Fundamental Theorem Of Calculus Example Problems

The fundamental theorem of calculus (FTC) is a cornerstone of calculus, connecting differentiation and integration. It simplifies evaluating definite integrals and provides a way to find the area under a curve by relating it to the antiderivative of the function. In essence, the FTC states that differentiation and integration are inverse processes.

Understanding the Fundamental Theorem of Calculus

The fundamental theorem of calculus comes in two parts:

• Part 1: If f is a continuous function on the interval [a, b], and F is defined by:

  F(x) = ∫ₐˣ f(t) dt

  Then F is continuous on [a, b] and differentiable on (a, b), and

  F'(x) = f(x)

• Part 2: If f is a continuous function on the interval [a, b], and F is any antiderivative of f on [a, b], then:

  ∫ₐᵇ f(x) dx = F(b) - F(a)

Part 1 essentially states that the derivative of the integral of a function is the original function itself. Part 2 provides a method to evaluate definite integrals using antiderivatives.

Let's delve into example problems that demonstrate the application of the fundamental theorem of calculus.

Example Problems: Part 1 of the Fundamental Theorem

Problem 1:

Find F'(x) if F(x) = ∫₀ˣ t³ dt

Solution:

According to Part 1 of the FTC, if F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x).

In this case, f(t) = t³, so:

F'(x) = x³

Problem 2:

Find G'(x) if G(x) = ∫₁ˣ cos(t) dt

Solution:

Here, f(t) = cos(t). Applying Part 1 of the FTC:

G'(x) = cos(x)

Problem 3:

Find H'(x) if H(x) = ∫₂ˣ eᵗ dt

Solution:

In this case, f(t) = eᵗ. Applying Part 1 of the FTC:

H'(x) = eˣ

Problem 4:

Find F'(x) if F(x) = ∫₀ˣ (t² + 3t + 1) dt

Solution:

Here, f(t) = t² + 3t + 1. Applying Part 1 of the FTC:

F'(x) = x² + 3x + 1

Problem 5:

Find G'(x) if G(x) = ∫₃ˣ √(1 + t²) dt

Solution:

Here, f(t) = √(1 + t²). Applying Part 1 of the FTC:

G'(x) = √(1 + x²)

Chain Rule Application with Part 1

When the upper limit of integration is a function of x, the chain rule must be applied.

Problem 6:

Find F'(x) if F(x) = ∫₀ˣ² t³ dt

Solution:

Let u = x², then F(x) = ∫₀ᵘ t³ dt. According to Part 1 of the FTC, d/du [∫₀ᵘ t³ dt] = u³. Now, apply the chain rule:

F'(x) = d/dx [∫₀ᵘ t³ dt] = (d/du [∫₀ᵘ t³ dt]) * (du/dx) = u³ * (du/dx)

Since u = x², du/dx = 2x. So,

F'(x) = (x²)³ * 2x = x⁶ * 2x = 2x⁷

Problem 7:

Find G'(x) if G(x) = ∫₁^(sin x) cos(t) dt

Solution:

Let u = sin x, then G(x) = ∫₁ᵘ cos(t) dt. According to Part 1 of the FTC, d/du [∫₁ᵘ cos(t) dt] = cos(u). Now, apply the chain rule:

G'(x) = d/dx [∫₁ᵘ cos(t) dt] = (d/du [∫₁ᵘ cos(t) dt]) * (du/dx) = cos(u) * (du/dx)

Since u = sin x, du/dx = cos x. So,

G'(x) = cos(sin x) * cos x

Problem 8:

Find H'(x) if H(x) = ∫₂^(x³) eᵗ dt

Solution:

Let u = x³, then H(x) = ∫₂ᵘ eᵗ dt. According to Part 1 of the FTC, d/du [∫₂ᵘ eᵗ dt] = eᵘ. Now, apply the chain rule:

H'(x) = d/dx [∫₂ᵘ eᵗ dt] = (d/du [∫₂ᵘ eᵗ dt]) * (du/dx) = eᵘ * (du/dx)

Since u = x³, du/dx = 3x². So,

H'(x) = e^(x³) * 3x² = 3x²e^(x³)

Problem 9:

Find F'(x) if F(x) = ∫₀^(√x) (t² + 3t + 1) dt

Solution:

Let u = √x, then F(x) = ∫₀ᵘ (t² + 3t + 1) dt. According to Part 1 of the FTC, d/du [∫₀ᵘ (t² + 3t + 1) dt] = u² + 3u + 1. Now, apply the chain rule:

