Question

Use the Exponential Rule to find the indefinite integral. $$ \int 3(x-4) e^{x^{2}-8 x} d x $$

Short answer

The indefinite integral is \( [3(x - 4)e^{x^{2} - 8x}] - \int 3e^{x^{2} - 8x}dx\)

Step-by-step solution

Re-write the integral

The integral can be rewritten as a simple multiplication of two functions, \(f(x) = 3(x - 4)\) and \(g(x) = e^{x^2 - 8x}\). So, \(\int f(x)g'(x)dx\).

Identify the derivative

Comparing \(\int f(x)g'(x)dx\) with the rule for the integration by parts: \(\int udv = uv - \int vdu\), we can identify each function and its derivative. Considering \(u = f(x) = 3(x - 4)\) and \(dv = g'(x)dx = e^{x^{2}-8x}dx\), we can find the derivative and the integral: \(du = f'(x)dx = 3dx\) and \(v = g(x) = \int e^{x^{2} - 8x}dx = e^{x^{2} - 8x}\).

Apply integration by parts formula

Now we substitute these into the formula for integration by parts \(\int udv = uv - \int vdu \), we get: \(\int 3(x - 4)e^{x^{2} - 8x} dx = [3(x - 4)e^{x^{2} - 8x}] - \int 3e^{x^{2} - 8x}dx\).

Key concepts

Exponential Rule

The Exponential Rule is a cornerstone of calculus, particularly when working with integrals involving exponential functions. Essentially, it guides us on how to handle integration when the exponent itself is a function of the variable of integration.

For an exponential function of the form \( e^{f(x)} \), where \( f(x) \) is a function of \( x \), the integral is generally not straightforward. However, if the function \( f(x) \) has a derivative that is present in the integrand, it becomes possible to solve the integral using substitution or integration by parts, depending on the complexity of \( f(x) \).

In our exercise, the presence of \( e^{x^2 - 8x} \) suggests that applying the Exponential Rule might not be direct because the derivative of \( x^2 - 8x \) is not exactly present in the integral. This is where we take help from the method of integration by parts to simplify the problem. This leads us nicely into exploring integration by parts as a method to tackle such problems.

Integration by Parts

Integration by parts is a technique derived from the product rule for derivatives and is used when an integral is a product of two functions. It is essentially a way to break down a complicated integral into simpler parts.

The rule is given by the formula: \( \int u dv = uv - \int v du \), where one function is designated as \( u \) and the other is differentiated to find \( du \), while \( dv \) is the differential of the second function, and \( v \) is its integral.

Choosing U and DV

One of the most vital skills in integration by parts is choosing which function in the integrand to designate as \( u \) and which as \( dv \). A common mnemonic for this is ILATE (Inverse trigonometric functions, Logarithmic functions, Algebraic functions, Trigonometric functions, Exponential functions) which suggests a priority order.

Please remember that sometimes applying the integration by parts formula once does not solve the integral; it may need to be applied multiple times or combined with other integration methods. In our initial exercise, after we selected \( u = 3(x - 4) \) and \( dv = e^{x^2 - 8x} dx \), we ended up with a simpler integral, which still required the evaluation of an exponential function.

Derivative

The concept of the derivative is fundamental in calculus. It represents the rate at which a function is changing at any point on its graph, and it is the basic building block for finding the gradient of a tangent to a curve at a given point.

The derivative of a function \( f(x) \) with respect to \( x \) is denoted as \( f'(x) \) or \( \frac{d}{dx}f(x) \). For example, if we have \( f(x) = x^2 \), then the derivative \( f'(x) \) or \( \frac{d}{dx}(x^2) \) would be \( 2x \). This means for every unit increase in \( x \), the function \( f(x) \) increases by an amount proportional to \( 2x \).

In the context of integration by parts, we find the derivative to determine the differential \( du \) of our selected \( u \). This is crucial because we use this derivative to simplify the integral further, as we did in the step-by-step solution of our exercise by finding that the derivative of \( 3(x - 4) \) is simply \( 3 \). Recognizing and accurately computing derivatives allows us to engage effectively with integration techniques.