Respuesta :
Answer:
[tex]f'(2)=4\sqrt{3}[/tex]
Step-by-step explanation:
Given : If f is the function given by [tex]f(x)=\int\limits^{2x}_{4} {\sqrt{t^2-t}} \, dt[/tex]
To find : [tex]f'(2)[/tex]?
Solution :
Function [tex]f(x)=\int\limits^{2x}_{4} {\sqrt{t^2-t}} \, dt[/tex]
Applying fundamental theorem of calculus,
If [tex]f(x)=\int\limits^{h_2(x)}_{h_1(x)} {g(t)} \, dt[/tex]
Then, [tex]f'(x)=g(h_2(x))\cdot h_2'(x)-g(h_1(x))\cdot h_1'(x)[/tex]
Comparing with our function,[tex]f(x)=\int\limits^{2x}_{4} {\sqrt{t^2-t}} \, dt[/tex]
[tex]h_1(x)=4\\h_1'(x)=0[/tex]
[tex]h_2(x)=2x\\h_2'(x)=2[/tex]
[tex]g(t)=\sqrt{t^2-t} \\g(h_1(x))=\sqrt{(4)^2-4}\\g(h_2(x))=\sqrt{(2x)^2-2x}\\[/tex]
Substituting the values we get,
[tex]f'(x)=\sqrt{(2x)^2-2x}\cdot 2-\sqrt{(4)^2-4}\cdot 0[/tex]
[tex]f'(x)=2\sqrt{4x^2-2x}-0[/tex]
Put x=2
[tex]f'(2)=2\sqrt{4(2)^2-2(2)}[/tex]
[tex]f'(2)=2\sqrt{16-4}[/tex]
[tex]f'(2)=2\sqrt{12}[/tex]
[tex]f'(2)=2\times 2\sqrt{3}[/tex]
[tex]f'(2)=4\sqrt{3}[/tex]
Therefore, If f is the function given by [tex]f(x)=\int\limits^{2x}_{4} {\sqrt{t^2-t}} \, dt[/tex] then [tex]f'(2)=4\sqrt{3}[/tex]
Using the integration property to solve the problem. Then f is the given function by [tex]f(x) = \int _4^{2x} \sqrt{t^2 - t} dt [/tex] then [tex] f'(2) = 4\sqrt{3} [/tex]
What is a function?
The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.
Given
The function f(x) is [tex] \rm \int^{2x}_4 \sqrt{t^2 - t} [/tex].
Then the function will be.
If
[tex] f(x) = \int ^{h_2(x)}_{h_1(x)} g(t) dt [/tex]
Then
[tex] f'(x) = g(h_2(x)) . h'_2(x) - g(h_1(x)). h'_1(x) [/tex]
Compared with our function
[tex] h_1 (x) = 4 [/tex]
[tex] h'_1 (x) = 0 [/tex]
[tex] h_2 (x) = 2x [/tex]
[tex] h'_2 (x) = 2 [/tex]
Then
[tex] \begin{aligned} g(t) &= \sqrt{t^2 - t} \\\\ g(h_1(x)) &= \sqrt{4^2 - 4} \\\\ g(h_2(x)) &= \sqrt{(2x)^2 - 2x } \end{aligned} [/tex]
Substituting the values we get
[tex] f'(x) = \sqrt{(2x)^2 - 2x } *2 - \sqrt{4^2 - 4} *0\\\\ f'(x) = 2\sqrt{4x^2 - 2x} [/tex]
Put x = 2, then
[tex] f'(2) = 2 \sqrt{2(2)^2 - 2 *2} \\\\f'(2) = 2 \sqrt{16-4}\\\\f'(2) = 4 \sqrt{3} [/tex]
Thus, if f is the given function by [tex] f(x) = \int _4^{2x} \sqrt{t^2 - t} dt [/tex] then [tex] f'(2) = 4\sqrt{3} [/tex]
More about the function link is given below.
https://brainly.com/question/5245372