F'(x) = d/dx [∫₀ᵘ (t² + 3t + 1) dt] = (d/du [∫₀ᵘ (t² + 3t + 1) dt]) * (du/dx) = (u² + 3u + 1) * (du/dx)

Since u = √x, du/dx = 1/(2√x). So,

F'(x) = ((√x)² + 3√x + 1) * (1/(2√x)) = (x + 3√x + 1) / (2√x)

Problem 10:

Find G'(x) if G(x) = ∫₃^(tan x) √(1 + t²) dt

Solution:

Let u = tan x, then G(x) = ∫₃ᵘ √(1 + t²) dt. According to Part 1 of the FTC, d/du [∫₃ᵘ √(1 + t²) dt] = √(1 + u²). Now, apply the chain rule:

G'(x) = d/dx [∫₃ᵘ √(1 + t²) dt] = (d/du [∫₃ᵘ √(1 + t²) dt]) * (du/dx) = √(1 + u²) * (du/dx)

Since u = tan x, du/dx = sec² x. So,

G'(x) = √(1 + tan² x) * sec² x = √(sec² x) * sec² x = sec x * sec² x = sec³ x

Example Problems: Part 2 of the Fundamental Theorem

Problem 1:

Evaluate ∫₁³ x² dx

Solution:

First, find the antiderivative of x², which is F(x) = (1/3)x³. Then, apply Part 2 of the FTC:

∫₁³ x² dx = F(3) - F(1) = (1/3)(3)³ - (1/3)(1)³ = (1/3)(27) - (1/3)(1) = 9 - 1/3 = 26/3

Problem 2:

Evaluate ∫₀^(π/2) cos(x) dx

Solution:

First, find the antiderivative of cos(x), which is F(x) = sin(x). Then, apply Part 2 of the FTC:

∫₀^(π/2) cos(x) dx = F(π/2) - F(0) = sin(π/2) - sin(0) = 1 - 0 = 1

Problem 3:

Evaluate ∫₁² (2x + 3) dx

Solution:

First, find the antiderivative of (2x + 3), which is F(x) = x² + 3x. Then, apply Part 2 of the FTC:

∫₁² (2x + 3) dx = F(2) - F(1) = (2² + 3(2)) - (1² + 3(1)) = (4 + 6) - (1 + 3) = 10 - 4 = 6

Problem 4:

Evaluate ∫₀¹ eˣ dx

Solution:

First, find the antiderivative of eˣ, which is F(x) = eˣ. Then, apply Part 2 of the FTC:

∫₀¹ eˣ dx = F(1) - F(0) = e¹ - e⁰ = e - 1

Problem 5:

Evaluate ∫₁(e) (1/x) dx

Solution:

First, find the antiderivative of (1/x), which is F(x) = ln|x|. Then, apply Part 2 of the FTC:

∫₁(e) (1/x) dx = F(e) - F(1) = ln(e) - ln(1) = 1 - 0 = 1

Problem 6:

Evaluate ∫₀² x³ dx

Solution:

First, find the antiderivative of x³, which is F(x) = (1/4)x⁴. Then, apply Part 2 of the FTC:

∫₀² x³ dx = F(2) - F(0) = (1/4)(2)⁴ - (1/4)(0)⁴ = (1/4)(16) - (1/4)(0) = 4 - 0 = 4

Problem 7:

Evaluate ∫₀^(π/4) sec²(x) dx

Solution:

First, find the antiderivative of sec²(x), which is F(x) = tan(x). Then, apply Part 2 of the FTC:

∫₀^(π/4) sec²(x) dx = F(π/4) - F(0) = tan(π/4) - tan(0) = 1 - 0 = 1

Problem 8:

Evaluate ∫(-2)¹ (3x² + 2x) dx

Solution:

First, find the antiderivative of (3x² + 2x), which is F(x) = x³ + x². Then, apply Part 2 of the FTC:

∫(-2)¹ (3x² + 2x) dx = F(1) - F(-2) = (1³ + 1²) - ((-2)³ + (-2)²) = (1 + 1) - (-8 + 4) = 2 - (-4) = 6

Problem 9:

Evaluate ∫₁(4) √x dx

Solution:

First, rewrite √x as x^(1/2). Then, find the antiderivative of x^(1/2), which is F(x) = (2/3)x^(3/2). Then, apply Part 2 of the FTC:

∫₁(4) √x dx = F(4) - F(1) = (2/3)(4)^(3/2) - (2/3)(1)^(3/2) = (2/3)(8) - (2/3)(1) = 16/3 - 2/3 = 14/3

