Differentiation
Integration
Integration Rule [Fundamental Theorem of Calculus 2]:
[tex]\displaystyle \frac{d}{dx}[\int\limits^x_a {f(t)} \, dt] = f(x)[/tex]
Let's organize what is given to us from the problem.
We are given an initial condition: f(0) = 5.420
We are given a derivative:
[tex]\displaystyle f'(x) = \sqrt{\sin^2 x + x}[/tex]
We are asked to find f(2π).
We can first formulate a rough function using integration rules, namely the Fundamental Theorem of Calculus (2nd part):
[tex]\displaystyle f(x) = \int \limits^{x}_{a} f(t) \, dt + C_0[/tex]
where f(t) is a derivative function and C₀ is an initial condition.
Now that we have a "baseline formula", we can substitute in our f(t) derivative function, our limits, and our initial condition:
[tex]\displaystyle f'(x) = \sqrt{\sin^2 x + x} \Rightarrow f(t) = \sqrt{\sin^2 t + t}[/tex]
[tex]\displaystyle \begin{aligned}f(x) & = \int \limits^{x}_{a} f(t) \, dt + C_0 \\& = \int \limits^{x}_{0} \sqrt{\sin^2 t + t} \, dt + f(0) \\& = \boxed{ \int \limits^{x}_{0} \sqrt{\sin^2 t + t} \, dt + 5.420 } \\\end{aligned}[/tex]
∴ our general function is equal to:
[tex]\displaystyle \boxed{ f(x) = \int \limits^{x}_{0} \sqrt{\sin^2 t + t} \, dt + 5.420 }[/tex]
*Note:
There isn't an antiderivative for the general function, so you will have to input the integral into a calculator.
In order to find the value of f at 2π, we substitute x = 2π into our general function:
[tex]\displaystyle \begin{aligned}f(2 \pi) & = \int \limits^{2 \pi}_{0} \sqrt{\sin^2 t + t} \, dt + 5.420 \\& = 11.44863485704886 + 5.420 \\& = 16.868634857 \\& \approx \boxed{ 16.868 } \\\end{aligned}[/tex]
∴ [tex]\displaystyle \boxed{ f(2 \pi) \approx 16.868 }[/tex]
∴ the value of f(2π) is approximately 16.868.
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Learn more about integration: https://brainly.com/question/28758158
Learn more about Calculus: https://brainly.com/question/13321245
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Topic: Calculus I
Unit: Integration
