Solved Exercises on Fundamental Theorem of Calculus Part 1: Accumulation Functions

Master FTC Part 1: accumulation functions, derivatives of integrals, and the relationship between differentiation and integration through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Accumulation Function
Exercise 1
Let \(F(x) = \int_1^x (t^2 + 2) dt\). Find \(F'(x)\) using the Fundamental Theorem of Calculus Part 1.
Definition:

Accumulation Function: \(F(x) = \int_a^x f(t) dt\) represents the accumulated area under the curve \(f(t)\) from a fixed point \(a\) to a variable point \(x\).

FTC Part 1 Method:
  1. Identify the accumulation function \(F(x) = \int_a^x f(t) dt\)
  2. Apply FTC Part 1: \(F'(x) = f(x)\)
  3. The derivative of the accumulation function equals the integrand evaluated at \(x\)
Accumulation
\(F(x) = \int_1^x (t^2 + 2) dt\)
Derivative
\(F'(x) = x^2 + 2\)
Step 1: Identify the accumulation function

We have \(F(x) = \int_1^x (t^2 + 2) dt\)

This is an accumulation function with integrand \(f(t) = t^2 + 2\)

Step 2: Apply FTC Part 1

According to FTC Part 1: If \(F(x) = \int_a^x f(t) dt\), then \(F'(x) = f(x)\)

This means the derivative of an accumulation function is the original integrand with the variable \(t\) replaced by \(x\)

Step 3: Find the derivative

Since the integrand is \(f(t) = t^2 + 2\), we have:

\(F'(x) = x^2 + 2\)

Step 4: Verification (optional)

We can verify by first evaluating the integral:

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

\(F(x) = \frac{x^3}{3} + 2x - \frac{7}{3}\)

\(F'(x) = x^2 + 2\) ✓

\(F'(x) = x^2 + 2\)
Final answer:

\(F'(x) = x^2 + 2\)

Applied rules:

• FTC Part 1: \(\frac{d}{dx}\int_a^x f(t) dt = f(x)\)

• Direct substitution: Replace the dummy variable with the upper limit

2 Variable Lower Limit
Exercise 2
Let \(G(x) = \int_x^4 (2t - 1) dt\). Find \(G'(x)\) using FTC Part 1 and the property of reversing limits.
Definition:

Reversing Limits Property: \(\int_a^b f(t) dt = -\int_b^a f(t) dt\). When the variable appears as the lower limit, we can rewrite the integral.

Original
\(G(x) = \int_x^4 (2t - 1) dt\)
Reversed
\(G(x) = -\int_4^x (2t - 1) dt\)
Derivative
\(G'(x) = -(2x - 1)\)
Step 1: Recognize the variable in the lower limit

We have \(G(x) = \int_x^4 (2t - 1) dt\), where the variable \(x\) appears as the lower limit

Step 2: Apply the property of reversing limits

\(\int_x^4 (2t - 1) dt = -\int_4^x (2t - 1) dt\)

So \(G(x) = -\int_4^x (2t - 1) dt\)

Step 3: Apply FTC Part 1 to the new form

Let \(H(x) = \int_4^x (2t - 1) dt\), then \(H'(x) = 2x - 1\)

Since \(G(x) = -H(x)\), we have \(G'(x) = -H'(x)\)

Step 4: Find the derivative

\(G'(x) = -(2x - 1) = -2x + 1\)

Step 5: Alternative approach using chain rule

We can also think of this as: \(G(x) = \int_x^4 (2t - 1) dt = -\int_4^x (2t - 1) dt\)

Using the chain rule: \(\frac{d}{dx}\int_{u(x)}^{v(x)} f(t) dt = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)\)

Here, \(v(x) = 4\) (so \(v'(x) = 0\)) and \(u(x) = x\) (so \(u'(x) = 1\))

\(G'(x) = f(4) \cdot 0 - f(x) \cdot 1 = -(2x - 1)\)

\(G'(x) = -2x + 1\)
Final answer:

\(G'(x) = -2x + 1\)

Applied rules:

• Limit reversal: \(\int_a^b f(t) dt = -\int_b^a f(t) dt\)

• FTC Part 1: \(\frac{d}{dx}\int_a^x f(t) dt = f(x)\)

• Chain rule: For variable limits in both positions

3 Chain Rule Application
Exercise 3
Let \(H(x) = \int_0^{x^2} \sin(t) dt\). Find \(H'(x)\) using FTC Part 1 and the chain rule.
Definition:

Chain Rule for Integrals: If \(F(x) = \int_a^{u(x)} f(t) dt\), then \(F'(x) = f(u(x)) \cdot u'(x)\). This combines FTC Part 1 with the chain rule.

