This is your dedicated extra practice set for every major Calculus 1 topic. Each problem below is designed to mirror what shows up most on finals — work through them on your own, then reveal each step one at a time. No shortcuts, just reps. 💪
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📚 Table of Contents
- Limits and Continuity
- Derivatives — Rules and Techniques
- Applications of Derivatives
- Integration Basics
1. 📌 Limits and Continuity — Extra Practice
These are the most commonly tested limit problems. Reveal each step only after you’ve attempted it yourself!
Problem 1. Find \( \lim_{x \to 5} \frac{x^2 – 25}{x – 5} \)
👉 Step 1 — Try direct substitution
Plug in x = 5:
$$\frac{25 – 25}{5 – 5} = \frac{0}{0}$$
⚠️ Indeterminate form — direct substitution fails. Move to Step 2.
👉 Step 2 — Factor the numerator
\( x^2 – 25 \) is a difference of squares:
$$x^2 – 25 = (x+5)(x-5)$$
👉 Step 3 — Cancel and simplify
$$\frac{(x+5)(x-5)}{x-5} = x + 5$$
👉 Step 4 — Apply the limit
$$\lim_{x \to 5}(x + 5) = 5 + 5 = 10$$
✅ Final Answer: 10
Problem 2. Find \( \lim_{x \to 0} \frac{\sin(3x)}{x} \)
👉 Step 1 — Try direct substitution
Plug in x = 0:
$$\frac{\sin(0)}{0} = \frac{0}{0}$$
⚠️ Indeterminate form. We need a trig limit strategy.
👉 Step 2 — Rewrite to match the known limit
Recall: \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \)
Multiply and divide by 3 to match that form:
$$\frac{\sin(3x)}{x} = 3 \cdot \frac{\sin(3x)}{3x}$$
👉 Step 3 — Apply the standard limit
As x → 0, we also have 3x → 0, so:
$$\lim_{x \to 0} \frac{\sin(3x)}{3x} = 1$$
👉 Step 4 — Multiply by the constant
$$3 \cdot 1 = 3$$
✅ Final Answer: 3
Problem 3. Find \( \lim_{x \to \infty} \frac{4x^2 – 3x + 1}{2x^2 + 5} \)
👉 Step 1 — Identify the highest power
The highest power in both numerator and denominator is \( x^2 \). Divide every term by \( x^2 \).
👉 Step 2 — Divide through by \( x^2 \)
$$\frac{4 – \frac{3}{x} + \frac{1}{x^2}}{2 + \frac{5}{x^2}}$$
👉 Step 3 — Apply the limit
As x → ∞, all terms with x in the denominator → 0:
$$\frac{4 – 0 + 0}{2 + 0} = \frac{4}{2} = 2$$
✅ Final Answer: 2
Problem 4. Find \( \lim_{x \to -2} \frac{x^2 + 5x + 6}{x + 2} \)
👉 Step 1 — Try direct substitution
$$\frac{4 – 10 + 6}{-2 + 2} = \frac{0}{0}$$
⚠️ Indeterminate — factor the numerator.
👉 Step 2 — Factor the numerator
$$x^2 + 5x + 6 = (x+2)(x+3)$$
👉 Step 3 — Cancel and simplify
$$\frac{(x+2)(x+3)}{x+2} = x + 3$$
👉 Step 4 — Apply the limit
$$\lim_{x \to -2}(x + 3) = -2 + 3 = 1$$
✅ Final Answer: 1
Problem 5. Find \( \lim_{x \to 0} \frac{x^2 + 3x}{x} \)
👉 Step 1 — Try direct substitution
$$\frac{0 + 0}{0} = \frac{0}{0}$$
⚠️ Indeterminate — factor the numerator.
👉 Step 2 — Factor x from the numerator
$$\frac{x(x + 3)}{x}$$
👉 Step 3 — Cancel and apply the limit
$$x + 3 \quad \Rightarrow \quad 0 + 3 = 3$$
✅ Final Answer: 3
Problem 6. Find \( \lim_{x \to \infty} \frac{7x^3 – 2x^2 + 4}{3x^3 + x} \)
👉 Step 1 — Identify the highest power
Highest power is \( x^3 \). Divide every term by \( x^3 \).
