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In each of Exercises \(29-34,\) calculate the derivative of \(F(x)\) with respect to \(x\) $$ F(x)=\int_{x}^{5} \sqrt{2-\sin ^{2}(t)} d t $$

Short Answer

Expert verified
The derivative is \( F'(x) = -\sqrt{2 - \sin^2(x)} \).

Step by step solution

01

Recognize It's an Integral with Variable Limit

The function given is \( F(x) = \int_{x}^{5} \sqrt{2 - \sin^2(t)} \, dt \). Notice that the lower limit of the integral is a variable \( x \). We can apply the Fundamental Theorem of Calculus part 1 to differentiate this integral with respect to \( x \).
02

Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). However, we have an integral from \( x \) to 5, so we need to account for the direction of integration.
03

Swap the Limits and Change the Sign

To use the Fundamental Theorem directly, rewrite the integral from \( 5 \) to \( x \) by changing the sign: \(F(x) = -\int_{5}^{x} \sqrt{2 - \sin^2(t)} \, dt\).
04

Differentiate Using the Theorem

Now apply the Fundamental Theorem of Calculus, differentiating this rewritten integral with respect to \( x \). \ The derivative \( F'(x) = -\sqrt{2 - \sin^2(x)} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) is key when dealing with integrals that include a variable limit. It links the concept of differentiation with that of integration.
Imagine the process like this: if you take the integral of a function and then differentiate the result, you essentially return to the original function. The FTC is usually stated as follows: if you have a function defined by an integral with an upper limit, say \( F(x) = \int_{a}^{x} f(t) \, dt \), then the derivative of this integral, \( F'(x) \), is simply \( f(x) \).
This powerful theorem allows us to move seamlessly between integrals and derivatives, greatly simplifying many problems in calculus. The important part? It emphasizes that the integral of a rate of change gives us the total change over an interval. So when you differentiate an integral, you're essentially peeling back layers to find the original rate of change or function beneath.
Integral with Variable Limit
An integral with a variable limit is slightly more complex than a typical definite integral. Here, the limit of integration is not a constant but depends on the variable itself. In our case, the integral goes from \( x \) to 5, denoted as \( F(x) = \int_{x}^{5} \sqrt{2 - \sin^2(t)} \, dt \).
To tackle integrals like this with variable limits, consider reversing the limits. This adjustment accounts for the integral running in the opposite direction, changing the sign when you swap limits. This change is essential to correctly apply the Fundamental Theorem of Calculus.
  • The limits reconfiguration switches the integral to \( -\int_{5}^{x} \sqrt{2 - \sin^2(t)} \, dt \).
  • Once the limits are reversed, you can utilize the FTC to differentiate the function by using the new limits.
  • Remember: reversing the limits flips the integral's sign, crucial for accuracy in differentiation.
Approaching the problem this way affirms how variable limit integrals are addressed using fundamental calculus principles.
Chain Rule
The Chain Rule becomes vital in differentiation when dealing with compositions of functions. It states that if you have two functions nested within each other, the derivative of the composite function involves the derivatives of both.
However, for our exercise, the function is quite simple, and the role of the Chain Rule is less pronounced. Despite this, its concept remains key in understanding how integrals and derivatives can intertwine, especially when involving more complicated functions or nested expressions.
While the FTC did most of the work in our solution, the Chain Rule's philosophy is echoed in reverse engineering any composed function's integral. Often, in calculus, many complex situations involve recognizing compositions within functions before differentiating—where the Chain Rule guides our intuition.
When learning calculus, always consider how compositions might appear unexpectedly. Differentiation tools like the Chain Rule are essential for breaking down complex problems, revealing the underlying simplicity just like we did with the integral to identify the correct solution form.

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Most popular questions from this chapter

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