Function
\(H(x) = \int_0^{x^2} \sin(t) dt\)
Upper limit
\(u(x) = x^2\)
Derivative
\(H'(x) = \sin(x^2) \cdot 2x\)
Step 1: Identify the composite structure

We have \(H(x) = \int_0^{x^2} \sin(t) dt\)

This is of the form \(\int_a^{u(x)} f(t) dt\) where \(u(x) = x^2\) and \(f(t) = \sin(t)\)

Step 2: Apply FTC Part 1 with the chain rule

When the upper limit is a function of \(x\), we use the generalized form:

If \(F(x) = \int_a^{u(x)} f(t) dt\), then \(F'(x) = f(u(x)) \cdot u'(x)\)

Step 3: Identify components

\(f(t) = \sin(t)\)

\(u(x) = x^2\), so \(u'(x) = 2x\)

Step 4: Apply the formula

\(H'(x) = f(u(x)) \cdot u'(x) = \sin(x^2) \cdot 2x = 2x\sin(x^2)\)

Step 5: Verification (conceptual)

The derivative represents the rate of change of the accumulated area under \(\sin(t)\) as the upper limit \(x^2\) changes

When \(x\) changes, the upper limit changes at rate \(2x\), and the integrand at that point is \(\sin(x^2)\)

\(H'(x) = 2x\sin(x^2)\)
Final answer:

\(H'(x) = 2x\sin(x^2)\)

Applied rules:

• FTC Part 1 with chain rule: \(\frac{d}{dx}\int_a^{u(x)} f(t) dt = f(u(x)) \cdot u'(x)\)

• Chain rule: For composite functions

• Trigonometric function: \(\sin(x^2)\) remains as is

Summary: FTC Part 1 - Accumulation Functions
\(\frac{d}{dx}\int_a^x f(t) dt = f(x)\)
Fundamental Theorem of Calculus Part 1
Basic FTC
\(\frac{d}{dx}\int_a^x f(t) dt = f(x)\)
Variable upper limit
Reversed Limits
\(\frac{d}{dx}\int_x^a f(t) dt = -f(x)\)
Variable lower limit
Chain Rule
\(\frac{d}{dx}\int_a^{u(x)} f(t) dt = f(u(x)) \cdot u'(x)\)
Composite upper limit
Key definitions:

Accumulation function: A function defined as \(F(x) = \int_a^x f(t) dt\) that represents the accumulated area under \(f(t)\) from a fixed point to a variable point.

Integrand: The function \(f(t)\) inside the integral that determines the rate of accumulation.

Antiderivative relationship: FTC Part 1 establishes that integration and differentiation are inverse operations.

Complete methodology:
  1. Identify the form: Determine if it's \(\int_a^x\), \(\int_x^a\), \(\int_{u(x)}^{v(x)}\), etc.
  2. Apply FTC Part 1: Recognize that the derivative "undoes" the integral
  3. Handle special cases: Apply chain rule for composite limits, negative signs for reversed limits
  4. Substitute appropriately: Replace the dummy variable with the appropriate limit
Tip 1: The dummy variable (t) inside the integral becomes the variable of the upper limit in the derivative.
Tip 2: When the variable is in the lower limit, add a negative sign.
Tip 3: For composite limits, multiply by the derivative of the limit function (chain rule).
Tip 4: The constant lower limit disappears in the derivative.
Key insight: FTC Part 1 shows that integration and differentiation are inverse operations.
Physical interpretation: The rate of accumulation equals the current rate of change.
Solution: Exercises 4 to 5
4 Complex Chain Rule Application
Exercise 4
Let \(K(x) = \int_{x^3}^{2x} e^{t^2} dt\). Find \(K'(x)\) using FTC Part 1 and the generalized chain rule.
Definition:

Generalized FTC: If \(F(x) = \int_{u(x)}^{v(x)} f(t) dt\), then \(F'(x) = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)\).

