dx $

To express the integral in terms of $ u $ and $ du $, we notice that our integral contains $ x^2 \, dx $ but lacks the factor of 3 from $ du $. To address this, we can multiply the integral by 3 and simultaneously divide by 3, effectively canceling it out. This allows us to rewrite the integral as follows:

$ \int x^2 e^{x^3} \, dx = \frac{1}{3} \int 3x^2 e^{x^3} \, dx = \frac{1}{3} \int e^u \, du $

Now, we can integrate with respect to $ u $. The antiderivative of $ e^u $ is simply $ e^u $, so we have:

$ \frac{1}{3} \int e^u \, du = \frac{1}{3} e^u + C $

Finally, we substitute back the original expression for $ u $ to express our answer in terms of $ x $. Since $ u = x^3 $, we find:

$ \frac{1}{3} e^{x^3} + C $

This gives us the final result for the indefinite integral:

$ \int x^2 e^{x^3} \, dx = \frac{1}{3} e^{x^3} + C $

Problem

Evaluate the indefinite integral.
$\int_{}^{}3t\sqrt{t^2+7}dt$

$t^{\frac{3}{2}} + C$

${(t^{2} + 7)}^{\frac{2}{3}} + C$

${(t^{2} + 7)}^{\frac{3}{2}} + C$

$t^{\frac{2}{3}} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{}\frac{1}{\left(3x+2\right)^5}dx$

$- \frac{1}{12 {(3 x + 2)}^{4}} + C$

$- \frac{1}{5 {(3 x + 2)}^{4}} + C$

$\frac{1}{12 {(3 x + 2)}^{4}} + C$

$\frac{1}{5 {(3 x + 2)}^{4}} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{} \frac{1}{2 x + 5} d x$

$\frac{1}{2} \ln x + C$

$\frac{1}{2} \ln {(2 x + 5)}^{2} + C$

$2 \ln (2 x + 5) + C$

$\frac{1}{2} \ln (2 x + 5) + C$

concept

Substitution With an Extra Variable

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Substitution With an Extra Variable Video Summary

In the process of evaluating integrals using substitution, it's essential to recognize that sometimes the integral may contain additional factors that need to be addressed. This summary illustrates how to effectively manage these extra variables during substitution.

Consider the integral of $ x \sqrt{x + 3} \, dx $. The first step is to choose a substitution. Here, we can let $ u = x + 3 $. Consequently, the differential $ du $ is equal to $ dx $ since the derivative of $ x + 3 $ is 1. This substitution simplifies our integral, but we still have the variable $ x $ present.

To express everything in terms of $ u $ and $ du $, we need to eliminate $ x $. Rearranging our substitution gives us $ x = u - 3 $. Substituting this back into the integral transforms it into:

\[\int (u - 3) \sqrt{u} \, du\]

Next, we simplify the integral. The square root of $ u $ can be expressed as $ u^{1/2} $, allowing us to distribute and combine the terms:

\[\int (u^{1} \cdot u^{1/2} - 3u^{1/2}) \, du = \int (u^{3/2} - 3u^{1/2}) \, du\]

Now, we apply the power rule for integration. For $ u^{3/2} $, we add 1 to the exponent, resulting in $ u^{5/2} $, and divide by $ 5/2 $, which is equivalent to multiplying by $ \frac{2}{5} $. For the term $ -3u^{1/2} $, we similarly add 1 to the exponent to get $ u^{3/2} $ and divide by $ 3/2 $, yielding $ -2u^{3/2} $. Thus, the integral becomes:

\[\frac{2}{5} u^{5/2} - 2 u^{3/2} + C\]

Finally, we substitute back $ u = x + 3 $ to express our answer in terms of $ x $:

\[\frac{2}{5} (x + 3)^{5/2} - 2 (x + 3)^{3/2} + C\]

This process highlights the importance of managing additional variables during substitution, ensuring that the integral is fully expressed in terms of the new variable before proceeding with integration. Mastering this technique allows for the evaluation of a wide range of integrals effectively.

