What is the integral of x^n where n ≠ -1?

What is the integral of x^n where n ≠ -1?

• A. (1/(n+1))x^(n+1) + C
• B. (1/n)x^n + C
• C. (1/(n-1))x^(n-1) + C
• D. x^(n+1) + C

Answer: (A)

The integral of \( x^n \) where \( n \neq -1 \) is derived from the power rule for integrals. According to this rule, when you integrate a function of the form \( x^n \), you increase the power by one and then divide by the new power. Thus, integrating \( x^n \) results in: \[ \int x^n \, dx = \frac{1}{n+1} x^{n+1} + C \] Here, \( n+1 \) is the new exponent after increasing \( n \) by one. Dividing by \( n+1 \) ensures that the area under the curve represented by \( x^n \) is accurately represented in the antiderivative. The \( C \) represents the constant of integration, which accounts for any vertical shifts in the family of antiderivative functions. The other choices do not follow this rule. For instance, the option suggesting \( \frac{1}{n} x^n + C \) does not apply since it does not reflect the appropriate adjustment in the exponent and division needed for finding the antiderivative of \( x^n \). Similarly, the choices involving

The integral of ( x^n ) where ( n \neq -1 ) is derived from the power rule for integrals. According to this rule, when you integrate a function of the form ( x^n ), you increase the power by one and then divide by the new power.

Thus, integrating ( x^n ) results in:

[

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

]

Here, ( n+1 ) is the new exponent after increasing ( n ) by one. Dividing by ( n+1 ) ensures that the area under the curve represented by ( x^n ) is accurately represented in the antiderivative. The ( C ) represents the constant of integration, which accounts for any vertical shifts in the family of antiderivative functions.

The other choices do not follow this rule. For instance, the option suggesting ( \frac{1}{n} x^n + C ) does not apply since it does not reflect the appropriate adjustment in the exponent and division needed for finding the antiderivative of ( x^n ). Similarly, the choices involving