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The functions \(y=e^{x^{3}}\) and \(y=x^{3} e^{x^{3}}\) do not have elementary anti- derivatives, but \(y=\left(1+3 x^{3}\right) e^{x^{3}}\) does. Evaluate $$\int\left(1+3 x^{3}\right) e^{x^{3}} d x$$

Short Answer

Expert verified
The integral evaluates to \((x^3 - \frac{2}{3}) e^{x^3} + C\).

Step by step solution

01

Identify the problem

We need to evaluate the integral \( \int (1 + 3x^3) e^{x^3} \, dx \).
02

Recognize the antiderivative pattern

The problem states that \( y = (1 + 3x^3) e^{x^3} \) has an elementary antiderivative. We can identify this expression as the derivative of a product of functions, particularly linked to the derivative of \( e^{x^3} \).
03

Set up the substitution

Let \( u = x^3 \). This implies \( du = 3x^2 \, dx \). Hence, \( x^2 \, dx = \frac{1}{3} du \).
04

Adjust the integral

Using the substitution, rewrite the original integral:\[\int (1 + 3x^3) e^{x^3} \, dx = \int (1 + 3u) e^{u} \cdot \frac{1}{3} \, du = \frac{1}{3} \int (1 + 3u) e^u \, du\]
05

Evaluate the integral

Recognize \( \int (1 + 3u) e^{u} \, du \). The integration of \( (1 + 3u) e^u \) is straightforward:\[\int e^{u} \, du + 3 \int u e^{u} \, du\]The integrals yield:\[e^u + 3 \left( u e^u - \int e^u \, du \right) \]\[=e^u + 3u e^u - 3e^u \]Factor out \( e^u \):\[= (1 + 3u - 3)e^u = (3u - 2)e^u\].
06

Substitute back

Substituting \( u = x^3 \) back in, the integral becomes:\[\frac{1}{3} ((3x^3 - 2) e^{x^3})\]Thus, the evaluated integral is:\[(x^3 - \frac{2}{3}) e^{x^3} + C\]where \( C \) is the constant of integration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Antiderivatives
Antiderivatives, also known as indefinite integrals, are functions that reverse the process of differentiation. In simpler terms, if you take the derivative of a function and then find its antiderivative, you should end up back at the original function. For example, if \(f(x)\) is a function, then an antiderivative of \(f(x)\) is a function \(F(x)\) such that \(F'(x) = f(x)\).
Antiderivatives are closely related to the concept of areas under curves and are key in solving integrals. However, not all functions have elementary antiderivatives, meaning they can't be expressed in terms of elementary functions like polynomials, exponentials, logarithms, etc. This is why recognizing patterns or using methods like substitution becomes vital in calculating integrals.
Substitution Method
The substitution method is a common technique used in integration to simplify the process by changing the variable of integration. This method is particularly useful when the integrand, the function being integrated, has a composition of functions that makes direct integration difficult.
In this method, we perform the following steps:
  • Identify a part of the integrand that you can substitute with a new variable. This is usually a function inside another function.
  • Define the new variable (\(u\) substitution) and compute its differential (\(du\)).
  • Replace the original variable in the integral with the new variable and its corresponding differential.
  • Perform the integration with respect to the new variable.
  • Finally, substitute back the original variable to express the result in terms of the original variable.
This method is akin to reversing the chain rule in differentiation, and it's a powerful technique to solve integrals that seem complex at first glance.
Exponential Functions
Exponential functions are functions of the form \( f(x) = a^x \), where \( a \) is a positive constant and \( x \) is the variable. A special type of exponential function is the natural exponential function, \( e^x \), where \( e \) is approximately equal to 2.71828. The natural exponential function is unique because it is its own derivative, which makes it significant in calculus.
When dealing with exponential functions and integration, especially if the exponent is a function of \( x \), substitution is often required. This is because directly integrating exponential functions with complex exponents can be challenging. Recognizing the pattern where an exponential function is multiplied by its derivative is crucial, as seen in the exercise provided. Such insights simplify the process, converting what might seem a daunting integral into a straightforward computation.

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