Problem

Evaluate the indefinite integral.
$\int_{}^{}x\left(5+x\right)^{79}dx$

$\frac{1}{81} {(5 + x)}^{81} - \frac{1}{16} {(5 + x)}^{80} + C$

$\frac{1}{81} x_{}^{81} - \frac{1}{16} x^{80} + C$

$\frac{1}{81} {(5 + x)}^{81} - \frac{1}{80} {(5 + x)}^{80} + C$

$\frac{1}{81} x^{81} - \frac{1}{80} x^{80} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{}\frac{t}{\sqrt{t-2}}dt$

$\frac{2}{3} t^{\frac{3}{2}} + 4 t^{\frac{1}{2}} + C$

$\frac{2}{3} {(t - 2)}^{\frac{2}{3}} + 2 + C$

$\frac{2}{3} {(t + 2)}^{\frac{3}{2}} + 4 {(t + 2)}^{\frac{1}{2}} + C$

$\frac{2}{3} {(t - 2)}^{\frac{3}{2}} + 4 {(t - 2)}^{\frac{1}{2}} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{}\frac{x}{\left(x-6\right)^5}dx$

$- \frac{x}{6 {(x - 6)}^{6}} - \frac{3 x}{6 {(x - 6)}^{6}} + C$

$- \frac{1}{3 {(x - 6)}^{6}} - \frac{3}{2 {(x - 6)}^{6}} + C$

$- \frac{1}{3 {(x - 6)}^{3}} - \frac{3}{2 {(x - 6)}^{4}} + C$

$\frac{1}{3 {(x - 6)}^{3}} + \frac{3}{2 {(x - 6)}^{4}} + C$

example

Substitution With an Extra Variable Example 3

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Substitution With an Extra Variable Example 3 Video Summary

To determine the total revenue function $ R(x) $ for a coffee truck selling lattes, we start with the marginal revenue function given by $ R'(x) = \frac{2x}{\sqrt{3x^2 - 50}} $. To find $ R(x) $, we need to integrate $ R'(x) $. This requires a substitution where we let $ u = 3x^2 - 50 $. Consequently, the derivative $ du = 6x \, dx $, which allows us to express $ dx $ in terms of $ du $. We adjust our integral accordingly, multiplying by a constant to match the form of $ du $.

The integral becomes:

\[R(x) = \frac{1}{3} \int \frac{1}{\sqrt{u}} \, du\]

Using the power rule for integration, we find:

\[R(x) = \frac{2}{3} \sqrt{3x^2 - 50} + C\]

To determine the constant $ C $, we use the information that selling 15 lattes yields a revenue of $75. Plugging in these values gives:

\[75 = \frac{2}{3} \sqrt{3(15^2) - 50} + C\]

Solving for $ C $ results in $ C \approx 58.33 $. Thus, the total revenue function is:

\[R(x) = \frac{2}{3} \sqrt{3x^2 - 50} + 58.33\]

Next, we need to find how many lattes must be sold to achieve at least $250 in revenue. Setting $ R(x) $ equal to 250, we have:

\[250 = \frac{2}{3} \sqrt{3x^2 - 50} + 58.33\]

Subtracting 58.33 from both sides gives:

\[191.67 = \frac{2}{3} \sqrt{3x^2 - 50}\]

Multiplying both sides by $ \frac{3}{2} $ leads to:

\[287.505 = \sqrt{3x^2 - 50}\]

Squaring both sides eliminates the square root:

\[82,659.125 = 3x^2 - 50\]

Adding 50 to both sides results in:

\[82,709.125 = 3x^2\]

Dividing by 3 gives:

\[27,569.708 = x^2\]

Taking the square root yields:

\[x \approx 166.04\]

Since we cannot sell a fraction of a latte, we round up to the nearest whole number, concluding that the coffee truck must sell at least 167 lattes to achieve a revenue of at least $250.

example

Substitution With an Extra Variable Example 4

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Substitution With an Extra Variable Example 4 Video Summary

To analyze the population growth of a species of fish in a lake, we start with the given population growth rate, represented as the derivative of the population function, $ p'(t) = 350 e^{\frac{t}{3}} $, where $ t $ is the time in years. Our goal is to find the population function $ p(t) $ given that there are 5,000 fish at $ t = 0 $.

