Problem 62
Question
In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{2}^{5} \frac{d x}{2 x-3}$$
Step-by-Step Solution
Verified Answer
The solution to the integral \(\int_{2}^{5} \frac{d x}{2 x-3}\) is \(\frac{1}{2} ln7\).
1Step 1: Define the Substitution
Choosing \(u = 2x-3\) will simplify our integral greatly. Calculate the derivative in terms of \(x\), resulting in \(du = 2dx\).
2Step 2: Adjust the Integral
Express the original integral in terms of \(u\). Replace \(2x - 3\) with \(u\), and replace \(dx\) with \(du/2\). This results in \((1/2)\int_{u(2)}^{u(5)} du/u\).
3Step 3: Adjust the limits of integration
The original limits were for \(x\), not \(u\). Therefore, substitute these \(x\) values into \(u = 2x - 3\) to get new limits. This results in \((u(2) = 2*2 - 3 = 1)\) and \((u(5) = 2*5 - 3 = 7)\). Thus, our integral becomes \((1/2)\int_{1}^{7} du/u\).
4Step 4: Integrate
Now we can carry out the integration. The integral of \(1/u\) with respect to \(u\) is \(ln|u|\). So, \((1/2)\int_{1}^{7} du/u = (1/2)[ln|7| - ln|1|]= \frac{1}{2} ln7\).
Key Concepts
Integration TechniquesDefinite IntegralsLimits of Integration
Integration Techniques
Mastering various integration techniques is essential in calculus, as they allow us to solve a variety of integrals. One of the most powerful methods is u-substitution, which simplifies the integral by changing the variable of integration. This technique is particularly useful when dealing with integrals where the integrand is a product of a function and its derivative, or close to it.
The process involves choosing a substitution that will make the integral easier to handle. In our example, setting u to 2x - 3 transformed a function with a variable in the denominator to one where the variable now stands alone. This can make an otherwise complex integral more straightforward, as seen by the simplified integral \(\frac{1}{2}\int du/u\).
Once the substitution is made, we replace all instances of the original variable, including differentials. For instance, if \(du = 2dx\), then \(dx = du/2\), as we saw in the exercise. Always ensure that every part of the original integral is accounted for in the substitution to avoid errors and maintain the integrity of the integral's value.
The process involves choosing a substitution that will make the integral easier to handle. In our example, setting u to 2x - 3 transformed a function with a variable in the denominator to one where the variable now stands alone. This can make an otherwise complex integral more straightforward, as seen by the simplified integral \(\frac{1}{2}\int du/u\).
Once the substitution is made, we replace all instances of the original variable, including differentials. For instance, if \(du = 2dx\), then \(dx = du/2\), as we saw in the exercise. Always ensure that every part of the original integral is accounted for in the substitution to avoid errors and maintain the integrity of the integral's value.
Definite Integrals
A definite integral is an integral with specific limits of integration, representing the area under the curve of a function on a given interval. It's a fundamental concept because it not only provides an exact value but also has practical applications in physics, engineering, and other fields.
Definite integrals are designated with lower and upper limits on the integral sign, such as \(\int_{a}^{b} f(x) dx\), and require evaluating the antiderivative at these limits. After performing integration, the result is found by subtracting the value of the antiderivative at the lower limit from the value at the upper limit.
In the provided exercise, we see the transition from an integral with variables as limits to one with numerical limits after performing u-substitution. After integrating \(\frac{1}{u}\), we apply the limits from \(1\) to \(7\) to \(ln|u|\) to obtain the result. The concept of a definite integral plays a crucial role here, ensuring that the result reflects the specific interval from the original problem. In our exercise, the value \(\frac{1}{2} ln7\) represents the area under the transformed function between the new limits after substitution.
Definite integrals are designated with lower and upper limits on the integral sign, such as \(\int_{a}^{b} f(x) dx\), and require evaluating the antiderivative at these limits. After performing integration, the result is found by subtracting the value of the antiderivative at the lower limit from the value at the upper limit.
In the provided exercise, we see the transition from an integral with variables as limits to one with numerical limits after performing u-substitution. After integrating \(\frac{1}{u}\), we apply the limits from \(1\) to \(7\) to \(ln|u|\) to obtain the result. The concept of a definite integral plays a crucial role here, ensuring that the result reflects the specific interval from the original problem. In our exercise, the value \(\frac{1}{2} ln7\) represents the area under the transformed function between the new limits after substitution.
Limits of Integration
Limits of integration define the interval over which we evaluate a definite integral. Correctly adjusting these limits is essential when performing a variable substitution, as the original interval represented x values, not u values.
When the substitution is made, such as \(u = 2x - 3\), we must calculate the corresponding u values for the original limits. This is crucial because evaluating the integral with respect to u requires limits that apply to u.
In our previous exercise, we began with the limits \(x = 2\) to \(x = 5\), which became \(u = 1\) to \(u = 7\) after plugging the x values into the substitution equation. Skipping or incorrectly adjusting these limits will lead to incorrect answers since the area calculated must match the interval laid out by the original function, as seen in the final integral \(\frac{1}{2}\int_{1}^{7} \frac{du}{u}\). The process of converting limits is a crucial step in using the u-substitution method for definite integrals.
When the substitution is made, such as \(u = 2x - 3\), we must calculate the corresponding u values for the original limits. This is crucial because evaluating the integral with respect to u requires limits that apply to u.
In our previous exercise, we began with the limits \(x = 2\) to \(x = 5\), which became \(u = 1\) to \(u = 7\) after plugging the x values into the substitution equation. Skipping or incorrectly adjusting these limits will lead to incorrect answers since the area calculated must match the interval laid out by the original function, as seen in the final integral \(\frac{1}{2}\int_{1}^{7} \frac{du}{u}\). The process of converting limits is a crucial step in using the u-substitution method for definite integrals.
Other exercises in this chapter
Problem 61
In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{0}^{7} \frac{d x}{x+2}$$
View solution Problem 61
Multiple Choice A slope field for the differential equation \(d y / d x=42-y\) will show (A) a line with slope \(-1\) and \(y\) -intercept 42 . (B) a vertical a
View solution Problem 62
Multiple Choice For which of the following differential equa- tions will a slope field show nothing but negative slopes in the fourth quadrant? (A) \(\frac{d y}
View solution Problem 63
In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{1}^{2} \frac{d t}{t-3}$$
View solution