Question

\(\int\left(x^{3}-2 x^{2}+5\right) e^{3 x} d x\)

Short answer

Question: Find the antiderivative of the function \(f(x) = (x^3 - 2x^2 + 5)e^{3x}\). Answer: The antiderivative of the function \(f(x) = (x^3 - 2x^2 + 5)e^{3x}\) is \(\frac{1}{3}e^{3x}(x^3 - x^2 + 2x - \frac{8}{9}) + C\), where C is the constant of integration.

Step-by-step solution

Apply integration by parts formula

Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). We will apply this formula 3 times, as the given function is a product of a cubic polynomial and an exponential function. We choose \(u\) as the polynomial part and \(dv\) as the exponential part. Then we differentiate \(u\) and integrate \(dv\).

Differentiate u and Integrate dv

Differentiate \(u\) and integrate \(dv\): \(\frac{d u}{dx} = \frac{d}{dx}(x^3 - 2x^2 + 5)\), leading to \(du = (3x^2 - 4x) dx\). \(\int dv = \int e^{3x} dx\), resulting in \(v = \frac{1}{3} e^{3x}\).

Apply integration by parts for the first time

Now apply the integration by parts formula: \(\int\left(x^{3}-2 x^{2}+5\right) e^{3 x} d x = (x^3 - 2x^2 + 5)\frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} (3x^2 - 4x) dx = \frac{1}{3}(x^3 - 2x^2 + 5)e^{3x} - \int (x^2 - \frac{4}{3}x) e^{3x} dx\). Notice that the remaining integral is a product of a quadratic function and an exponential function, we should apply integration by parts again.

Differentiate u and Integrate dv again

Differentiate \(u\) and integrate \(dv\) for the remaining integral: \(\frac{d u}{dx} = \frac{d}{dx}(x^2 - \frac{4}{3}x)\), leading to \(du = (2x - \frac{4}{3}) dx\). \(\int dv = \int e^{3x} dx\), resulting in \(v = \frac{1}{3} e^{3x}\).

Apply integration by parts for the second time

Now apply the integration by parts formula for the remaining integral: \(\int (x^2 - \frac{4}{3}x) e^{3x} dx = (x^2 - \frac{4}{3}x) \frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} (2x - \frac{4}{3}) dx = \frac{1}{3}(x^2 - \frac{4}{3}x)e^{3x} - \int (\frac{2}{3}x - \frac{4}{9}) e^{3x} dx\). The remaining integral involves a product of a linear function and an exponential function. We should apply the integration by parts one final time.

Differentiate u and Integrate dv for the last time

Differentiate \(u\) and integrate \(dv\) for the remaining integral: \(\frac{d u}{dx} = \frac{d}{dx}(\frac{2}{3}x - \frac{4}{9})\), leading to \(du = \frac{2}{3} dx\). \(\int dv = \int e^{3x} dx\), resulting in \(v = \frac{1}{3} e^{3x}\).

Apply integration by parts for the third and final time

Now apply the integration by parts formula for the remaining integral: \(\int (\frac{2}{3}x - \frac{4}{9}) e^{3x} dx = (\frac{2}{3}x - \frac{4}{9}) \frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} (\frac{2}{3}) dx = \frac{1}{3}(\frac{2}{3}x - \frac{4}{9})e^{3x} - \frac{2}{9}\int e^{3x} dx\). The remaining integral is simply an exponential function, which can be straightforwardly integrated.

Integrate the final exponential function

Integrate the remaining exponential function: \(\int e^{3x} dx = \frac{1}{3}e^{3x} + C\).

Combine all terms

Now we combine all the terms we obtained from the integration by parts: \(\int\left(x^{3}-2 x^{2}+5\right) e^{3 x} d x = \frac{1}{3}(x^{3} - 2x^{2} + 5)e^{3x} - \frac{1}{3}(x^{2} - \frac{4}{3}x)e^{3x} + \frac{1}{3}(\frac{2}{3}x - \frac{4}{9})e^{3x} + \frac{2}{9}(\frac{1}{3}e^{3x}) + C\). Simplify the expression: \(\int\left(x^{3}-2 x^{2}+5\right) e^{3 x} d x = \frac{1}{3}e^{3x}(x^3 - x^2 + 2x - \frac{8}{9}) + C\). Thus, the antiderivative of the given function is \(\frac{1}{3}e^{3x}(x^3 - x^2 + 2x - \frac{8}{9}) + C\).

Key concepts

Integral Calculus

Integral calculus is one of the two main branches of calculus, the other being differential calculus. It is concerned with the accumulation of quantities, such as areas under curves and the cumulative effect of continuously changing rates. To solve complex integrals, various techniques can be applied, and one of them is integration by parts.

This method is particularly useful when the integral involves a product of functions with different characteristics, such as polynomial and exponential functions. The process of integration by parts allows us to break down such complex integrals into simpler ones, ultimately leading to a solution. It's a powerful tool that leverages the product rule for differentiation and the fundamental theorem of calculus to find antiderivatives or evaluate definite integrals.

Exponential Functions

Exponential functions are mathematical expressions that describe growth or decay processes, having variables in the exponent. The base of an exponential function can be any positive number, and its power is variable. They are commonly written in the form \(e^{rx}\), where \(e\) is Euler's number (approximately 2.71828), \(r\) is the growth or decay rate, and \(x\) is the variable.

In the context of integration, exponential functions are unique because their derivative is proportional to the function itself. This makes integrating exponential functions relatively straightforward once isolated, as was demonstrated in the final steps of the provided exercise.

Polynomial Functions

Polynomial functions consist of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. A polynomial function of degree three, for example, will have the general form \(ax^3+bx^2+cx+d\) where \(a, b, c\), and \(d\) are constants and \(a \eq 0\).

Polynomials are straightforward to differentiate, which makes them a good candidate for the \(u\) part in the integration by parts technique (as seen in the exercise). With each differentiation, they reduce in degree, eventually reaching a constant, which simplifies the process of integration.

Antiderivatives

An antiderivative, also known as an indefinite integral, is a function that reverses the process of differentiation. In other words, given a function \(f(x)\), its antiderivative is a function \(F(x)\) such that \(F'(x) = f(x)\).

When calculating antiderivatives, a constant of integration (usually denoted as \(C\)) is added because the derivative of a constant is zero, making the antiderivative not unique. Antiderivatives are crucial in evaluating the areas under curves and solving differential equations. The integration by parts technique, as demonstrated in the exercise, helps find antiderivatives for functions that are products of other functions.