Question

Find the integral by using the appropriate formula. $$ \int x^{3} e^{2 x} d x $$

Short answer

The definite integral of the given function is \(\frac{1}{2}x^{3}e^{2x} - \frac{3}{4}x^{2}e^{2x} + \frac{15}{8}xe^{2x} - \frac{15}{16}e^{2x} + C\).

Step-by-step solution

Set up Integration by Parts

To start, you will need to apply the integration by parts formula, which is \(\int u dv = u v - \int v du\). The trickiest part is usually determining what parts of your integral to assign as \(u\) and \(dv\). Here, select \(u = x^3\) and \(dv = e^{2x} dx\). Now, calculate \(du\) and \(v\). So, \(du = 3x^2 dx\) and \(v = \frac{1}{2}e^{2x}\).

Apply Integration by Parts Formula

With \(u\), \(v\), \(du\) and \(dv\) known, you can apply the integration by parts formula. You get: \(\int x^{3} e^{2 x} dx = x^{3} * \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x} * 3x^{2} dx = \frac{1}{2}x^{3}e^{2x} - \frac{3}{2}\int x^{2} e^{2x} dx\)

Apply Integration by Parts Again

You will notice that the integral \(\int x^{2} e^{2x} dx\) obtained in step 2, is still a product of two functions and can again be solved by repeating the method of integration by parts. Now, take the new parts as \(u = x^2\) and \(dv = e^{2x} dx\). Therefore, \(du = 2x dx\) and \(v = \frac{1}{2}e^{2x}\). Apply the integration by parts formula again to the new integral.

Simplify and Repeat Until Solved

Continue to apply the integration by parts formula until the integral is a form one can integrate directly. Simplify the expression at each step until you reach the final result: \(\frac{1}{2}x^{3}e^{2x} - \frac{3}{2}\int x^{2} e^{2x} dx = \frac{1}{2}x^{3}e^{2x} - \frac{3}{2}(x^2 * \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x} * 2x dx) = \frac{1}{2}x^{3}e^{2x} - \frac{3}{4}x^{2}e^{2x} + \frac{3}{4}\int xe^{2x} dx\). Again, apply the same steps to solve integral \(\int xe^{2x} dx\). Continue the procedure until all integrals are solved. Then combine the results and simplify to obtain the final solution.

Add Constant of Integration

The antiderivative of a function is determined up to a constant of integration. Therefore, after solving the integral, always add the constant of integration \(C\) to the result.