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Calculus A: Problems with Solutions

Welcome to this comprehensive guide on introductory Calculus A problems. This page is designed to help students master the fundamental concepts of limits, derivatives, and basic integrals through step-by-step solutions. Each section breaks down the methodology required to solve common calculus problems effectively.

Part 1: Limits and Continuity

Understanding limits is the foundation of calculus. A limit describes the value that a function approaches as the input approaches a certain point. Below is a typical problem involving factoring to solve a limit that initially results in an indeterminate form.

Problem 1:
Evaluate the following limit: \[ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \]
Solution:
Step 1: Direct Substitution
First, try plugging in \( x = 3 \) directly into the expression.
\[ \frac{3^2 - 9}{3 - 3} = \frac{9 - 9}{0} = \frac{0}{0} \]
Since direct substitution yields the indeterminate form \( \frac{0}{0} \), we must use algebraic simplification.
Step 2: Factor the Numerator
Recognize that the numerator is a difference of squares.
\[ x^2 - 9 = (x + 3)(x - 3) \]
Step 3: Simplify the Expression
Rewrite the limit with the factored numerator.
\[ \lim_{x \to 3} \frac{(x + 3)(x - 3)}{x - 3} \]
Cancel the common factor \( (x - 3) \) from the numerator and the denominator.
\[ \lim_{x \to 3} (x + 3) \]
Step 4: Evaluate the Simplified Limit
Now, substitute \( x = 3 \) into the simplified expression.
\[ 3 + 3 = 6 \]
Answer: The limit is 6.

Part 2: Derivatives (The Power Rule and Chain Rule)

Derivatives measure the rate at which a function changes. The Power Rule is one of the most basic tools, while the Chain Rule is essential for differentiating composite functions (functions within functions).

Problem 2:
Find the derivative of the function: \[ f(x) = (3x^2 + 1)^5 \]
Solution:
Step 1: Identify the Structure
This function is a composition of two functions. The outer function is \( u^5 \) and the inner function is \( u = 3x^2 + 1 \). This requires the Chain Rule.
Step 2: Apply the Chain Rule Formula
The Chain Rule states:
\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]
Step 3: Differentiate the Outer Function
Differentiate \( u^5 \) with respect to \( u \), keeping the inner function the same.
\[ 5(3x^2 + 1)^4 \]
Step 4: Differentiate the Inner Function
Differentiate \( 3x^2 + 1 \) with respect to \( x \).
\[ \frac{d}{dx}(3x^2 + 1) = 6x \]
Step 5: Multiply the Results
Multiply the derivative of the outer function by the derivative of the inner function.
\[ f'(x) = 5(3x^2 + 1)^4 \cdot 6x \]
Step 6: Simplify
Coefficient: \( 5 \cdot 6x = 30x \).
\[ f'(x) = 30x(3x^2 + 1)^4 \]
Answer: The derivative is \( 30x(3x^2 + 1)^4 \).

Part 3: Product Rule and Quotient Rule

When functions are multiplied or divided by one another, standard differentiation rules are not enough. We must employ the Product Rule for multiplication and the Quotient Rule for division.

Problem 3:
Find the derivative of the function: \[ y = \frac{x^3}{x^2 + 4} \]
Solution:
Step 1: Identify the Top and Bottom
Numerator \( u = x^3 \)
Denominator \( v = x^2 + 4 \)
Step 2: Apply the Quotient Rule Formula
The formula is:
\[ y' = \frac{vu' - uv'}{v^2} \]
Step 3: Calculate Individual Derivatives
Find \( u' \): \( \frac{d}{dx}(x^3) = 3x^2 \)
Find \( v' \): \( \frac{d}{dx}(x^2 + 4) = 2x \)
Step 4: Substitute into the Formula
Substitute \( u, v, u', \) and \( v' \) into the quotient rule equation.
\[ y' = \frac{(x^2 + 4)(3x^2) - (x^3)(2x)}{(x^2 + 4)^2} \]
Step 5: Expand and Simplify the Numerator
Distribute the terms in the numerator:
\[ (x^2 \cdot 3x^2 + 4 \cdot 3x^2) - (2x^4) = (3x^4 + 12x^2) - 2x^4 \]
Combine like terms:
\[ 3x^4 - 2x^4 + 12x^2 = x^4 + 12x^2 \]
Step 6: Final Expression
Place the simplified numerator over the denominator.
\[ y' = \frac{x^4 + 12x^2}{(x^2 + 4)^2} \]
Optionally, factor out \( x^2 \) from the numerator:
\[ y' = \frac{x^2(x^2 + 12)}{(x^2 + 4)^2} \]
Answer: The derivative is \( \frac{x^2(x^2 + 12)}{(x^2 + 4)^2} \).

Part 4: Basic Integration

Integration is the reverse process of differentiation. It is used to find areas under curves and to accumulate quantities. The Power Rule for Integration is essential here.

Problem 4:
Evaluate the definite integral: \[ \int_{1}^{2} (4x^3 + 2x) \, dx \]
Solution:
Step 1: Find the Antiderivative
We apply the Power Rule for integration, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \), to each term separately.
For the term \( 4x^3 \):
\[ \int 4x^3 \, dx = 4 \cdot \frac{x^{3+1}}{3+1} = 4 \cdot \frac{x^4}{4} = x^4 \]
For the term \( 2x \):
\[ \int 2x \, dx = 2 \cdot \frac{x^{1+1}}{1+1} = 2 \cdot \frac{x^2}{2} = x^2 \]
So, the indefinite integral is \( x^4 + x^2 + C \).
Step 2: Apply the Fundamental Theorem of Calculus
Evaluate the antiderivative at the upper limit (2) and subtract the antiderivative evaluated at the lower limit (1).
\[ [x^4 + x^2]_{1}^{2} = ( (2)^4 + (2)^2 ) - ( (1)^4 + (1)^2 ) \]
Step 3: Calculate the Values
Upper limit calculation:
\[ 2^4 + 2^2 = 16 + 4 = 20 \]
Lower limit calculation:
\[ 1^4 + 1^2 = 1 + 1 = 2 \]
Step 4: Subtract
\[ 20 - 2 = 18 \]
Answer: The value of the definite integral is 18.

Conclusion

By practicing these fundamental problems in limits, differentiation, and integration, students build the necessary skills to tackle more complex mathematical challenges. Consistent practice with these mechanics ensures a solid understanding of Calculus A principles.

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