Function
\(K(x) = \int_{x^3}^{2x} e^{t^2} dt\)
Limits
\(u(x) = x^3, v(x) = 2x\)
Derivative
\(K'(x) = e^{(2x)^2} \cdot 2 - e^{(x^3)^2} \cdot 3x^2\)
Step 1: Identify the form

We have \(K(x) = \int_{x^3}^{2x} e^{t^2} dt\)

This is of the form \(\int_{u(x)}^{v(x)} f(t) dt\) where:

\(u(x) = x^3\) (lower limit), \(v(x) = 2x\) (upper limit), \(f(t) = e^{t^2}\)

Step 2: Apply the generalized FTC formula

For \(F(x) = \int_{u(x)}^{v(x)} f(t) dt\), the derivative is:

\(F'(x) = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)\)

Step 3: Find the derivatives of the limits

\(v(x) = 2x\), so \(v'(x) = 2\)

\(u(x) = x^3\), so \(u'(x) = 3x^2\)

Step 4: Evaluate the integrand at the limits

\(f(v(x)) = f(2x) = e^{(2x)^2} = e^{4x^2}\)

\(f(u(x)) = f(x^3) = e^{(x^3)^2} = e^{x^6}\)

Step 5: Apply the formula

\(K'(x) = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)\)

\(K'(x) = e^{4x^2} \cdot 2 - e^{x^6} \cdot 3x^2\)

\(K'(x) = 2e^{4x^2} - 3x^2e^{x^6}\)

Step 6: Conceptual verification

This represents the rate of change of the accumulated area between \(x^3\) and \(2x\) as \(x\) changes

The positive term accounts for the expanding upper limit, the negative term for the expanding lower limit

\(K'(x) = 2e^{4x^2} - 3x^2e^{x^6}\)
Final answer:

\(K'(x) = 2e^{4x^2} - 3x^2e^{x^6}\)

Applied rules:

• Generalized FTC: \(\frac{d}{dx}\int_{u(x)}^{v(x)} f(t) dt = f(v(x))v'(x) - f(u(x))u'(x)\)

• Chain rule: For composite functions in both limits

• Power rule: For derivatives of polynomial limits

5 Physical Application
Exercise 5
The velocity of a particle is given by \(v(t) = 3t^2 + 2t\). If its position at time \(t = 1\) is \(s(1) = 5\), find the position function \(s(t)\) and interpret using FTC Part 1.
Definition:

Position-Accumulation: Since \(v(t) = s'(t)\), we have \(s(t) = s(a) + \int_a^t v(u) du\). This shows position as an accumulation of velocity.

Velocity
\(v(t) = 3t^2 + 2t\)
Initial condition
\(s(1) = 5\)
Position function
\(s(t) = \int_1^t (3u^2 + 2u) du + 5\)
Step 1: Set up the relationship

Since velocity is the derivative of position: \(v(t) = s'(t)\)

By FTC Part 1, position is the accumulation of velocity: \(s(t) = s(a) + \int_a^t v(u) du\)

Step 2: Use the given initial condition

We know \(s(1) = 5\), so we can write:

\(s(t) = s(1) + \int_1^t v(u) du = 5 + \int_1^t (3u^2 + 2u) du\)

Step 3: Evaluate the integral

\(\int_1^t (3u^2 + 2u) du = \left[u^3 + u^2\right]_1^t = (t^3 + t^2) - (1^3 + 1^2)\)

\(= t^3 + t^2 - 2\)

Step 4: Find the position function

\(s(t) = 5 + (t^3 + t^2 - 2) = t^3 + t^2 + 3\)

Step 5: Verify using FTC Part 1

To verify: \(s'(t) = \frac{d}{dt}[t^3 + t^2 + 3] = 3t^2 + 2t = v(t)\) ✓

Also: \(s(1) = 1^3 + 1^2 + 3 = 5\) ✓

Step 6: Physical interpretation

The position at time \(t\) is the initial position plus the accumulated displacement from time 1 to time \(t\)

FTC Part 1 tells us that the rate of change of this accumulated position equals the current velocity

\(s(t) = t^3 + t^2 + 3\)
Final answer:

The position function is \(s(t) = t^3 + t^2 + 3\).

This demonstrates that position is an accumulation of velocity, and velocity is the rate of change of position.

Applied rules:

• FTC Part 1: Position is the accumulation of velocity

• Initial value problem: Use given condition to find constant

• Integration: Power rule for polynomial functions

Key Concepts, Laws, Methods, and Formulas
\(\frac{d}{dx}\int_{u(x)}^{v(x)} f(t) dt = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)\)
Generalized FTC Part 1
Key definitions:

Accumulation function: \(F(x) = \int_a^x f(t) dt\) represents the accumulated change from a fixed reference point to a variable point.

Rate of accumulation: The derivative of an accumulation function equals the rate function being accumulated.

Antiderivative: If \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\).

