If f(x) is a continuous function such that `f(x) gt 0` for all `x gt 0` and `(f(x))^(2020)=1+int_(0)^(x) f(t) dt`, then the value of `{f(2020)}^(2019)` is equal to

Options

A. 2019
B. 2020
C. 2021
D. 2018

Explanation

To solve the problem, we start with the given equation:

\[ f(x)^{2020} = 1 + \int_0^x f(t) \, dt \]

### Step 1: Differentiate both sides
We differentiate both sides of the equation with respect to \( x \).

Using the chain rule on the left side, we have:

\[ \frac{d}{dx} \left( f(x)^{2020} \right) = 2020 f(x)^{2019} f'(x) \]

For the right side, we apply the Fundamental Theorem of Calculus:

\[ \frac{d}{dx} \left( 1 + \int_0^x f(t) \, dt \right) = f(x) \]

Thus, we get the equation:

\[ 2020 f(x)^{2019} f'(x) = f(x) \]

### Step 2: Simplify the equation
Assuming \( f(x) > 0 \) for all \( x > 0 \), we can divide both sides by \( f(x) \):

\[ 2020 f(x)^{2018} f'(x) = 1 \]

### Step 3: Rearranging and integrating
Rearranging gives us:

\[ f'(x) = \frac{1}{2020 f(x)^{2018}} \]

Now we can integrate both sides. We rewrite it as:

\[ f'(x) f(x)^{2018} = \frac{1}{2020} \]

Integrating both sides with respect to \( x \):

\[ \int f'(x) f(x)^{2018} \, dx = \frac{1}{2020} \int dx \]

The left side can be integrated using the substitution \( u = f(x) \):

\[ \frac{f(x)^{2019}}{2019} = \frac{x}{2020} + C \]

### Step 4: Solve for the constant \( C \)
To find the constant \( C \), we can use the initial condition. We substitute \( x = 0 \):

From the original equation:

\[ f(0)^{2020} = 1 + \int_0^0 f(t) \, dt \implies f(0)^{2020} = 1 \implies f(0) = 1 \]

Substituting \( x = 0 \) into our integrated equation gives:

\[ \frac{f(0)^{2019}}{2019} = \frac{0}{2020} + C \implies \frac{1^{2019}}{2019} = C \implies C = \frac{1}{2019} \]

### Step 5: Substitute back to find \( f(x) \)
Now substituting \( C \) back into our integrated equation:

\[ \frac{f(x)^{2019}}{2019} = \frac{x}{2020} + \frac{1}{2019} \]

Multiplying through by 2019 gives:

\[ f(x)^{2019} = \frac{2019x}{2020} + 1 \]

### Step 6: Find \( f(2020) \)
Now we substitute \( x = 2020 \):

\[ f(2020)^{2019} = \frac{2019 \cdot 2020}{2020} + 1 = 2019 + 1 = 2020 \]

### Step 7: Final answer
Thus, we have:

\[ f(2020)^{2019} = 2020 \]

So, the value of \( f(2020)^{2019} \) is equal to:

\[ \boxed{2020} \]