Second Fundamental Theorem Of Calculus Examples

The second fundamental theorem of calculus establishes a profound connection between differentiation and integration, revealing that they are essentially inverse processes. This theorem provides a powerful tool for evaluating definite integrals and understanding the behavior of functions defined as integrals.

Understanding the Second Fundamental Theorem of Calculus

The theorem comes in two primary forms. Let's explore each with clarity:

• Form 1: If f is a continuous function on an open interval I containing a, then for every x in I, we define a function F as follows:

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

  The theorem states that F is differentiable on I, and its derivative is f(x). In mathematical notation:

  $F'(x) = \frac{d}{dx} \int_a^x f(t) , dt = f(x)$

  In simpler terms, differentiating an integral with a variable upper limit gives you back the original function evaluated at that upper limit.

• Form 2: If f is a continuous function on the closed interval [a, b], and F is any antiderivative of f (i.e., F'(x) = f(x)), then:

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

  This form is what we typically use to evaluate definite integrals. Find an antiderivative, plug in the limits of integration, and subtract.

Let's delve into some examples to solidify our understanding.

Example 1: Applying Form 1 - Differentiating an Integral

Consider the function:

$G(x) = \int_0^x \cos(t^2) , dt$

We want to find G'(x).

Here, f(t) = cos(t²), and the lower limit is a constant (0). Directly applying the first form of the second fundamental theorem, we have:

$G'(x) = \frac{d}{dx} \int_0^x \cos(t^2) , dt = \cos(x^2)$

That's it! We didn't need to actually evaluate the integral to find its derivative. The theorem gives us a shortcut.

Example 2: A Slight Variation on Form 1

Let's look at a slightly more complicated example:

$H(x) = \int_1^{x^3} e^{-t^2} , dt$

Now, the upper limit is x³, not just x. This requires a slight modification using the chain rule. We can think of this as a composite function. Let u = x³. Then,

$H(x) = \int_1^u e^{-t^2} , dt$

By the chain rule:

$\frac{dH}{dx} = \frac{dH}{du} \cdot \frac{du}{dx}$

We know that dH/du = e^-u² (from the second fundamental theorem, form 1). And du/dx = 3x². Therefore:

$\frac{dH}{dx} = e^{-u^2} \cdot 3x^2 = e^{-(x^3)^2} \cdot 3x^2 = 3x^2 e^{-x^6}$

So, the derivative of H(x) is 3x²e^-x⁶.

Example 3: Applying Form 2 - Evaluating a Definite Integral

Evaluate the definite integral:

$\int_1^3 (3x^2 + 2x - 1) , dx$

Here, f(x) = 3x² + 2x - 1. We need to find an antiderivative, F(x), such that F'(x) = f(x). We can find this by using the power rule for integration:

$F(x) = \int (3x^2 + 2x - 1) , dx = x^3 + x^2 - x + C$

Notice the "+ C". Remember that the antiderivative is not unique. However, the constant of integration will cancel out when we apply the second form of the fundamental theorem, so we can ignore it for the purpose of evaluating definite integrals. Let's choose C = 0.

Now, we apply Form 2:

$\int_1^3 (3x^2 + 2x - 1) , dx = F(3) - F(1)$

$= (3^3 + 3^2 - 3) - (1^3 + 1^2 - 1)$

$= (27 + 9 - 3) - (1 + 1 - 1)$

$= 33 - 1$

$= 32$

Therefore, the value of the definite integral is 32.

Example 4: Dealing with Discontinuities

The second fundamental theorem requires the function f to be continuous on the interval of integration. What happens if there's a discontinuity? We need to be careful! Let's consider an example:

$\int_{-1}^2 \frac{1}{x^2} , dx$

It might be tempting to find the antiderivative of 1/x², which is -1/x, and then evaluate it at the limits of integration. However, f(x) = 1/x² has a discontinuity at x = 0, which lies within the interval [-1, 2]. Therefore, we cannot directly apply the second fundamental theorem.