Complete methodology:
  1. Identify the integral form: Determine if it's basic, reversed, or composite limits
  2. Apply appropriate FTC version: Use basic, reversed, or generalized form
  3. Handle chain rule: Multiply by derivatives of composite limits
  4. Simplify: Combine terms and reduce to final form
Tip 1: Always check the position of the variable: upper limit (+), lower limit (-).
Tip 2: The integrand becomes the derivative, but with the variable limit substituted.
Tip 3: Don't forget the derivative of the limit function when it's composite.
Tip 4: FTC Part 1 connects the concepts of area (integration) and slope (differentiation).
Foundational importance: FTC Part 1 is one of the most important theorems in calculus.
Computational power: Allows us to find derivatives of integrals without evaluating the integral first.
Formulas to know by heart:

• Basic FTC: \(\frac{d}{dx}\int_a^x f(t) dt = f(x)\)

• Reversed limits: \(\frac{d}{dx}\int_x^a f(t) dt = -f(x)\)

• Composite upper: \(\frac{d}{dx}\int_a^{u(x)} f(t) dt = f(u(x)) \cdot u'(x)\)

• Generalized: \(\frac{d}{dx}\int_{u(x)}^{v(x)} f(t) dt = f(v(x))v'(x) - f(u(x))u'(x)\)

Visualization: Accumulation and Its Derivative
Exercise 6: Graphical Relationship
Compare the function \(f(x) = x^2 + 1\) with its accumulation function \(F(x) = \int_0^x (t^2 + 1) dt\) and their relationship through FTC Part 1.
\(F(x) = \frac{x^3}{3} + x\) and \(F'(x) = x^2 + 1 = f(x)\)

Analysis: The chart shows how the accumulation function \(F(x)\) grows according to the rate given by \(f(x)\).

  • Where \(f(x)\) is large, \(F(x)\) grows rapidly
  • Where \(f(x)\) is small, \(F(x)\) grows slowly
  • The slope of \(F(x)\) at any point equals the value of \(f(x)\) at that point

Questions & Answers

Question: Why does FTC Part 1 say that differentiation "undoes" integration? Isn't that counterintuitive?

Answer: It's actually quite intuitive once you think about it:

  • Integration: Accumulates area under a curve (finds total change)
  • Differentiation: Finds the rate of change at a point
  • The connection: The rate at which accumulated area changes equals the height of the curve at that point

Think of it like filling a tank: the rate of filling (integrand) determines how fast the water level rises (accumulation), and the rate of rise at any moment equals the current filling rate.

Question: How do I remember when to add a negative sign when the variable is in the lower limit?

Answer: Remember the limit swapping property:

  • \(\int_a^b f(t) dt = -\int_b^a f(t) dt\)
  • So if you have \(\int_x^a f(t) dt\), rewrite it as \(-\int_a^x f(t) dt\)
  • Then apply FTC: \(\frac{d}{dx}[-\int_a^x f(t) dt] = -f(x)\)

Alternatively, think of it as: "variable in lower limit = negative of the integrand evaluated at the variable."

Question: What's the physical interpretation of FTC Part 1?

Answer: FTC Part 1 has beautiful physical interpretations:

  • Position-Velocity: Position is accumulation of velocity, velocity is rate of change of position
  • Accumulated Cost: Total cost is accumulation of marginal cost, marginal cost is rate of change of total cost
  • Population Growth: Population is accumulation of birth/death rates, growth rate is rate of change of population

In general: "What you accumulate" and "the rate at which you accumulate" are mathematically inverse processes.

Question: Can FTC Part 1 be applied to any function? Are there restrictions?

Answer: Yes, there are important restrictions:

  • Continuity: The integrand \(f(t)\) must be continuous at the point where you're taking the derivative
  • Integrability: The function must be integrable on the interval containing the point of interest
  • Well-defined limits: The upper limit function must be differentiable

For example, FTC Part 1 doesn't apply directly if the integrand has jump discontinuities at the point of evaluation, though there are extensions for piecewise continuous functions.

Question: How does FTC Part 1 relate to the concept of antiderivatives?

Answer: FTC Part 1 establishes that accumulation functions are antiderivatives:

  • If \(F(x) = \int_a^x f(t) dt\), then \(F'(x) = f(x)\)
  • This means \(F(x)\) is an antiderivative of \(f(x)\)
  • FTC Part 1 proves that every continuous function has an antiderivative

This is profound because it guarantees that for any continuous function, we can construct an antiderivative by forming an accumulation function. This bridges the gap between definite integrals and antiderivatives.