Problem 10:

Evaluate ∫₀^(π) sin(x) dx

Solution:

First, find the antiderivative of sin(x), which is F(x) = -cos(x). Then, apply Part 2 of the FTC:

∫₀^(π) sin(x) dx = F(π) - F(0) = -cos(π) - (-cos(0)) = -(-1) - (-1) = 1 + 1 = 2

More Complex Example Problems

Problem 11:

Evaluate ∫₀² |x - 1| dx

Solution:

The function |x - 1| changes its behavior at x = 1. Therefore, split the integral into two parts:

∫₀² |x - 1| dx = ∫₀¹ (1 - x) dx + ∫₁² (x - 1) dx

First, find the antiderivatives:

• For (1 - x), F₁(x) = x - (1/2)x²
• For (x - 1), F₂(x) = (1/2)x² - x

Now, apply Part 2 of the FTC:

∫₀¹ (1 - x) dx = F₁(1) - F₁(0) = (1 - (1/2)(1)²) - (0 - (1/2)(0)²) = 1 - 1/2 = 1/2 ∫₁² (x - 1) dx = F₂(2) - F₂(1) = ((1/2)(2)² - 2) - ((1/2)(1)² - 1) = (2 - 2) - (1/2 - 1) = 0 - (-1/2) = 1/2

Add the results:

∫₀² |x - 1| dx = 1/2 + 1/2 = 1

Problem 12:

Evaluate ∫₀^(π/2) sin(2x) dx

Solution:

First, find the antiderivative of sin(2x), which is F(x) = -(1/2)cos(2x). Then, apply Part 2 of the FTC:

∫₀^(π/2) sin(2x) dx = F(π/2) - F(0) = -(1/2)cos(2(π/2)) - (-(1/2)cos(2(0))) = -(1/2)cos(π) + (1/2)cos(0) = -(1/2)(-1) + (1/2)(1) = 1/2 + 1/2 = 1

Problem 13:

Evaluate ∫(-1)¹ (x² + 1) dx

Solution:

First, find the antiderivative of (x² + 1), which is F(x) = (1/3)x³ + x. Then, apply Part 2 of the FTC:

∫(-1)¹ (x² + 1) dx = F(1) - F(-1) = ((1/3)(1)³ + 1) - ((1/3)(-1)³ + (-1)) = (1/3 + 1) - (-1/3 - 1) = 4/3 - (-4/3) = 4/3 + 4/3 = 8/3

Problem 14:

Evaluate ∫₁(e²) (1/x) dx

Solution:

First, find the antiderivative of (1/x), which is F(x) = ln|x|. Then, apply Part 2 of the FTC:

∫₁(e²) (1/x) dx = F(e²) - F(1) = ln(e²) - ln(1) = 2ln(e) - 0 = 2(1) = 2

Problem 15:

Evaluate ∫₀¹ (x√(1 - x²)) dx

Solution:

Use substitution. Let u = 1 - x², so du = -2x dx. Then, x dx = -1/2 du. When x = 0, u = 1 - 0² = 1. When x = 1, u = 1 - 1² = 0. Rewrite the integral:

∫₀¹ (x√(1 - x²)) dx = ∫₁⁰ √(u) * (-1/2) du = -1/2 ∫₁⁰ u^(1/2) du = 1/2 ∫₀¹ u^(1/2) du

First, find the antiderivative of u^(1/2), which is G(u) = (2/3)u^(3/2). Then, apply Part 2 of the FTC:

1/2 ∫₀¹ u^(1/2) du = 1/2 [G(1) - G(0)] = 1/2 [(2/3)(1)^(3/2) - (2/3)(0)^(3/2)] = 1/2 [2/3 - 0] = 1/2 * 2/3 = 1/3

Applications and Significance

The fundamental theorem of calculus is essential in numerous areas, including:

• Physics: Calculating displacement from velocity functions, or work done by a force.
• Engineering: Designing structures and systems by understanding rates of change.
• Economics: Modeling marginal cost and revenue.
• Statistics: Calculating probabilities and distributions.

By relating derivatives and integrals, the FTC provides a powerful tool for solving problems involving rates of change and accumulation.

Conclusion

The fundamental theorem of calculus is a cornerstone of calculus, bridging the concepts of differentiation and integration. Through various examples, we've demonstrated how to apply both Part 1 and Part 2 of the theorem to solve problems ranging from simple derivatives and integrals to more complex applications involving chain rule and substitution. Mastering the FTC is crucial for anyone studying calculus and its applications in science, engineering, and beyond.