Problem 38
Question
Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{2}^{10} \sqrt{y-1} d y $$
Step-by-Step Solution
Verified Answer
The value of the integral is \( \frac{52}{3} \).
1Step 1: Choose a Substitution
For the integral \( \int_{2}^{10} \sqrt{y-1} \, dy \), we choose the substitution \( u = y - 1 \). This substitution simplifies the expression under the square root to \( \sqrt{u} \).
2Step 2: Differentiate and Express dy in terms of du
Differentiate the substitution \( u = y - 1 \) to find \( du \). We get \( du = dy \). Therefore, \( dy = du \).
3Step 3: Change the Limits of Integration
Substitute the limits using \( u = y - 1 \):- When \( y = 2 \), \( u = 2 - 1 = 1 \).- When \( y = 10 \), \( u = 10 - 1 = 9 \).The new limits of integration are from \( u = 1 \) to \( u = 9 \).
4Step 4: Rewrite the Integral in Terms of u
Substituting into the integral, we get:\[ \int_{1}^{9} \sqrt{u} \, du \]
5Step 5: Integrate with Respect to u
To integrate \( \sqrt{u} = u^{1/2} \), apply the power rule:\[ \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} + C \]
6Step 6: Evaluate the Definite Integral
Now, evaluate \( \frac{2}{3} u^{3/2} \) from \( u = 1 \) to \( u = 9 \):\[ \left. \frac{2}{3} u^{3/2} \right|_1^9 = \frac{2}{3} (9^{3/2}) - \frac{2}{3} (1^{3/2}) \]Calculating further:\[ 9^{3/2} = (\sqrt{9})^3 = 3^3 = 27 \]Thus:\[ \frac{2}{3} \times 27 - \frac{2}{3} \times 1 = 18 - \frac{2}{3} \]\[ = \frac{54}{3} - \frac{2}{3} = \frac{52}{3} \].
Key Concepts
CalculusDefinite IntegralsIntegration Techniques
Calculus
Calculus is a branch of mathematics that deals with change and motion. It is primarily concerned with understanding how quantities vary and understanding the behavior of functions. This branch of mathematics includes two main parts: differential calculus and integral calculus. Differential calculus focuses on the rate at which quantities change, while integral calculus involves accumulation of quantities and areas under curves.
In the context of definite integrals, like the integral \( \int_{2}^{10} \sqrt{y-1} \ dy \), calculus allows us to calculate the area under the curve represented by the function \(\sqrt{y-1}\) between the limits \(y=2\) and \(y=10\). The concept of substitution helps simplify complicated functions to make them easier to integrate, as seen in the solution provided.
In the context of definite integrals, like the integral \( \int_{2}^{10} \sqrt{y-1} \ dy \), calculus allows us to calculate the area under the curve represented by the function \(\sqrt{y-1}\) between the limits \(y=2\) and \(y=10\). The concept of substitution helps simplify complicated functions to make them easier to integrate, as seen in the solution provided.
Definite Integrals
Definite integrals form the backbone of integral calculus, providing a method to compute the area under a curve over a specific interval. They are termed "definite" because they result in a number, unlike indefinite integrals that result in a function.
In the original exercise, \( \int_{2}^{10} \sqrt{y-1} \ dy \) represents a definite integral with limits 2 and 10. The definite integral is evaluated using the Fundamental Theorem of Calculus, which connects differentiation and integration, ensuring the calculation is precise. When evaluating, we often substitute variables to simplify the integrand, changing both the function and limits accordingly. This results in a streamlined problem, converting complex integrals into a more manageable form, as demonstrated when substituting \( u = y - 1 \). This approach simplifies the integral into a basic form \( \int_{1}^{9} \sqrt{u} \ du \).
Furthermore, by substituting the limits of integration to match the new variable, we ensure the area calculated matches that of the original problem, maintaining the integrity of the definite integral.
In the original exercise, \( \int_{2}^{10} \sqrt{y-1} \ dy \) represents a definite integral with limits 2 and 10. The definite integral is evaluated using the Fundamental Theorem of Calculus, which connects differentiation and integration, ensuring the calculation is precise. When evaluating, we often substitute variables to simplify the integrand, changing both the function and limits accordingly. This results in a streamlined problem, converting complex integrals into a more manageable form, as demonstrated when substituting \( u = y - 1 \). This approach simplifies the integral into a basic form \( \int_{1}^{9} \sqrt{u} \ du \).
Furthermore, by substituting the limits of integration to match the new variable, we ensure the area calculated matches that of the original problem, maintaining the integrity of the definite integral.
Integration Techniques
Integration techniques are essential in solving definite integrals, providing methods to find the antiderivative of functions. One widely used technique is the substitution rule, ideal for integrands that can be simplified by changing variables.
In our exercise, the substitution \( u = y - 1 \) was used. This changed the function \( \sqrt{y-1} \) into \( \sqrt{u} \), and correspondingly, the differentials transformed as \( dy = du \). This technique is powerful because it can transform a complex-looking integral into a basic form that is much easier to evaluate.
Once in the simpler form, the integral \( \int_{1}^{9} \sqrt{u} \ du \) is solved using the power rule for integration. This rule states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) for any real number \( n e -1 \). With this substitution, the integral becomes \( \frac{2}{3} u^{3/2} \), evaluated between the limits 1 and 9, illustrating the definite nature of the problem.
Substitution and power rule are fundamental techniques that make solving integrals both precise and efficient in calculus.
In our exercise, the substitution \( u = y - 1 \) was used. This changed the function \( \sqrt{y-1} \) into \( \sqrt{u} \), and correspondingly, the differentials transformed as \( dy = du \). This technique is powerful because it can transform a complex-looking integral into a basic form that is much easier to evaluate.
Once in the simpler form, the integral \( \int_{1}^{9} \sqrt{u} \ du \) is solved using the power rule for integration. This rule states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) for any real number \( n e -1 \). With this substitution, the integral becomes \( \frac{2}{3} u^{3/2} \), evaluated between the limits 1 and 9, illustrating the definite nature of the problem.
Substitution and power rule are fundamental techniques that make solving integrals both precise and efficient in calculus.
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