Problem 4 - AP Calculus with 8 Practice Tests 15th

Question

\(\int(2-3 x)^{5} d x=\) (A) \(\frac{1}{6}(2-3 x)^{6}+C\) (B) \(-\frac{1}{2}(2-3 x)^{6}+C\) (C) \(\frac{1}{2}(2-3 x)^{6}+C\) (D) \(-\frac{1}{18}(2-3 x)^{6}+C\)

Step-by-Step Solution

Answer

The answer is option (D): \(-\frac{1}{18}(2-3 x)^{6}+C\).

Step 1: Identify the Type of Integral

This is an integral of the form \( \int (ax + b)^n \, dx \), where \( a = -3 \), \( b = 2 \), and \( n = 5 \). We will use the power rule for integration, which requires substitution.

Step 2: Apply Substitution

Let \( u = 2 - 3x \). Then, \( \frac{du}{dx} = -3 \) or \( du = -3 \, dx \). Rearranging gives \( dx = -\frac{1}{3} du \). Substitute \( u \) for \( 2 - 3x \) and \( dx \) in the integral: \[\int (2 - 3x)^5 \, dx = \int u^5 \left(-\frac{1}{3}\right) du.\]

Step 3: Simplify the Integral Expression

The integral now becomes \[-\frac{1}{3} \int u^5 \, du.\]Factor out the constant and apply the power rule for integrals.

Step 4: Integrate using the Power Rule

Integrate using the power rule, \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). In this case, \[-\frac{1}{3} \int u^5 \, du = -\frac{1}{3} \cdot \frac{u^{6}}{6} + C = -\frac{1}{18} u^6 + C.\]

Step 5: Substitute Back for Original Variable

Substitute back \( u = 2 - 3x \) to express the answer in terms of \( x \):\[-\frac{1}{18} (2 - 3x)^6 + C.\]

Step 6: Match the Solution with Given Options

The solution matches option (D): \(-\frac{1}{18}(2-3 x)^{6}+C\).

Key Concepts

Substitution Method

The substitution method is a useful technique for simplifying the process of integration, especially when dealing with composite functions like the one in the given exercise. In this method, you typically choose a substitution that simplifies the integrand (the function to be integrated).

Here's how it works:

• Identify a portion of the integrand that can be replaced with a single variable, often chosen as u.
• Calculate the derivative of your chosen substitution component with respect to x.
• Use this derivative to express dx in terms of du.

In our original exercise, the substitution was u = 2 - 3x. By substituting u into the integral, we change the variable in our integral which often makes it easier to solve. The goal here is to transform the integral into a simpler form that can readily be integrated. After integration, we replace u with the original expression to find the integral in terms of x again.

Power Rule for Integration

The power rule for integration is a fundamental tool when integrating expressions of the form \(x^n\). It states that the integral of \(x^n\) with respect to x is given by:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

where C is the constant of integration. This rule is quite straightforward when n is not equal to -1.

In the exercise provided, once we made the substitution, we ended up with an integral of the form \(u^5\). Applying the power rule, we integrated \(u^5\) to obtain \(\frac{u^{6}}{6}\). Because of our substitution method, we initially factor the constant \(-\frac{1}{3}\) outside the integral, leading to a final integrated expression of \(-\frac{1}{18}u^6 + C\).

The power rule simplifies the task of calculating antiderivatives when applied correctly, and in this problem, it serves as the key to progressing from the substituted integral back to our variable of interest—x.

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is crucial when tackling integration problems.

• Indefinite Integrals: These include the constant of integration C and represent a family of functions. When you see the notation \(\int f(x) \, dx\), you are dealing with an indefinite integral. This type of integral tells us about the accumulator function without specifying the limits.
• Definite Integrals: These have upper and lower limits, providing the area under the curve between these two points. The result is a specific numerical value, not a function, as it indicates bounded area.

The current exercise involves an indefinite integral, as indicated by the presence of C in all the solution options. Indefinite integrals emphasize finding the function that differentiates back to the original integrand, reflecting the general solution shape. This concept helps ensure that after solving, we know how any derived function can be shifted vertically to match specific conditions or initial values.