Solving Problems Using the Fundamental Theorem of Calculus

Introduction

Key Concepts

Understanding the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the gap between the two main operations in calculus: differentiation and integration. It consists of two parts:

1. First Part of the FTC: Connects the concept of the integral to antiderivatives. It states that if $f$ is a continuous function on $[a, b]$ and $F$ is an antiderivative of $f$ on $[a, b]$, then: $$\int_{a}^{b} f(x) dx = F(b) - F(a)$$
2. Second Part of the FTC: Provides a way to compute derivatives of integral functions. It states that if $f$ is continuous on an open interval $I$ containing $a$, then the function $F$ defined by: $$F(x) = \int_{a}^{x} f(t) dt$$ is differentiable on $I$, and $F'(x) = f(x)$.

Applications of the FTC

The FTC is instrumental in various applications, including finding the area under curves, solving motion problems, and computing average values of functions. By providing a link between derivatives and integrals, the FTC simplifies the process of solving complex calculus problems.

Evaluating Definite Integrals

To evaluate a definite integral using the FTC, follow these steps:

• Find an antiderivative $F$ of the integrand $f$.
• Compute $F(b) - F(a)$, where $a$ and $b$ are the limits of integration.

Example: Evaluate $$\int_{1}^{3} (2x) dx$$.

Solution:

1. Find an antiderivative of $2x$, which is $x^2$.
2. Compute $x^2 \biggr|_{1}^{3} = 3^2 - 1^2 = 9 - 1 = 8$.

Solving Area Between Curves

The FTC is also used to find the area between two curves. Suppose $f(x)$ and $g(x)$ are continuous on $[a, b]$ and $f(x) \geq g(x)$ for all $x$ in $[a, b]$. The area $A$ between the curves is given by: $$A = \int_{a}^{b} [f(x) - g(x)] dx$$

Example: Find the area between $f(x) = x^2$ and $g(x) = x$ from $x = 0$ to $x = 1$.

Solution:

1. Determine which function is on top. At $x=0$, $f(0)=0$ and $g(0)=0$. At $x=1$, $f(1)=1$ and $g(1)=1$. Test $x=0.5$: $f(0.5)=0.25$, $g(x)=0.5$. Thus, $g(x) \geq f(x)$ on $[0,1]$.
2. Set up the integral: $$A = \int_{0}^{1} [g(x) - f(x)] dx = \int_{0}^{1} [x - x^2] dx$$.
3. Find the antiderivative: $$\int [x - x^2] dx = \frac{1}{2}x^2 - \frac{1}{3}x^3$$.
4. Evaluate from $0$ to $1$: $$\left[\frac{1}{2}(1)^2 - \frac{1}{3}(1)^3\right] - \left[\frac{1}{2}(0)^2 - \frac{1}{3}(0)^3\right] = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.$$

Average Value of a Function

The FTC is also used to determine the average value of a continuous function $f$ on the interval $[a, b]$. The average value $f_{avg}$ is given by: $$f_{avg} = \frac{1}{b - a} \int_{a}^{b} f(x) dx$$.

Example: Find the average value of $f(x) = \sin(x)$ on $[0, \pi]$.

Solution:

1. Set up the average value formula: $$f_{avg} = \frac{1}{\pi - 0} \int_{0}^{\pi} \sin(x) dx = \frac{1}{\pi} \left[ -\cos(x) \right]_{0}^{\pi}.$$
2. Evaluate the integral: $$-\cos(\pi) + \cos(0) = -(-1) + 1 = 2.$$
3. Thus, $f_{avg} = \frac{2}{\pi}$.

Application in Physics: Motion Along a Line

In physics, the FTC is used to determine the displacement of an object when its velocity function is known. If $v(t)$ is the velocity of an object at time $t$, then the displacement from time $a$ to $b$ is: $$\text{Displacement} = \int_{a}^{b} v(t) dt$$.

Example: If an object's velocity is given by $v(t) = 3t^2$ m/s, find its displacement from $t=1$ s to $t=3$ s.

Solution:

1. Set up the integral: $$\int_{1}^{3} 3t^2 dt.$$
2. Find the antiderivative of $3t^2$, which is $t^3$.
3. Evaluate from $1$ to $3$: $$3^3 - 1^3 = 27 - 1 = 26 \text{ meters}.$$

Inverse Applications: Finding Functions from Their Integrals

The FTC can also be used to recover a function $f$ from its integral if additional conditions are given. For example, if $F(x) = \int_{a}^{x} f(t) dt$, then $F'(x) = f(x)$.

Example: If $F(x) = \int_{2}^{x} (4t - 3) dt$, find $F'(x)$.

Solution: $$F'(x) = 4x - 3.$$

Techniques for Evaluating Complex Integrals

Sometimes, integrals may not be straightforward and require techniques such as substitution or integration by parts before applying the FTC. Proper manipulation of the integrand can simplify the process of finding antiderivatives.

Example: Evaluate $$\int_{0}^{\pi/2} \cos^2(x) dx$$.

Solution:

1. Use the power-reduction identity: $$\cos^2(x) = \frac{1 + \cos(2x)}{2}.$$
2. Set up the integral: $$\int_{0}^{\pi/2} \frac{1 + \cos(2x)}{2} dx = \frac{1}{2} \int_{0}^{\pi/2} 1 + \cos(2x) dx.$$
3. Find the antiderivative: $$\frac{1}{2} \left[ x + \frac{1}{2} \sin(2x) \right]_{0}^{\pi/2}.$$
4. Evaluate from $0$ to $\pi/2$: $$\frac{1}{2} \left[ \frac{\pi}{2} + \frac{1}{2} \sin(\pi) - (0 + \frac{1}{2} \sin(0)) \right] = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}.$$

Comparison Table

Summary and Key Takeaways