👉 Step 2 — Divide through by \( x^3 \)
$$\frac{7 – \frac{2}{x} + \frac{4}{x^3}}{3 + \frac{1}{x^2}}$$
👉 Step 3 — Apply the limit
$$\frac{7 – 0 + 0}{3 + 0} = \frac{7}{3}$$
✅ Final Answer: 7/3
Problem 7. Find \( \lim_{x \to 1} \frac{x^3 – 1}{x – 1} \)
👉 Step 1 — Try direct substitution
$$\frac{1 – 1}{1 – 1} = \frac{0}{0}$$
⚠️ Indeterminate — use the difference of cubes formula.
👉 Step 2 — Apply difference of cubes
Formula: \( a^3 – b^3 = (a – b)(a^2 + ab + b^2) \)
$$x^3 – 1 = (x-1)(x^2 + x + 1)$$
👉 Step 3 — Cancel and simplify
$$\frac{(x-1)(x^2+x+1)}{x-1} = x^2 + x + 1$$
👉 Step 4 — Apply the limit
$$1 + 1 + 1 = 3$$
✅ Final Answer: 3
Problem 8. Find \( \lim_{x \to 2} \frac{x^2 – 4}{x^2 – x – 2} \)
👉 Step 1 — Try direct substitution
$$\frac{4 – 4}{4 – 2 – 2} = \frac{0}{0}$$
⚠️ Indeterminate — factor both numerator and denominator.
👉 Step 2 — Factor the numerator
$$x^2 – 4 = (x+2)(x-2)$$
👉 Step 3 — Factor the denominator
$$x^2 – x – 2 = (x-2)(x+1)$$
👉 Step 4 — Cancel and apply the limit
$$\frac{(x+2)(x-2)}{(x-2)(x+1)} = \frac{x+2}{x+1} \quad \Rightarrow \quad \frac{4}{3}$$
✅ Final Answer: 4/3
Problem 9. Find \( \lim_{x \to \infty} \frac{6x + 1}{x^2 – 3} \)
👉 Step 1 — Identify the highest power
Highest power is \( x^2 \) (in the denominator). Divide every term by \( x^2 \).
👉 Step 2 — Divide through by \( x^2 \)
$$\frac{\frac{6}{x} + \frac{1}{x^2}}{1 – \frac{3}{x^2}}$$
👉 Step 3 — Apply the limit
$$\frac{0 + 0}{1 – 0} = 0$$
✅ Final Answer: 0
Problem 10. Find \( \lim_{x \to 0} \frac{1 – \cos(x)}{x} \)
👉 Step 1 — Try direct substitution
$$\frac{1 – \cos(0)}{0} = \frac{1 – 1}{0} = \frac{0}{0}$$
⚠️ Indeterminate — this is a standard trig limit.
👉 Step 2 — Recall the standard trig limit
This is one of the two fundamental trig limits you must memorize:
$$\lim_{x \to 0} \frac{1 – \cos(x)}{x} = 0$$
The numerator approaches 0 faster than the denominator.
👉 Step 3 — State the answer
✅ Final Answer: 0
2. 📌 Derivatives — Extra Practice
Power rule, chain rule, product rule, quotient rule — reveal each step only after you’ve made your attempt!
Problem 1. Find f'(x) if \( f(x) = 6x^4 – 5x^3 + 2x – 8 \)
👉 Step 1 — Identify the rule
Each term is a power of x — use the Power Rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \)
👉 Step 2 — Differentiate each term
$$\frac{d}{dx}[6x^4] = 24x^3$$
$$\frac{d}{dx}[-5x^3] = -15x^2$$
$$\frac{d}{dx}[2x] = 2$$
$$\frac{d}{dx}[-8] = 0$$
👉 Step 3 — Combine
$$f'(x) = 24x^3 – 15x^2 + 2$$
✅ Final Answer: \( 24x^3 – 15x^2 + 2 \)
Problem 2. Find f'(x) if \( f(x) = \sqrt{x} + \frac{1}{x^2} \)
👉 Step 1 — