Fundamental Theorem Of Calculus With Chain Rule

The fundamental theorem of calculus, a cornerstone of mathematical analysis, intricately links the concepts of differentiation and integration, revealing them as inverse processes. When combined with the chain rule, this theorem becomes an even more potent tool for solving complex calculus problems involving composite functions.

Unveiling the Fundamental Theorem of Calculus

At its core, the fundamental theorem of calculus is presented in two parts, each offering a unique perspective on the relationship between differentiation and integration.

Part 1: The Derivative of an Integral

This part elucidates how differentiation can "undo" integration. Formally, it states:

If f is a continuous function on the interval [a, b], and we define a function F as:

F(x) = ∫ₐˣ f(t) dt

Then, for x in the interval (a, b), the derivative of F is:

F'(x) = d/dx ∫ₐˣ f(t) dt = f(x)

In simpler terms, if you integrate a function f from a constant a to a variable x, and then differentiate the result with respect to x, you end up with the original function f(x).

Example:

Let f(t) = t². Then F(x) = ∫₀ˣ t² dt = [⅓t³]₀ˣ = ⅓x³.

Now, differentiate F(x): F'(x) = d/dx (⅓x³) = x². As you can see, F'(x) = f(x).

Part 2: Evaluating Definite Integrals

This part provides a method for computing definite integrals, which represent the area under a curve between two specific points. It states:

If f is a continuous function on the interval [a, b], and F is any antiderivative of f (i.e., F'(x) = f(x)), then:

∫ₐᵇ f(x) dx = F(b) - F(a)

In essence, to evaluate the definite integral of f from a to b, find an antiderivative F of f, and then subtract the value of F at a from the value of F at b.

Example:

Let f(x) = x². An antiderivative of f(x) is F(x) = ⅓x³.

To evaluate ∫₁³ x² dx, we calculate F(3) - F(1) = (⅓ * 3³) - (⅓ * 1³) = 9 - ⅓ = 26/3.

The Chain Rule: A Quick Refresher

The chain rule is a fundamental concept in differential calculus that allows us to find the derivative of a composite function. A composite function is a function that is composed of another function, such as f(g(x)).

The chain rule states:

If we have a composite function y = f(g(x)), then the derivative of y with respect to x is:

dy/dx = dy/du * du/dx = f'(g(x)) * g'(x)

Where u = g(x). In other words, we take the derivative of the outer function f evaluated at the inner function g(x), and then multiply it by the derivative of the inner function g'(x).

Example:

Let y = sin(x²). Here, f(u) = sin(u) and u = g(x) = x².

Then dy/du = cos(u) and du/dx = 2x.

Applying the chain rule: dy/dx = cos(u) * 2x = cos(x²) * 2x = 2xcos(x²).

Combining the Fundamental Theorem of Calculus and the Chain Rule

The true power of the fundamental theorem of calculus is unleashed when combined with the chain rule. This combination is particularly useful when dealing with integrals where the upper limit of integration is a function of x.

Let's consider a function defined as:

G(x) = ∫ₐᵘ⁽ˣ⁾ f(t) dt

Where a is a constant and u(x) is a differentiable function of x. Our goal is to find the derivative of G(x) with respect to x, i.e., G'(x).

Here's how we apply the fundamental theorem of calculus and the chain rule:

1. Define an intermediate function: Let F(u) = ∫ₐᵘ f(t) dt. According to the fundamental theorem of calculus (Part 1), F'(u) = f(u).

2. Express G(x) as a composite function: Notice that G(x) = F(u(x)). This means that G is a composite function where F is the outer function and u(x) is the inner function.

3. Apply the Chain Rule: Now we can use the chain rule to find the derivative of G(x) with respect to x:

   G'(x) = d/dx [F(u(x))] = F'(u(x)) * u'(x)

4. Substitute using the Fundamental Theorem: We know that F'(u) = f(u), so we can substitute this into the equation:

   G'(x) = f(u(x)) * u'(x)

5. Final Result: Substituting u(x) back into the equation, we get:

   G'(x) = f(u(x)) * u'(x)

Therefore, the derivative of ∫ₐᵘ⁽ˣ⁾ f(t) dt with respect to x is f(u(x)) multiplied by the derivative of u(x).

Example 1:

Let G(x) = ∫₀ˣ² sin(t) dt. Here, f(t) = sin(t) and u(x) = x².

Then u'(x) = 2x.

Applying the formula: G'(x) = f(u(x)) * u'(x) = sin(x²) * 2x = 2xsin(x²).

Example 2:

Let G(x) = ∫₁ˢⁱⁿ⁽ˣ⁾ t³ dt. Here, f(t) = t³ and u(x) = sin(x).

Then u'(x) = cos(x).

Applying the formula: G'(x) = f(u(x)) * u'(x) = (sin(x))³ * cos(x) = sin³(x)cos(x).

Example 3:

Let G(x) = ∫₂ˣ³ + ¹ cos(t²) dt. Here, f(t) = cos(t²) and u(x) = x³ + 1.

Then u'(x) = 3x².

Applying the formula: G'(x) = f(u(x)) * u'(x) = cos((x³ + 1)²) * 3x² = 3x²cos((x³ + 1)²).

Advanced Applications and Considerations

The combination of the fundamental theorem of calculus and the chain rule has far-reaching implications and applications in various fields, including physics, engineering, and economics.

