Derivative Of Integral Function F(x): Step-by-Step Solution

Have you ever stumbled upon a function defined as an integral and wondered how to find its derivative? It might seem daunting at first, but with the Fundamental Theorem of Calculus, it becomes a straightforward process. In this guide, we'll break down how to find the derivative of an integral function, using the example $F(x) = \int_x^2 \frac{6}{4t+7} dt$. Let's dive in!

Understanding the Fundamental Theorem of Calculus

Before we jump into the specifics, let's refresh our understanding of the Fundamental Theorem of Calculus (FTC). This theorem is the cornerstone of calculus, linking differentiation and integration. It comes in two parts, but the one we're most interested in here is the first part, often referred to as FTC Part 1.

FTC Part 1 states that if we have a function $F(x)$ defined as an integral with a variable upper limit:

$F(x) = \int_a^x f(t) dt$

where $a$ is a constant, then the derivative of $F(x)$ is simply the integrand evaluated at $x$:

$F'(x) = f(x)$

In simpler terms, differentiating an integral "undoes" the integration, and we're left with the original function inside the integral, but with the variable of integration ($t$) replaced by the variable of differentiation ($x$).

This powerful theorem provides a direct way to find the derivative of many integral functions. However, our example function, $F(x) = \int_x^2 \frac{6}{4t+7} dt$, has a slight twist: the variable limit of integration is in the lower bound, not the upper bound. Don't worry, we can handle this with a simple property of definite integrals.

Dealing with Reversed Limits of Integration

One crucial property of definite integrals is that reversing the limits of integration changes the sign of the integral. Mathematically, this means:

$\int_a^b f(t) dt = -\int_b^a f(t) dt$

This property is essential for aligning our function with the form required by FTC Part 1. In our case, we have $F(x) = \int_x^2 \frac{6}{4t+7} dt$. To apply FTC Part 1, we need the variable limit ($x$) to be the upper limit. So, we reverse the limits and change the sign:

$F(x) = -\int_2^x \frac{6}{4t+7} dt$

Now our function is in the correct form for applying the Fundamental Theorem of Calculus!

Applying the Fundamental Theorem of Calculus

With our function rewritten as $F(x) = -\int_2^x \frac{6}{4t+7} dt$, we can now directly apply FTC Part 1. The integrand is $f(t) = \frac{6}{4t+7}$, and we want to find $F'(x)$.

According to FTC Part 1, we simply replace $t$ with $x$ in the integrand and multiply by the constant factor (in this case, -1 from reversing the limits):

$F'(x) = -\frac{6}{4x+7}$

And that's it! We've found the derivative of our integral function.

Key Takeaway: The Fundamental Theorem of Calculus provides a powerful tool for differentiating functions defined as integrals. By understanding FTC Part 1 and the properties of definite integrals, you can tackle a wide range of problems involving derivatives of integral functions.

A Deeper Dive into the Concepts

To truly master this concept, let's delve a bit deeper into the underlying principles.

The Fundamental Theorem of Calculus connects the seemingly disparate concepts of differentiation and integration. It reveals that these two operations are, in a sense, inverses of each other. Thinking about the derivative as the instantaneous rate of change and the integral as the accumulation of a quantity over an interval, we can begin to see this connection intuitively.

The first part of the theorem, which we used in our example, focuses on how to differentiate an integral. It tells us that the rate at which the integral accumulates is precisely the value of the function being integrated at the upper limit. This is a powerful result that simplifies the process of finding derivatives of integral functions.

The second part of the Fundamental Theorem of Calculus provides a method for evaluating definite integrals. It states that if we can find an antiderivative $G(x)$ of a function $f(x)$, then the definite integral of $f(x)$ from $a$ to $b$ is simply the difference in the values of $G(x)$ at $b$ and $a$:

$\int_a^b f(x) dx = G(b) - G(a)$

This part of the theorem is crucial for evaluating definite integrals, as it allows us to bypass the often cumbersome process of finding limits of Riemann sums.

Together, the two parts of the Fundamental Theorem of Calculus form the bedrock of calculus, providing the tools and understanding necessary to solve a vast array of problems in mathematics, physics, engineering, and other fields.

Common Mistakes to Avoid

While applying the Fundamental Theorem of Calculus is relatively straightforward, there are a few common mistakes to watch out for:

• Forgetting to reverse the limits of integration: As we saw in our example, if the variable limit is in the lower bound, you must reverse the limits and change the sign before applying FTC Part 1. Failing to do so will result in the wrong answer.
• Incorrectly applying the chain rule: If the upper limit of integration is a function of $x$ (e.g., $\int_a^{g(x)} f(t) dt$), you'll need to use the chain rule in conjunction with FTC Part 1. The derivative will be $F'(x) = f(g(x)) * g'(x)$.
• Confusing the variable of integration: Remember that the variable of integration ($t$ in our example) is a dummy variable. It disappears when you evaluate the integral or take the derivative. Don't accidentally include it in your final answer.
• Not simplifying the result: After applying FTC Part 1, always simplify your answer as much as possible. This will make it easier to work with in subsequent calculations.

By being mindful of these common pitfalls, you can ensure that you're applying the Fundamental Theorem of Calculus correctly and efficiently.

Example Problems and Solutions

Let's solidify our understanding with a few more examples.

Example 1:

Find $F'(x)$ if $F(x) = \int_0^x (t^2 + 1) dt$.

Solution:

This is a direct application of FTC Part 1. The upper limit is $x$, and the integrand is $f(t) = t^2 + 1$. Therefore,

$F'(x) = x^2 + 1$

Example 2:

Find $F'(x)$ if $F(x) = \int_x^1 \sqrt{t^3 + 1} dt$.

Solution:

First, we reverse the limits of integration:

$F(x) = -\int_1^x \sqrt{t^3 + 1} dt$

Now we can apply FTC Part 1. The integrand is $f(t) = \sqrt{t^3 + 1}$, so

$F'(x) = -\sqrt{x^3 + 1}$

Example 3:

Find $F'(x)$ if $F(x) = \int_0^{x^2} \sin(t) dt$.

Solution:

This example requires the chain rule. The upper limit is $g(x) = x^2$, and the integrand is $f(t) = \sin(t)$. Applying FTC Part 1 and the chain rule, we get

$F'(x) = \sin(x^2) * (x^2)' = 2x\sin(x^2)$

By working through these examples, you'll gain confidence in applying the Fundamental Theorem of Calculus in various situations.

Conclusion

Finding the derivative of an integral function might seem intimidating at first, but with the Fundamental Theorem of Calculus, it becomes a manageable task. Remember to check the limits of integration, apply the theorem carefully, and simplify your result. With practice, you'll master this essential calculus skill. By understanding the Fundamental Theorem of Calculus and its applications, you've unlocked a powerful tool for solving a wide range of calculus problems. Keep practicing, and you'll become even more proficient in this area. For further exploration of calculus concepts, visit a trusted resource like Khan Academy's Calculus section.