If F F F Is The Function Given By F ( X ) = ∫ 4 2 X T 2 − T D T F(x)=\int_4^{2x} \sqrt{t^2-t} \, Dt F ( X ) = ∫ 4 2 X ​ T 2 − T ​ D T , Then F ′ ( 2 ) = F^{\prime}(2)= F ′ ( 2 ) = A. 0 B. 7 2 12 \frac{7}{2 \sqrt{12}} 2 12 ​ 7 ​ C. 2 \sqrt{2} 2 ​ D. 12 \sqrt{12} 12 ​ E. 2 12 2 \sqrt{12} 2 12 ​

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Introduction

In this article, we will explore the concept of finding the derivative of a function that involves an integral. The function given is $f(x)=\int_4^{2x} \sqrt{t^2-t} \, dt$. We will use the fundamental theorem of calculus to find the derivative of this function and then evaluate it at $x=2$.

The Fundamental Theorem of Calculus

The fundamental theorem of calculus states that if $f$ is a continuous function on the interval $[a,b]$, then the function $F(x)=\int_{a}^{x} f(t) \, dt$ is differentiable and its derivative is given by $F^{\prime}(x)=f(x)$.

Applying the Fundamental Theorem of Calculus

In this case, we have $f(x)=\int_4^{2x} \sqrt{t^2-t} \, dt$. We can apply the fundamental theorem of calculus by letting $F(x)=f(x)$ and $a=4$. Then, we have $F(x)=\int_{4}^{2x} \sqrt{t^2-t} \, dt$.

Finding the Derivative

Using the fundamental theorem of calculus, we can find the derivative of $F(x)$ by differentiating the integral. We have:

$$F^{\prime}(x)=\frac{d}{dx} \int_{4}^{2x} \sqrt{t^2-t} \, dt$$

$$F^{\prime}(x)=\sqrt{(2x)^2-(2x)} \cdot \frac{d}{dx}(2x)$$

$$F^{\prime}(x)=\sqrt{4x^2-2x} \cdot 2$$

$$F^{\prime}(x)=2\sqrt{4x^2-2x}$$

Evaluating the Derivative at $x=2$

Now that we have found the derivative of $F(x)$, we can evaluate it at $x=2$. We have:

$$F^{\prime}(2)=2\sqrt{4(2)^2-2(2)}$$

$$F^{\prime}(2)=2\sqrt{16-4}$$

$$F^{\prime}(2)=2\sqrt{12}$$

Conclusion

In this article, we used the fundamental theorem of calculus to find the derivative of a function that involves an integral. We then evaluated the derivative at $x=2$ and found that $F^{\prime}(2)=2\sqrt{12}$. This is the correct answer.

Final Answer

The final answer is $\boxed{2\sqrt{12}}$.

Discussion

The discussion of this problem involves understanding the concept of the fundamental theorem of calculus and how it can be used to find the derivative of a function that involves an integral. It also involves evaluating the derivative at a specific value of $x$.

Related Problems

• Find the derivative of the function $f(x)=\int_{0}^{x^2} \sin(t) \, dt$.
• Evaluate the derivative of the function $f(x)=\int_{0}^{x^2} \sin(t) \, dt$ at $x=1$.
• Find the derivative of the function $f(x)=\int_{0}^{x^3} \cos(t) \, dt$.

Solutions to Related Problems

• The derivative of the function $f(x)=\int_{0}^{x^2} \sin(t) \, dt$ is given by $f^{\prime}(x)=2x\sin(x^2)$.
• The derivative of the function $f(x)=\int_{0}^{x^2} \sin(t) \, dt$ evaluated at $x=1$ is given by $f^{\prime}(1)=2\sin(1)$.
• The derivative of the function $f(x)=\int_{0}^{x^3} \cos(t) \, dt$ is given by $f^{\prime}(x)=3x^2\cos(x^3)$.

Conclusion

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Q&A

Q: What is the fundamental theorem of calculus?

A: The fundamental theorem of calculus states that if $f$ is a continuous function on the interval $[a,b]$, then the function $F(x)=\int_{a}^{x} f(t) \, dt$ is differentiable and its derivative is given by $F^{\prime}(x)=f(x)$.

Q: How can we apply the fundamental theorem of calculus to find the derivative of the function $f(x)=\int_4^{2x} \sqrt{t^2-t} \, dt$?

A: We can apply the fundamental theorem of calculus by letting $F(x)=f(x)$ and $a=4$. Then, we have $F(x)=\int_{4}^{2x} \sqrt{t^2-t} \, dt$.

Q: What is the derivative of the function $F(x)=\int_{4}^{2x} \sqrt{t^2-t} \, dt$?

A: Using the fundamental theorem of calculus, we can find the derivative of $F(x)$ by differentiating the integral. We have:

$$F^{\prime}(x)=\frac{d}{dx} \int_{4}^{2x} \sqrt{t^2-t} \, dt$$

$$F^{\prime}(x)=\sqrt{(2x)^2-(2x)} \cdot \frac{d}{dx}(2x)$$

$$F^{\prime}(x)=\sqrt{4x^2-2x} \cdot 2$$

$$F^{\prime}(x)=2\sqrt{4x^2-2x}$$

Q: How can we evaluate the derivative of the function $F(x)=\int_{4}^{2x} \sqrt{t^2-t} \, dt$ at $x=2$?

A: We can evaluate the derivative of the function $F(x)=\int_{4}^{2x} \sqrt{t^2-t} \, dt$ at $x=2$ by substituting $x=2$ into the derivative. We have:

$$F^{\prime}(2)=2\sqrt{4(2)^2-2(2)}$$

$$F^{\prime}(2)=2\sqrt{16-4}$$

$$F^{\prime}(2)=2\sqrt{12}$$

Q: What is the final answer to the problem?

A: The final answer to the problem is $\boxed{2\sqrt{12}}$.

Related Questions

• Find the derivative of the function $f(x)=\int_{0}^{x^2} \sin(t) \, dt$.
• Evaluate the derivative of the function $f(x)=\int_{0}^{x^2} \sin(t) \, dt$ at $x=1$.
• Find the derivative of the function $f(x)=\int_{0}^{x^3} \cos(t) \, dt$.

Solutions to Related Questions

• The derivative of the function $f(x)=\int_{0}^{x^2} \sin(t) \, dt$ is given by $f^{\prime}(x)=2x\sin(x^2)$.
• The derivative of the function $f(x)=\int_{0}^{x^2} \sin(t) \, dt$ evaluated at $x=1$ is given by $f^{\prime}(1)=2\sin(1)$.
• The derivative of the function $f(x)=\int_{0}^{x^3} \cos(t) \, dt$ is given by $f^{\prime}(x)=3x^2\cos(x^3)$.

Conclusion

In this article, we used the fundamental theorem of calculus to find the derivative of a function that involves an integral. We then evaluated the derivative at $x=2$ and found that $F^{\prime}(2)=2\sqrt{12}$. This is the correct answer.