Question

In Exercises \(25-28\) , evaluate the integral analytically. Support your answer using NINT. $$\int_{-3}^{2} e^{-2 x} \sin 2 x d x$$

Short answer

The value of the integral \(\int_{-3}^{2} e^{-2x} \sin 2x dx\) is given by \(\frac{1}{3} ( -1/4 \cos{4} + 1/4 \sin{4}) - \frac{1}{3} ( -e^6 \cos{-6} + e^6 \sin{-6})\).

Step-by-step solution

Set Up the Integral

To begin, recognize that this is a definite integral from -3 to 2 of the function \(e^{-2x} \sin 2x\). To integrate this, it will be useful to use the method of integration by parts, which states that the integral of \(u dv = u v - \int v du \). Now we need to identify the two parts of the function to apply this method. Let's select \(u = e^{-2x}\) and \(dv = \sin 2x\). From these choices, we differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\). Doing these we get \(du = -2e^{-2x} dx\) and \(v = -\frac{1}{2} \cos{2x}\).

Apply Integration by Parts

Now we will substitute our parts into the integration by parts formula \(\int udv = uv - \int v du \). This gives the integral as \( e^{-2x}(-\frac{1}{2}\cos{2x}) - \int_{-3}^{2}-\frac{1}{2}\cos{2x}(-2e^{-2x} dx)\). Simplifying the integral and factoring out constants where possible, we get, \(-\frac{1}{2} e^{-2x} \cos{2x} - \int_{-3}^{2} e^{-2x} \cos{2x} dx\). Now, ironically, this integral is also a product of two functions. So, we have to apply integration by parts again, this time choosing \(u = e^{-2x}\) and \(dv = \cos{2x} dx\), when we differentiate \(u\) and integrate \(dv\) we get \(du = -2e^{-2x} dx\) and \(v = \frac{1}{2} \sin{2x}\). We apply integration by parts on the integral we got above.

Evaluate the Resulting Integral

Substitute the new parts into the formula. This gives us the integral as \( -\frac{1}{2} e^{-2x} \sin{2x} - \int_{-3}^{2} -2e^{-2x} \sin{2x} dx\). Combined with the expression we got from the previous step we get, \( -\frac{1}{2} e^{-2x} \cos{2x} + \frac{1}{2} e^{-2x} \sin{2x} + 2 \int_{-3}^{2} e^{-2x} \sin{2x} dx\). Now, we notice that the integral we just evaluated is similar to the integral we started with. This generally happens when we apply integration by parts for function containing product of transcendental functions like exponential, trigonometric or logarithm functions. To solve for the original integral, we combine like terms and solve for the integral, call the original integral as \(I\) and moving terms we get, \(I = -\frac{1}{2} e^{-2x} \cos{2x} + \frac{1}{2} e^{-2x} \sin{2x} + 2I\), solving for \(I\) we get, \(I = \frac{1}{3} ( -e^{-2x} \cos{2x} + e^{-2x} \sin{2x})\)

Evaluate the Definite Integral

Now that we have a function for \(I\), we can substitute the limits of the integral, which are -3 and 2. So \(I = \frac{1}{3} ( -e^{-2*2} \cos{2*2} + e^{-2*2} \sin{2*2}) - \frac{1}{3} ( -e^{-2*-3} \cos{2*-3} + e^{-2*-3} \sin{2*-3}), which simplifies to \(\frac{1}{3} ( -1/4 \cos{4} + 1/4 \sin{4}) - \frac{1}{3} ( -e^6 \cos{-6} + e^6 \sin{-6})\).