Integrate $\int (4x^{-3} - 7x^2 + 5x - 6) \mathrm d x$.

To solve this integral, we use the Power Rule for Integration.

1. Analyze the Concept

The power rule states that for any term $ax^n$ (where $n \neq -1$):

Essentially, you add 1 to the exponent and then divide the term by that new exponent. We apply this rule to each term in the expression individually.

2. Step-by-Step Solution

We are integrating: $\int (4x^{-3} - 7x^2 + 5x^1 - 6x^0) \, dx$

• Term 1: $4x^{-3}$ Add 1 to the exponent: $-3 + 1 = -2$. Divide by the new exponent: $\frac{4x^{-2}}{-2} = -2x^{-2}$

• Term 2: $-7x^2$ Add 1 to the exponent: $2 + 1 = 3$. Divide by the new exponent: $-\frac{7x^3}{3} = -\frac{7}{3}x^3$

• Term 3: $5x^1$ Add 1 to the exponent: $1 + 1 = 2$. Divide by the new exponent: $\frac{5x^2}{2} = \frac{5}{2}x^2$

• Term 4: $-6$ (which is $-6x^0$) Add 1 to the exponent: $0 + 1 = 1$. Divide by the new exponent: $-\frac{6x^1}{1} = -6x$

3. Combine the results

Putting these together, we get:

Comparing this result to your given options, it matches Option A.

Exam Tip

When integrating, always check your work by differentiating your answer. If you differentiate $-2x^{-2} - \frac{7}{3}x^3 + \frac{5}{2}x^2 - 6x$, you will arrive back at the original expression $(4x^{-3} - 7x^2 + 5x - 6)$. This is a great way to guarantee you haven't made a sign error during the exam!