Instead, we must recognize that this is an improper integral and needs to be treated accordingly. We need to split the integral into two parts:

$\int_{-1}^2 \frac{1}{x^2} , dx = \int_{-1}^0 \frac{1}{x^2} , dx + \int_0^2 \frac{1}{x^2} , dx$

Now we evaluate each improper integral separately as a limit:

$\int_{-1}^0 \frac{1}{x^2} , dx = \lim_{b \to 0^-} \int_{-1}^b \frac{1}{x^2} , dx = \lim_{b \to 0^-} \left[ -\frac{1}{x} \right]{-1}^b = \lim{b \to 0^-} \left( -\frac{1}{b} - 1 \right) = \infty$

Since the first integral diverges, the entire integral diverges. The second fundamental theorem cannot be applied directly because the function is not continuous on the entire interval.

Example 5: Combining Form 1 and Form 2

Let's define a function:

$F(x) = \int_0^x t \cdot \sin(t) , dt$

And we want to find F''(x).

First, we find F'(x) using the first form of the second fundamental theorem:

$F'(x) = \frac{d}{dx} \int_0^x t \cdot \sin(t) , dt = x \cdot \sin(x)$

Now, we need to find the derivative of F'(x), which is F''(x). We use the product rule:

$F''(x) = \frac{d}{dx} (x \cdot \sin(x)) = x \cdot \cos(x) + \sin(x) \cdot 1 = x \cos(x) + \sin(x)$

Therefore, F''(x) = x cos(x) + sin(x). In this example, we used Form 1 to find the first derivative and then standard differentiation techniques to find the second derivative.

Example 6: A Piecewise Defined Function

Let

$f(x) = \begin{cases} x^2 & \text{if } x \le 1 \ 2x-1 & \text{if } x > 1 \end{cases}$

and let

$F(x) = \int_0^x f(t) , dt$

Find F'(x).

Since f(x) is piecewise defined, we need to consider two cases:

• Case 1: x ≤ 1

  In this case, f(t) = t² for all t between 0 and x. Therefore,

  $F(x) = \int_0^x t^2 , dt = \left[ \frac{1}{3}t^3 \right]_0^x = \frac{1}{3}x^3$

  And F'(x) = x², which is equal to f(x).

• Case 2: x > 1

  In this case, we need to split the integral into two parts:

  $F(x) = \int_0^x f(t) , dt = \int_0^1 t^2 , dt + \int_1^x (2t-1) , dt$

  We already know that $\int_0^1 t^2 , dt = \frac{1}{3}$ So,

  $F(x) = \frac{1}{3} + \int_1^x (2t-1) , dt = \frac{1}{3} + \left[ t^2 - t \right]_1^x = \frac{1}{3} + (x^2 - x) - (1^2 - 1) = \frac{1}{3} + x^2 - x$

  Therefore, F'(x) = 2x - 1, which is equal to f(x).

In both cases, F'(x) = f(x), as expected by the second fundamental theorem. This example shows how to handle piecewise defined functions when working with the theorem.

Example 7: The Leibniz Rule (A Generalization)

The second fundamental theorem can be generalized to the Leibniz rule, which handles cases where both the upper and lower limits of integration are functions of x. Specifically, if

$H(x) = \int_{a(x)}^{b(x)} f(t) , dt$

Then

$H'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$

Let's apply this to an example:

$H(x) = \int_{x^2}^{x^3} \sin(t) , dt$

Here, f(t) = sin(t), a(x) = x², and b(x) = x³. Therefore, a'(x) = 2x and b'(x) = 3x². Applying the Leibniz rule:

$H'(x) = \sin(x^3) \cdot 3x^2 - \sin(x^2) \cdot 2x = 3x^2 \sin(x^3) - 2x \sin(x^2)$

This shows how the Leibniz rule, a generalization of the second fundamental theorem, allows us to differentiate integrals with variable upper and lower limits.

