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Magazine Chapter 4: Problem 33
$$ \int 3 t \sqrt[3]{2 t^{2}-11} d t $$
Short Answer
Expert verified
The integral is \( \frac{9}{16} (2t^{2} - 11)^{4/3} + C \).
Step by step solution
01
Identify a Suitable Substitution
To make this integral easier to solve, let's use substitution. We identify the expression inside the cube root sign as a potential variable substitution. Let: \[ u = 2t^{2} - 11 \] Then, compute the derivative of \( u \) with respect to \( t \): \[ \frac{du}{dt} = 4t \] So, \( du = 4t \, dt \) and we can solve for \( dt \): \[ dt = \frac{du}{4t} \]
02
Apply the Substitution
Substitute the variable \( u \) into the original integral. The integral becomes: \[ \int 3t \sqrt[3]{u} \cdot \frac{du}{4t} \] This simplifies to: \[ \int \frac{3}{4} \sqrt[3]{u} \, du \]
03
Integrate with Respect to the New Variable
Now, we integrate \( \frac{3}{4} \sqrt[3]{u} \) with respect to \( u \): Using the power rule for integration, where \( \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \), we notice that \( \sqrt[3]{u} = u^{1/3} \). Thus: \[ \int u^{1/3} \, du = \frac{u^{4/3}}{4/3} + C = \frac{3}{4}u^{4/3} + C \] Hence, the integral changes to: \[ \frac{3}{4} \times \frac{3}{4} u^{4/3} + C = \frac{9}{16} u^{4/3} + C \]
04
Back-Substitute the Original Variables
Finally, let's revert back to the variable \( t \) by substituting \( u = 2t^{2} - 11 \) back into the integrated expression: \[ \frac{9}{16} (2t^{2} - 11)^{4/3} + C \] This yields the antiderivative in terms of \( t \).
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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 in integration is a powerful technique used to simplify problems that might otherwise be difficult to evaluate directly. The idea is to transform the given integral into a simpler form by substituting a new variable in place of an existing one.
- **Identify the Substitution**: Start by spotting expressions in the integral that appear complex, particularly ones inside other functions like square roots or cube roots.- **Choose a New Variable**: Assign a new variable, generally denoted as \( u \), to represent the complex part. For our given problem with \( \int 3t \sqrt[3]{2t^2 - 11} \, dt \), we set \( u = 2t^2 - 11 \).- **Differentiate**: Compute the derivative of the new variable with respect to the original variable, \( t \). This leads to \( \frac{du}{dt} = 4t \), and enables calculating \( dt \) in terms of \( du \).- **Rewrite the Integral**: Replace both the expression and \( dt \) in the original integral with their substitutions, transforming it into an integral in terms of \( u \).By using the substitution method, we can simplify an intricate integral, making it manageable and easier to solve.
- **Identify the Substitution**: Start by spotting expressions in the integral that appear complex, particularly ones inside other functions like square roots or cube roots.- **Choose a New Variable**: Assign a new variable, generally denoted as \( u \), to represent the complex part. For our given problem with \( \int 3t \sqrt[3]{2t^2 - 11} \, dt \), we set \( u = 2t^2 - 11 \).- **Differentiate**: Compute the derivative of the new variable with respect to the original variable, \( t \). This leads to \( \frac{du}{dt} = 4t \), and enables calculating \( dt \) in terms of \( du \).- **Rewrite the Integral**: Replace both the expression and \( dt \) in the original integral with their substitutions, transforming it into an integral in terms of \( u \).By using the substitution method, we can simplify an intricate integral, making it manageable and easier to solve.
Definite and Indefinite Integrals
Indefinite integrals and definite integrals represent two main types of integrals in calculus. They are used to find antiderivatives and calculate areas, respectively.
**Indefinite Integrals**- **Lack Limits**: Unlike definite integrals, indefinite integrals do not have upper and lower limits.- **Result in Functions**: Solving an indefinite integral results in a function plus a constant, denoted by \( C \), representing an infinite number of antiderivatives. For instance, solving \( \int f(t) \, dt = F(t) + C \).**Definite Integrals**- **Include Limits**: Feature specific upper and lower boundaries a and b, respectively.- **Result in Numbers**: Compute the integral over a certain interval and result in a numerical value, often representing areas under a curve.In our context, solving the indefinite integral from the exercise without set limits results in an expression for the antiderivative. When resolved, it gives \( \frac{9}{16} (2t^2 - 11)^{4/3} + C \). Understanding these types helps in selecting proper techniques for specific problems.
**Indefinite Integrals**- **Lack Limits**: Unlike definite integrals, indefinite integrals do not have upper and lower limits.- **Result in Functions**: Solving an indefinite integral results in a function plus a constant, denoted by \( C \), representing an infinite number of antiderivatives. For instance, solving \( \int f(t) \, dt = F(t) + C \).**Definite Integrals**- **Include Limits**: Feature specific upper and lower boundaries a and b, respectively.- **Result in Numbers**: Compute the integral over a certain interval and result in a numerical value, often representing areas under a curve.In our context, solving the indefinite integral from the exercise without set limits results in an expression for the antiderivative. When resolved, it gives \( \frac{9}{16} (2t^2 - 11)^{4/3} + C \). Understanding these types helps in selecting proper techniques for specific problems.
Integration by Parts
Integration by parts is another crucial technique in calculus, useful for tackling integral cases involving products of functions. It is particularly applicable when substitution alone does not simplify the integral enough.
**The Formula**: Derived from the product rule for differentiation, \( \int u \, dv = uv - \int v \, du \), where you identify parts of the integrand that correspond to \( u \) and \( dv \).**Choose Wisely**- Identify components that when differentiated, simplify the integral. Choose \( u \) to be the function that becomes simpler when differentiated.- Use \( dv \) for the remaining part of the function, requiring integration.Even though integration by parts was not needed for the original problem, it is pivotal for tackling integrals involving the product of polynomial and exponential, logarithmic, or trigonometric functions. Practicing these methods lets you decide on the right strategy depending on the integral's structure.
**The Formula**: Derived from the product rule for differentiation, \( \int u \, dv = uv - \int v \, du \), where you identify parts of the integrand that correspond to \( u \) and \( dv \).**Choose Wisely**- Identify components that when differentiated, simplify the integral. Choose \( u \) to be the function that becomes simpler when differentiated.- Use \( dv \) for the remaining part of the function, requiring integration.Even though integration by parts was not needed for the original problem, it is pivotal for tackling integrals involving the product of polynomial and exponential, logarithmic, or trigonometric functions. Practicing these methods lets you decide on the right strategy depending on the integral's structure.
