Fundamental Theorem Of Calculus Practice Problems

The fundamental theorem of calculus connects the two central operations of calculus: differentiation and integration. Essentially, it states that differentiation and integration are inverse processes. This theorem has two parts, each offering a distinct yet related perspective on this relationship.

Understanding the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) comes in two parts:

• Part 1: Deals with the derivative of an integral function.
• Part 2: Provides a method to compute definite integrals using antiderivatives.

This article will cover the FTC in detail and provide numerous practice problems to help you master the concepts.

Part 1: The Derivative of an Integral Function

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

F(x) = ∫ₐˣ f(t) dt

Then F is continuous on [a, b], differentiable on the open interval (a, b), and:

F'(x) = d/dx ∫ₐˣ f(t) dt = f(x)

In simpler terms, Part 1 tells us that if we define a new function F(x) as the integral of another function f(t) from a constant a to a variable x, then the derivative of F(x) is simply f(x).

Part 2: Evaluating Definite Integrals

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:

∫ₐᵇ f(x) dx = F(b) - F(a)

This part of the theorem is incredibly useful for evaluating definite integrals. It states that if you can find an antiderivative F(x) of the integrand f(x), you can evaluate the definite integral by simply subtracting the value of F(x) at the lower limit of integration (a) from the value of F(x) at the upper limit of integration (b).

Practice Problems: Part 1 of the FTC

Let's start with problems focusing on Part 1 of the Fundamental Theorem of Calculus.

Problem 1:

Find F'(x) if F(x) = ∫₀ˣ t³ dt

Solution:

Using Part 1 of the FTC, we directly substitute x for t in the integrand:

F'(x) = x³

Problem 2:

Find G'(x) if G(x) = ∫₁ˣ cos(t²) dt

Solution:

Applying Part 1, we have:

G'(x) = cos(x²)

Problem 3:

Find H'(x) if H(x) = ∫ₓ² sin(t) dt

Solution:

Here, the variable x is in the lower limit. We can reverse the limits of integration by changing the sign:

H(x) = -∫₀ˣ² sin(t) dt

Now, we need to use the chain rule. Let u = x². Then, H(x) = -∫₀ᵘ sin(t) dt.

dH/dx = dH/du * du/dx = -sin(u) * 2x = -sin(x²) * 2x = -2x sin(x²)

Problem 4:

Find K'(x) if K(x) = ∫√ₓ ⁵ e^(t²) dt

Solution:

Let u = √x. Then, K(x) = ∫ᵤ⁵ e^(t²) dt = -∫₅ᵘ e^(t²) dt.

dK/dx = dK/du * du/dx = -e^(u²) * (1 / (2√x)) = -e^(x) / (2√x)

Problem 5:

Find L'(x) if L(x) = ∫ₓ^(x²) (t² + 1) dt

Solution:

We need to split the integral into two parts using a constant, say 0:

L(x) = ∫ₓ⁰ (t² + 1) dt + ∫₀^(x²) (t² + 1) dt

L(x) = -∫₀ˣ (t² + 1) dt + ∫₀^(x²) (t² + 1) dt

Now, let u = x². Then, the second integral becomes ∫₀ᵘ (t² + 1) dt.

L'(x) = -(x² + 1) + (u² + 1) * (du/dx) = -(x² + 1) + ((x²)² + 1) * 2x = -(x² + 1) + (x⁴ + 1) * 2x = -x² - 1 + 2x⁵ + 2x = 2x⁵ - x² + 2x - 1

Practice Problems: Part 2 of the FTC

Now, let's work through problems focusing on Part 2 of the Fundamental Theorem of Calculus, which deals with evaluating definite integrals.

Problem 1:

Evaluate ∫₁³ (x² + 2x) dx

Solution:

First, find the antiderivative of x² + 2x:

F(x) = (x³/3) + x²

Now, evaluate F(3) - F(1):

F(3) = (3³/3) + 3² = 9 + 9 = 18 F(1) = (1³/3) + 1² = 1/3 + 1 = 4/3

∫₁³ (x² + 2x) dx = F(3) - F(1) = 18 - 4/3 = 54/3 - 4/3 = 50/3

Problem 2:

Evaluate ∫₀^(π/2) cos(x) dx

Solution:

The antiderivative of cos(x) is sin(x):

F(x) = sin(x)

Now, evaluate F(π/2) - F(0):

F(π/2) = sin(π/2) = 1 F(0) = sin(0) = 0

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

Problem 3:

Evaluate ∫₋₂¹ (3x² - 4x + 5) dx

Solution:

Find the antiderivative:

F(x) = x³ - 2x² + 5x

Evaluate F(1) - F(-2):

F(1) = 1³ - 2(1)² + 5(1) = 1 - 2 + 5 = 4 F(-2) = (-2)³ - 2(-2)² + 5(-2) = -8 - 8 - 10 = -26

∫₋₂¹ (3x² - 4x + 5) dx = F(1) - F(-2) = 4 - (-26) = 30

Problem 4:

Evaluate ∫₁⁴ (√x + 1/x) dx

Solution:

Rewrite √x as x^(1/2):

∫₁⁴ (x^(1/2) + 1/x) dx

Find the antiderivative:

F(x) = (2/3)x^(3/2) + ln|x|

Evaluate F(4) - F(1):

F(4) = (2/3)(4)^(3/2) + ln(4) = (2/3)(8) + ln(4) = 16/3 + ln(4) F(1) = (2/3)(1)^(3/2) + ln(1) = 2/3 + 0 = 2/3

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

Problem 5:

Evaluate ∫₀^(π/4) sec²(x) dx

Solution:

The antiderivative of sec²(x) is tan(x):

F(x) = tan(x)

Evaluate F(π/4) - F(0):

F(π/4) = tan(π/4) = 1 F(0) = tan(0) = 0

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

More Challenging Problems

Let's increase the difficulty with some more involved problems that require a deeper understanding of the Fundamental Theorem of Calculus.

