91Ó°ÊÓ

In calculus, finding an antiderivative involves determining a function whose derivative matches a given function. Given a function like \( f(x) = 12(2x+1)^5 \), you're tasked with identifying a function \( F(x) \) such that \( F'(x) = f(x) \). The process of finding such a function is called integration.

An antiderivative is not unique; it includes an arbitrary constant because the derivative of a constant is zero. This is represented as \( + C \) in the solution. For example, if the antiderivative of a function is \( (2x+1)^6 \), then the complete solution is \( (2x+1)^6 + C \).

This means that there's a family of functions that are all antiderivatives of the given function, differing only by a constant. Understanding this concept is vital as it forms the basis of integral calculus, which is concerned with finding the areas, volumes, and totals.

Integration by substitution is a crucial technique in calculus that simplifies the integration process by changing variables. In our problem, we aim to integrate \( 12(2x+1)^5 \).

Notice the expression inside the power. Set \( u = 2x + 1 \); hence, \( du = 2dx \) or \( dx = \frac{1}{2}du \). This substitution makes the integration simpler.

By substituting these new variables into our integral, the problem becomes integrating \( 6u^5 \) with respect to \( u \). Substitution is handy especially when dealing with compositions of functions or more complex algebraic expressions.

• The purpose is to transform a tricky integral into a standard form that is easier to handle.
• Remember to always replace all \( x \) terms with \( u \), and \( dx \) with the corresponding \( du \) expression you find earlier.
• Finally, once you find the antiderivative in terms of \( u \), substitute back the original variable \( x \).

This method is advantageous because it makes integration look more straightforward by converting it into a form already known to be integrable.

The power rule is one of the simplest and most commonly used techniques in integration. It states that if you have \( \int x^n \, dx \), the result is \( \frac{x^{n+1}}{n+1} + C \), provided that \( n eq -1 \).

In our problem, after substitution the integral becomes \( \int 6u^5 \, du \). Applying the power rule gives us \( 6 \cdot \frac{u^{6}}{6} + C = u^6 + C \).

This is straightforward because:

• You increase the power of \( u \) by one, making it \( u^6 \).
• Then divide by this new power, which in this case, simplifies the coefficient of 6 perfectly.

The power rule is vital for integrating polynomial functions, making it a powerful tool in calculus.

Once you have the result in terms of \( u \), don’t forget to back-substitute to express the final answer in terms of the original variable \( x \). So our final antiderivative is \( (2x+1)^6 + C \). This simple rule significantly helps solve integration problems predictably and consistently.