For the function [tex]$f$[/tex] given below, find the antiderivative [tex]$F$[/tex] that satisfies [tex][tex]$F(1)=1$[/tex][/tex].

[tex]
f(x) = x^5 - 4x^{-2} - 5
[/tex]

The antiderivative that satisfies the given condition is [tex]$F(x)=$[/tex] [tex] \square[/tex].



Answer :

To solve for the antiderivative [tex]\( F(x) \)[/tex] of the function [tex]\( f(x) = x^5 - 4x^{-2} - 5 \)[/tex] that satisfies the condition [tex]\( F(1) = 1 \)[/tex], let's go through the steps systematically.

### Step 1: Find the Indefinite Integral
First, we need to find the indefinite integral (antiderivative) of [tex]\( f(x) \)[/tex].

[tex]\[ F(x) = \int (x^5 - 4x^{-2} - 5) \, dx \][/tex]

We can integrate the function term by term:

1. [tex]\(\int x^5 \, dx = \frac{x^6}{6}\)[/tex]
2. [tex]\(\int -4x^{-2} \, dx = -4 \cdot \int x^{-2} \, dx = -4 \cdot \left( -\frac{1}{x} \right) = \frac{4}{x}\)[/tex]
3. [tex]\(\int -5 \, dx = -5x\)[/tex]

So, combining these results, we get:

[tex]\[ F(x) = \frac{x^6}{6} + \frac{4}{x} - 5x + C \][/tex]

Where [tex]\( C \)[/tex] is the constant of integration.

### Step 2: Solve for the Constant [tex]\( C \)[/tex]
We are given the condition [tex]\( F(1) = 1 \)[/tex]. We use this condition to solve for [tex]\( C \)[/tex].

[tex]\[ F(1) = \frac{1^6}{6} + \frac{4}{1} - 5 \cdot 1 + C = 1 \][/tex]

Simplifying this equation:

[tex]\[ \frac{1}{6} + 4 - 5 + C = 1 \][/tex]

[tex]\[ \frac{1}{6} - 1 + C = 1 \][/tex]

[tex]\[ C - \frac{5}{6} = 1 \][/tex]

[tex]\[ C = 1 + \frac{5}{6} \][/tex]

[tex]\[ C = \frac{6}{6} + \frac{5}{6} \][/tex]

[tex]\[ C = \frac{11}{6} \][/tex]

### Step 3: Write the Final Antiderivative
Substitute [tex]\( C = \frac{11}{6} \)[/tex] back into the antiderivative:

[tex]\[ F(x) = \frac{x^6}{6} + \frac{4}{x} - 5x + \frac{11}{6} \][/tex]

So the antiderivative [tex]\( F(x) \)[/tex] that satisfies [tex]\( F(1) = 1 \)[/tex] is:

[tex]\[ F(x) = \frac{x^6}{6} + \frac{4}{x} - 5x + \frac{11}{6} \][/tex]

Thus, the final answer is:

[tex]\[ F(x) = \boxed{\frac{x^6}{6} + \frac{4}{x} - 5x + \frac{11}{6}} \][/tex]