Question

Is this computation correct? \(\int_{-2}^{1} \frac{\mathrm{dx}}{2 \mathrm{x}+1}=\left.\frac{1}{2} \ln |2 \mathrm{x}+1|\right|_{-2} ^{1}=\frac{1}{2} \ln 3-\frac{1}{2} \ln 3=0\)

Short answer

Question: Verify if the given computation for the integral \(\int_{-2}^{1} \frac{1}{2x+1} dx = \frac{1}{2} \ln 3 - \frac{1}{2} \ln 3\) is correct. Answer: After following the steps to evaluate the definite integral, we arrived at the same solution \(\frac{1}{2} \ln 3 - \frac{1}{2} \ln 3 = 0\). Therefore, the given computation is correct.

Step-by-step solution

Compute the indefinite integral

To begin, let's compute the indefinite integral of the given function first. The function is \(\frac{1}{2x+1}\). To find the indefinite integral, we will use the formula for integrating functions of the form \(\frac{1}{ax + b}\), which is \(\frac{1}{a} \ln |ax + b| + C\), where \(C\) is the constant of integration: $$\int \frac{\mathrm{dx}}{2x+1} = \frac{1}{2} \ln |2x+1| + C$$ Step 2: Compute the definite integral using the limits of integration

Evaluate the definite integral using the limits

Now we are ready to evaluate the definite integral by applying the limits of integration. We substitute the upper limit (\(1\)) and the lower limit (\(-2\)) into the indefinite integral: $$\left.\frac{1}{2} \ln |2x+1|\right|_{-2}^{1} = \frac{1}{2} \ln |2(1) + 1| - \frac{1}{2} \ln |2(-2) + 1|$$ In the next step, we'll calculate the values inside the absolute values and logarithms. Step 3: Compute the definite integral

Calculate the definite integral

To compute the definite integral, we first perform the arithmetic inside the absolute values and logarithms: $$\frac{1}{2} \ln |3| - \frac{1}{2} \ln |-3| = \frac{1}{2} \ln 3 - \frac{1}{2} \ln 3$$ Now, we have the same expression as the given solution, so we can proceed with simplification. Step 4: Simplify the expression

Simplify the definite integral

We can see that the given solution already performed the last step, which is to subtract one expression from the other: $$\frac{1}{2} \ln 3 - \frac{1}{2} \ln 3 = 0$$ From these calculations, we can confirm that the given computation is correct.

Key concepts

Indefinite Integral

An indefinite integral represents the general form of antiderivatives for a given function—it’s essentially the reverse of taking a derivative. When calculating an indefinite integral, or integrating a function, we're searching for a new function whose derivative would yield the original function. This process is known as 'anti-differentiation' or 'integration'.

For example, consider the indefinite integral \( \int \frac{\mathrm{dx}}{2x+1} \). To solve this, we look for a function whose derivative is \( \frac{1}{2x+1} \). Applying a formula specific for integrating functions of form \( \frac{1}{ax + b} \), we get \( \frac{1}{a} \ln |ax + b| + C \), where \( C \) is the constant of integration. This constant represents the family of all antiderivatives of the function, since the derivative of a constant is zero and does not change the slope of the tangent on the graph of a function.

Integration Limits

When we introduce integration limits to an integral, it becomes a definite integral. The limits define the interval over which we are summing up the area under the curve of the function. A definite integral has both an upper and a lower limit, turning our result into a specific number rather than a range of values.

For the definite integral of \( \frac{1}{2x+1} \) from \( -2 \) to \( 1 \) as seen in our example, \( -2 \) and \( 1 \) are the lower and upper limits of integration, respectively. The integral, then, not only finds the family of functions but pinpoints the exact area under the curve between \( x = -2 \) and \( x = 1 \) for the function given.

Integral Calculus

Integral calculus is a branch of calculus focused on finding the integral of functions. It plays a vital role in mathematics and is central to areas of physics, engineering, and economics, among others.

Understanding integral calculus involves mastering various techniques to find both indefinite and definite integrals. Indefinite integrals provide a general solution (the antiderivative), while definite integrals give precise values that represent areas under curves, accumulation of quantities, and other context-dependent concepts. The process showcased in our exercise, where we transitioned from the indefinite integral to evaluating with the limits, is a prime example of integral calculus at work.

Logarithmic Integration

Logarithmic integration is a technique used for integrating functions that result in a logarithmic form. It's particularly useful when dealing with integrals of the form \( \int \frac{1}{x} dx \), where the antiderivative is \( \ln|x| + C \).

In our example exercise, logarithmic integration comes into play when we evaluate the indefinite integral of \( \frac{1}{2x+1} \) to get \( \frac{1}{2} \ln |2x+1| + C \). This integration results in a natural logarithm function because we're integrating a function that is the derivative of a log function. The properties of logarithms are then utilized in simplifying the computation in definite integrals, as shown when the constant \( \frac{1}{2} \) is factored out and the logarithms of the same base are subtracted from each other.