Problem 1:

Find d/dx ∫_(x²) ^(x³) sin(t²) dt

Solution:

Let F(x) = ∫_(x²) ^(x³) sin(t²) dt. We can rewrite this as:

F(x) = ∫(0) ^(x³) sin(t²) dt - ∫(0) ^(x²) sin(t²) dt

Now, let u = x³ and v = x². Then,

F(x) = ∫(0) ^(u) sin(t²) dt - ∫(0) ^(v) sin(t²) dt

Using the chain rule:

dF/dx = (d/du ∫(0) ^(u) sin(t²) dt) * (du/dx) - (d/dv ∫(0) ^(v) sin(t²) dt) * (dv/dx)

dF/dx = sin(u²) * (3x²) - sin(v²) * (2x)

Substitute back u = x³ and v = x²:

dF/dx = sin((x³)² ) * (3x²) - sin((x²)²) * (2x)

dF/dx = 3x² sin(x⁶) - 2x sin(x⁴)

Problem 2:

If g(x) = ∫(0) ^(x) f(t) dt, where f(t) = ∫(1) ^(t²) sqrt(1 + u⁴) du, find g''(x).

Solution:

First, find g'(x) using Part 1 of the FTC:

g'(x) = f(x) = ∫_(1) ^(x²) sqrt(1 + u⁴) du

Now, find g''(x). To do this, we need to differentiate f(x). Let v = x². Then, f(x) = ∫_(1) ^(v) sqrt(1 + u⁴) du.

Using the chain rule:

g''(x) = f'(x) = (d/dv ∫_(1) ^(v) sqrt(1 + u⁴) du) * (dv/dx)

g''(x) = sqrt(1 + v⁴) * (2x)

Substitute back v = x²:

g''(x) = 2x * sqrt(1 + (x²)⁴) = 2x * sqrt(1 + x⁸)

Problem 3:

Let F(x) = ∫_(0) ^(x) x*f(t) dt, where f is a continuous function. Express F'(x) in terms of f(x).

Solution:

Notice that the variable x is outside the integral with respect to t. We can pull x out of the integral:

F(x) = x * ∫_(0) ^(x) f(t) dt

Now, we need to use the product rule to find F'(x):

F'(x) = (d/dx x) * ∫(0) ^(x) f(t) dt + x * (d/dx ∫(0) ^(x) f(t) dt)

F'(x) = 1 * ∫_(0) ^(x) f(t) dt + x * f(x)

F'(x) = ∫_(0) ^(x) f(t) dt + x*f(x)

Problem 4:

Evaluate lim_(h->0) (1/h) ∫_(x) ^(x+h) t² cos(t) dt

Solution:

This limit represents the derivative of the function F(x) = ∫_(a) ^(x) t² cos(t) dt, where 'a' is a constant.

By Part 1 of the FTC:

F'(x) = x² cos(x)

Therefore,

lim_(h->0) (1/h) ∫_(x) ^(x+h) t² cos(t) dt = x² cos(x)

Problem 5:

Given that ∫(0) ^(3) f(x) dx = 6 and ∫(0) ^(3) g(x) dx = 4, evaluate ∫_(0) ^(3) [2f(x) - 5g(x) + 3] dx

Solution:

Using the properties of definite integrals:

∫(0) ^(3) [2f(x) - 5g(x) + 3] dx = 2 ∫(0) ^(3) f(x) dx - 5 ∫(0) ^(3) g(x) dx + ∫(0) ^(3) 3 dx

Substitute the given values:

= 2 * (6) - 5 * (4) + [3x]_(0) ^(3)

= 12 - 20 + (33 - 30)

= 12 - 20 + 9

= 1

Tips and Tricks for Solving FTC Problems

• Always Identify the Form: Determine whether the problem requires Part 1 or Part 2 of the FTC. Part 1 involves derivatives of integrals, while Part 2 involves evaluating definite integrals.
• Chain Rule is Your Friend: When the limits of integration are functions of x, remember to apply the chain rule.
• Simplify When Possible: Before applying the FTC, simplify the integrand or the limits of integration if possible.
• Reverse Limits with Caution: Remember that reversing the limits of integration changes the sign of the integral: ∫ₐᵇ f(x) dx = -∫ᵇₐ f(x) dx.
• Practice Regularly: The more you practice, the more comfortable you will become with the various types of problems and techniques.

Common Mistakes to Avoid

• Forgetting the Chain Rule: This is a common mistake when dealing with Part 1 of the FTC and the limits of integration are functions of x.
• Incorrect Antiderivatives: Make sure you find the correct antiderivative when using Part 2 of the FTC. A mistake here will lead to an incorrect answer.
• Ignoring the Order of Subtraction: When using Part 2, remember to subtract the value of the antiderivative at the lower limit from the value at the upper limit: F(b) - F(a).
• Not Simplifying: Sometimes, simplifying the integrand or the limits of integration can make the problem much easier.
• Mixing Up the Parts: Be clear on which part of the FTC you are using. Applying the wrong part will lead to incorrect results.

Conclusion

The Fundamental Theorem of Calculus is a cornerstone of calculus, linking differentiation and integration in a profound way. Mastering the FTC is essential for success in calculus and related fields. Through understanding the two parts of the theorem and working through a variety of practice problems, you can develop a strong foundation in this critical concept. Remember to pay attention to details, apply the chain rule when necessary, and always double-check your work. Keep practicing, and you'll find yourself confidently tackling even the most challenging FTC problems.