• Variable Lower and Upper Limits: The formula G'(x) = f(u(x)) * u'(x) is derived assuming the lower limit of integration is a constant. If both the upper and lower limits are functions of x, say u(x) and v(x) respectively, then the formula becomes:

  d/dx [∫ᵥ⁽ˣ⁾ᵘ⁽ˣ⁾ f(t) dt] = f(u(x)) * u'(x) - f(v(x)) * v'(x)

  This is derived by splitting the integral into two parts, each with a constant limit, and applying the formula derived above.

• Leibniz Integral Rule: The general form of differentiating an integral with variable limits of integration is known as the Leibniz Integral Rule. This rule provides a more comprehensive framework for handling integrals where the integrand also depends on x:

  d/dx [∫ₐ⁽ˣ⁾ᵇ⁽ˣ⁾ f(x, t) dt] = f(x, b(x)) * b'(x) - f(x, a(x)) * a'(x) + ∫ₐ⁽ˣ⁾ᵇ⁽ˣ⁾ ∂/∂x f(x, t) dt

  This rule incorporates the chain rule and also accounts for the case where the function being integrated explicitly depends on the variable with respect to which we are differentiating. The fundamental theorem of calculus with the chain rule is a special case of the Leibniz Integral Rule where f does not explicitly depend on x.

• Applications in Differential Equations: The fundamental theorem of calculus is crucial in solving differential equations, particularly those involving integral equations. By differentiating both sides of an integral equation using the fundamental theorem and the chain rule, one can often transform the integral equation into a more manageable differential equation.

• Physics and Engineering: In physics, this combination is used to solve problems related to work, energy, and motion. For example, calculating the work done by a force that varies with position often involves evaluating an integral with variable limits, requiring the application of these concepts. Similarly, in engineering, it is used in analyzing systems with time-varying inputs and outputs.

Common Mistakes to Avoid

When applying the fundamental theorem of calculus with the chain rule, several common mistakes can lead to incorrect results. Being aware of these pitfalls can help ensure accuracy:

1. Forgetting the Derivative of the Inner Function: The most common error is forgetting to multiply by u'(x), the derivative of the upper limit function. Remember that the chain rule requires you to differentiate the outer function and multiply by the derivative of the inner function.

2. Incorrectly Identifying the Integrand: Ensure that you correctly identify the function being integrated, f(t), and that you are substituting u(x) into f(t) correctly.

3. Ignoring Constant Lower Limits: The fundamental theorem (Part 1) applies directly when the lower limit of integration is a constant. If the lower limit is also a function of x, you need to use the more general form (as discussed earlier), which involves subtracting a term corresponding to the lower limit.

4. Misapplying the Leibniz Rule: When the integrand f explicitly depends on x, remember to include the additional integral term in the Leibniz rule. Failing to do so will lead to an incorrect result.

5. Confusing Variables: Be careful not to confuse the variable of integration (t in our notation) with the variable with respect to which you are differentiating (x).

Practice Problems and Solutions

To solidify your understanding, let's work through a few more practice problems:

Problem 1:

Find G'(x) if G(x) = ∫₀ᶜᵒˢ⁽ˣ⁾ eᵗ dt.

Solution:

Here, f(t) = eᵗ and u(x) = cos(x). Then u'(x) = -sin(x).

Applying the formula: G'(x) = f(u(x)) * u'(x) = eᶜᵒˢ⁽ˣ⁾ * (-sin(x)) = -sin(x)eᶜᵒˢ⁽ˣ⁾.

Problem 2:

Find G'(x) if G(x) = ∫ₓ²ˣ t² + 1 dt.

Solution:

Here we have both upper and lower limits as functions of x. Let u(x) = 2x and v(x) = x. Then u'(x) = 2 and v'(x) = 1. Also, f(t) = t² + 1.

Then G'(x) = f(u(x)) * u'(x) - f(v(x)) * v'(x) = ((2x)² + 1) * 2 - (x² + 1) * 1 = (4x² + 1) * 2 - (x² + 1) = 8x² + 2 - x² - 1 = 7x² + 1

Problem 3:

Find G'(x) if G(x) = ∫₀ˣ sin(xt) dt.

Solution: This example requires the full Leibniz integral rule since the integrand depends explicitly on x.

Here, f(x, t) = sin(xt). Then ∂/∂x f(x, t) = tcos(xt). Also, a(x) = 0, b(x) = x, a'(x) = 0, and b'(x) = 1.

Using the Leibniz rule: G'(x) = f(x, b(x)) * b'(x) - f(x, a(x)) * a'(x) + ∫ₐ⁽ˣ⁾ᵇ⁽ˣ⁾ ∂/∂x f(x, t) dt = sin(xx) * 1 - sin(x0) * 0 + ∫₀ˣ tcos(xt) dt = sin(x²) + ∫₀ˣ tcos(xt) dt

The remaining integral ∫₀ˣ tcos(xt) dt can be solved using integration by parts, but this problem illustrates how the Leibniz rule is applied when the integrand explicitly depends on x.

Conclusion

The fundamental theorem of calculus, when coupled with the chain rule, provides a powerful framework for differentiating integrals with variable limits. This combination is not only a theoretical cornerstone of calculus but also a practical tool for solving problems in various scientific and engineering disciplines. By understanding the underlying principles, practicing with examples, and avoiding common mistakes, you can effectively leverage this powerful technique to tackle complex calculus problems. Mastery of these concepts unlocks a deeper understanding of the interwoven relationship between differentiation and integration, solidifying your grasp on the fundamentals of calculus.