Question

assignment 10: problem (1 point) let $f(x) = \frac{16}{\sqrt{1 - x^2}}$. enter an antiderivative of $f(x)$.

Explanation:

Step1: Recall the antiderivative formula

The antiderivative of $\frac{1}{\sqrt{1 - x^2}}$ is $\arcsin(x)+C$ (where $C$ is the constant of integration). For a function $af(x)$ (where $a$ is a constant), the antiderivative is $a$ times the antiderivative of $f(x)$. Here, $a = 16$ and $f(x)=\frac{1}{\sqrt{1 - x^2}}$.

Step2: Apply the constant multiple rule

Multiply the antiderivative of $\frac{1}{\sqrt{1 - x^2}}$ by 16. So the antiderivative of $\frac{16}{\sqrt{1 - x^2}}$ is $16\arcsin(x)+C$. We can choose $C = 0$ for a specific antiderivative.

Answer:

$16\arcsin(x)$ (or $16\arcsin(x)+C$ where $C$ is any constant)