Example 8: Using Symmetry

Sometimes, recognizing symmetry can simplify the evaluation of definite integrals. Consider:

$\int_{-\pi}^{\pi} x^3 \cos(x) , dx$

The function f(x) = x³cos(x) is an odd function because f(-x) = (-x)³cos(-x) = -x³cos(x) = -f(x). The integral of an odd function over a symmetric interval (like [-π, π]) is always zero. Therefore:

$\int_{-\pi}^{\pi} x^3 \cos(x) , dx = 0$

We didn't even need to find an antiderivative! Recognizing the symmetry saved us a lot of work. While this doesn't directly use the second fundamental theorem in the traditional sense, it demonstrates a valuable technique for simplifying integral calculations.

Example 9: Integration by Substitution First

Consider the integral:

$\int_0^2 x \sqrt{4 - x^2} , dx$

Directly finding the antiderivative might be tricky. However, we can use u-substitution to simplify it. Let u = 4 - x². Then du = -2x dx, or x dx = -1/2 du. Also, we need to change the limits of integration:

• When x = 0, u = 4 - 0² = 4
• When x = 2, u = 4 - 2² = 0

So, the integral becomes:

$\int_4^0 \sqrt{u} \cdot \left( -\frac{1}{2} \right) , du = -\frac{1}{2} \int_4^0 u^{1/2} , du = \frac{1}{2} \int_0^4 u^{1/2} , du$

Now, we can easily find the antiderivative:

$\frac{1}{2} \int_0^4 u^{1/2} , du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_0^4 = \frac{1}{3} \left[ u^{3/2} \right]_0^4 = \frac{1}{3} (4^{3/2} - 0^{3/2}) = \frac{1}{3} (8 - 0) = \frac{8}{3}$

Therefore, the value of the integral is 8/3. This example highlights the importance of using other integration techniques, like u-substitution, in conjunction with the second fundamental theorem.

Example 10: Area Between Curves

The second fundamental theorem is also crucial when calculating the area between two curves. Suppose we want to find the area between the curves y = f(x) and y = g(x) from x = a to x = b, where f(x) ≥ g(x) on the interval [a, b]. The area is given by:

$A = \int_a^b [f(x) - g(x)] , dx$

For example, let's find the area between y = x² and y = √x from x = 0 to x = 1. On this interval, √x ≥ x². Therefore:

$A = \int_0^1 (\sqrt{x} - x^2) , dx = \int_0^1 (x^{1/2} - x^2) , dx = \left[ \frac{2}{3} x^{3/2} - \frac{1}{3} x^3 \right]_0^1 = \left( \frac{2}{3} - \frac{1}{3} \right) - (0 - 0) = \frac{1}{3}$

The area between the curves is 1/3. The second fundamental theorem allows us to calculate this area by finding the antiderivatives of the functions and evaluating them at the limits of integration.

Common Mistakes and How to Avoid Them

• Forgetting the Constant of Integration (when it matters): When finding antiderivatives for indefinite integrals, always remember the "+ C". While it cancels out in definite integrals, it's crucial for finding general antiderivatives.

• Ignoring Discontinuities: The second fundamental theorem only applies when the function is continuous on the interval of integration. Always check for discontinuities and, if present, treat the integral as an improper integral.

• Incorrectly Applying the Chain Rule: When the upper limit of integration is a function of x, remember to use the chain rule when differentiating (as in Example 2).

• Not Recognizing Opportunities for Simplification: Look for symmetry, u-substitution opportunities, or other techniques that can simplify the integral before blindly applying the theorem.

• Confusing the Two Forms: Understand the difference between the two forms of the theorem. Form 1 is for differentiating integrals, and Form 2 is for evaluating definite integrals.

Conclusion

The second fundamental theorem of calculus is a cornerstone of calculus, providing a direct link between differentiation and integration. Mastering this theorem and its applications is crucial for success in calculus and related fields. By understanding the underlying principles and practicing with various examples, you can confidently tackle a wide range of integration problems. Remember to be mindful of the conditions required for the theorem to apply and to explore techniques like u-substitution and recognizing symmetry to simplify your calculations. With practice, you'll become proficient in using this powerful tool to solve complex problems in mathematics, physics, and engineering.