a. The value of \displaystyle \int_{-2}^{-1} \frac{14}{ 4 x } dx is

b. The value of \displaystyle \int_{1}^{2} \frac{14}{ 4 x } dx is

a. \displaystyle \int_{-2}^{-1} \frac{14}{ 4 x } dx = \frac{7}{2} \ln|-2| - \frac{7}{2} \ln|-1| = \frac{7}{2} \ln(2) - \frac{7}{2} \ln(1) = \frac{7}{2} \ln(2)

b. \displaystyle \int_{1}^{2} \frac{14}{ 4 x } dx = \frac{7}{2} \ln|2| - \frac{7}{2} \ln|1| = \frac{7}{2} \ln(2) - \frac{7}{2} \ln(1) = \frac{7}{2} \ln(2)

To find the value of the integral \(\displaystyle \int_{-2}^{-1} \frac{14}{4x} \, dx\), we can follow these steps:

Step 1: Simplify the integrand
Since \(\frac{14}{4x} = \frac{7}{2x}\), the integral becomes \(\displaystyle \int_{-2}^{-1} \frac{7}{2x} \, dx\).

Step 2: Find the antiderivative
To find the antiderivative of \(\frac{7}{2x}\), we can use the logarithmic property of integrals. Recall that \(\int \frac{1}{x} \, dx = \ln|x|\) (natural logarithm of the absolute value of \(x\)). Therefore, the antiderivative of \(\frac{7}{2x}\) is \(\frac{7}{2} \ln|x|\).

Step 3: Evaluate the antiderivative at the limits of integration
To find the value of the integral between the given limits of -2 and -1, we can evaluate the antiderivative at these points. Plug in -2 and -1 into the antiderivative expression:

\(\frac{7}{2} \ln|-1| - \frac{7}{2} \ln|-2|\)
\((-2\) and \(-1\) will become positive inside the absolute values.)

Step 4: Compute the final value
\(\frac{7}{2} \ln(1) - \frac{7}{2} \ln(2)\)
Since \(\ln(1) = 0\), the first term becomes zero. Therefore, the value of the integral is \(-\frac{7}{2} \ln(2)\).

Now let's move on to the second question.

To find the value of the integral \(\displaystyle \int_{1}^{2} \frac{14}{4x} \, dx\), the steps are similar:

Step 1: Simplify the integrand
Since \(\frac{14}{4x} = \frac{7}{2x}\), the integral becomes \(\displaystyle \int_{1}^{2} \frac{7}{2x} \, dx\).

Step 2: Find the antiderivative
The antiderivative of \(\frac{7}{2x}\) is still \(\frac{7}{2} \ln|x|\).

Step 3: Evaluate the antiderivative at the limits of integration
Plug in 1 and 2 into the antiderivative expression:

\(\frac{7}{2} \ln|2| - \frac{7}{2} \ln|1|\)
Since \(\ln|1| = 0\), the second term becomes zero.

Step 4: Compute the final value
\(\frac{7}{2} \ln(2)\)

Therefore, the value of the integral is \(\frac{7}{2} \ln(2)\).

a. To find the value of \(\displaystyle \int_{-2}^{-1} \frac{14}{4x} dx\), we can use the property of definite integrals that states \(\displaystyle \int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx\).

Rewriting the integral, we have \(\displaystyle \int_{-2}^{-1} \frac{14}{4x} dx = -\int_{-1}^{-2} \frac{14}{4x} dx\).

Next, we can simplify the fraction by factoring out a constant: \(\displaystyle \frac{14}{4x} = \frac{7}{2x}\).

Thus, the integral becomes \(\displaystyle -\int_{-1}^{-2} \frac{7}{2x} dx\).

To evaluate this integral, we can use the fact that \(\displaystyle \int \frac{1}{x} dx = \ln|x|\).

Applying this, the integral becomes \(\displaystyle -\int_{-1}^{-2} \frac{7}{2} \cdot \frac{1}{x} dx = -\frac{7}{2} \ln|x| \Big|_{-1}^{-2}\).

Now, substituting the limits, we have \(\displaystyle -\frac{7}{2} \ln|-1| - (-\frac{7}{2} \ln|-2|)\).

Since the natural logarithm of a negative number is undefined, \(\displaystyle \ln|-1|\) does not exist. However, \(\displaystyle \ln|-2|\) does exist.

Therefore, the value of \(\displaystyle \int_{-2}^{-1} \frac{14}{4x} dx\) is undefined.

b. Similarly, to find the value of \(\displaystyle \int_{1}^{2} \frac{14}{4x} dx\), we can use the property of definite integrals to rewrite the integral as \(\displaystyle \int_{2}^{1} \frac{14}{4x} dx\).

Simplifying the fraction by factoring out a constant, we have \(\displaystyle \int_{2}^{1} \frac{7}{2x} dx\).

Using the fact that \(\displaystyle \int \frac{1}{x} dx = \ln|x|\), the integral becomes \(\displaystyle \int_{2}^{1} \frac{7}{2} \cdot \frac{1}{x} dx = \frac{7}{2} \ln|x| \Big|_{2}^{1}\).

Now substituting the limits, we have \(\displaystyle \frac{7}{2} \ln|1| - \frac{7}{2} \ln|2|\).

Since \(\displaystyle \ln|1| = 0\), the value of the integral is \(\displaystyle -\frac{7}{2} \ln|2|\).