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For the following function [tex]\( f \)[/tex], find the antiderivative [tex]\( F \)[/tex] that satisfies the given condition.

[tex]\[
\begin{array}{l}
f(v) = \frac{3}{4} \sec v \tan v, \quad F(0) = 3, \quad -\frac{\pi}{2} \ \textless \ v \ \textless \ \frac{\pi}{2} \\
F(v) = \square
\end{array}
\][/tex]


Sagot :

Sure, let's go through the steps to find the antiderivative [tex]\( F(v) \)[/tex] of the function [tex]\( f(v) = \frac{3}{4} \sec(v) \tan(v) \)[/tex] that satisfies the condition [tex]\( F(0) = 3 \)[/tex].

### Step 1: Find the Antiderivative

First, we need to find the indefinite integral of the function [tex]\( f(v) \)[/tex]:
[tex]\[ \int \frac{3}{4} \sec(v) \tan(v) \, dv \][/tex]

We know that the derivative of [tex]\( \sec(v) \)[/tex] is [tex]\( \sec(v) \tan(v) \)[/tex], so integrating [tex]\( \sec(v) \tan(v) \)[/tex] gives [tex]\( \sec(v) \)[/tex]. Thus:

[tex]\[ \int \sec(v) \tan(v) \, dv = \sec(v) \][/tex]

Considering the constant multiplier [tex]\(\frac{3}{4}\)[/tex], the integral becomes:
[tex]\[ \int \frac{3}{4} \sec(v) \tan(v) \, dv = \frac{3}{4} \sec(v) \][/tex]

Thus, the antiderivative [tex]\( F(v) \)[/tex] can be written as:
[tex]\[ F(v) = \frac{3}{4} \sec(v) + C \][/tex]

### Step 2: Determine the Constant of Integration

To determine the constant [tex]\( C \)[/tex], we use the condition [tex]\( F(0) = 3 \)[/tex]:
[tex]\[ F(0) = \frac{3}{4} \sec(0) + C = 3 \][/tex]

We know that [tex]\( \sec(0) = 1 \)[/tex], so:
[tex]\[ \frac{3}{4} \cdot 1 + C = 3 \][/tex]

This simplifies to:
[tex]\[ \frac{3}{4} + C = 3 \][/tex]

Subtract [tex]\(\frac{3}{4}\)[/tex] from both sides to solve for [tex]\( C \)[/tex]:
[tex]\[ C = 3 - \frac{3}{4} = \frac{12}{4} - \frac{3}{4} = \frac{9}{4} = 2.25 \][/tex]

### Step 3: Write the Final Antiderivative

Now that we have determined [tex]\( C = 2.25 \)[/tex], we can write the final expression for the antiderivative [tex]\( F(v) \)[/tex]:
[tex]\[ F(v) = \frac{3}{4} \sec(v) + 2.25 \][/tex]

Thus, the antiderivative [tex]\( F \)[/tex] that satisfies the given condition is:
[tex]\[ F(v) = 2.25 + \frac{3}{4} \sec(v) \][/tex]

So we fill in the blank:
[tex]\[ F(v) = \boxed{2.25 + \frac{3}{4} \sec(v)} \][/tex]