To derive $ p(t) $, we need to integrate the growth rate. The integral we need to solve is:

\[p(t) = \int 350 e^{\frac{t}{3}} \, dt\]

To simplify the integration, we perform a substitution. Let $ u = \frac{t}{3} $, which implies $ du = \frac{1}{3} dt $ or $ dt = 3 du $. Substituting these into the integral gives:

\[p(t) = 350 \int e^{u} \cdot 3 \, du = 1050 \int e^{u} \, du\]

The antiderivative of $ e^{u} $ is $ e^{u} $, so we have:

\[p(t) = 1050 e^{u} + C\]

Substituting back for $ u $, we find:

\[p(t) = 1050 e^{\frac{t}{3}} + C\]

To determine the constant $ C $, we use the initial condition $ p(0) = 5000 $. Plugging in $ t = 0 $:

\[5000 = 1050 e^{0} + C \implies 5000 = 1050 + C \implies C = 3950\]

Thus, the population function is:

\[p(t) = 1050 e^{\frac{t}{3}} + 3950\]

Next, we want to find the population after 6 years, $ p(6) $. We substitute $ t = 6 $ into our population function:

\[p(6) = 1050 e^{\frac{6}{3}} + 3950 = 1050 e^{2} + 3950\]

Calculating $ e^{2} $ and then substituting gives:

\[p(6) \approx 1050 \cdot 7.389 + 3950 \approx 7758.45 + 3950 \approx 11708.45\]

Rounding this to the nearest whole number, we find that the population of fish after 6 years is approximately:

\[\boxed{11709}\]

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Here’s what students ask on this topic:

What is the substitution method for indefinite integrals?

The substitution method for indefinite integrals is a technique used to simplify the integration of composite functions. It involves choosing a new variable, $u$ , which is typically the inner function of the composite function. The differential, $d u$ , is then determined. This substitution transforms the integral into a simpler form, often allowing the use of the power rule to find the antiderivative. After integrating, the result is expressed back in terms of the original variable. This method is particularly useful for integrals that resemble the derivative of a composite function, making it easier to evaluate complex integrals efficiently.

How do you choose $u$ in substitution for integrals?

Choosing $u$ in substitution for integrals involves identifying the inner function of a composite function. Typically, $u$ is selected as the expression inside parentheses, under a radical, or in an exponent. The goal is to simplify the integral by transforming it into a function of $u$ and $d u$ . Once $u$ is chosen, its derivative is calculated to find $d u$ . This choice should make the integral easier to evaluate, often allowing the use of the power rule. It's important to ensure that the substitution accounts for all parts of the original integral, including any constants or extra variables.

What is the role of the power rule in substitution for indefinite integrals?

The power rule plays a crucial role in substitution for indefinite integrals by providing a straightforward method to find antiderivatives. Once the integral is rewritten in terms of $u$ and $d u$ , the power rule can be applied to integrate functions of the form $u^{n}$ . The power rule states that the antiderivative of $u^{n}$ is $\frac{u}{n+1}$ divided by $n+1$ , plus a constant of integration. This simplifies the integration process, allowing for efficient evaluation of complex integrals. After applying the power rule, the result is converted back to the original variable to complete the integration.

How do you handle constants in substitution for indefinite integrals?

Handling constants in substitution for indefinite integrals involves adjusting the integral to match the differential $d u$ . If $d u$ is off by a constant factor, you can multiply the integral by that constant and its reciprocal to maintain the integral's value. For example, if $d u$ is $4 d x$ but the integral has $d x$ , multiply by $4$ and $\frac{1}{4}$ . This ensures the substitution works without altering the integral's value. This technique is essential for correctly applying substitution and simplifying the integration process.

What are the steps to perform substitution in indefinite integrals?

The steps to perform substitution in indefinite integrals are as follows: 1) Choose $u$ , the inner function of the composite function, and find $d u$ by differentiating $u$ . 2) Rewrite the integral in terms of $u$ and $d u$ . If necessary, adjust for constants by multiplying by the constant and its reciprocal. 3) Integrate the new expression using the power rule or other appropriate methods. 4) Substitute back the original variable to express the final result in terms of the original variable. These steps simplify the integration of complex functions, making it more manageable.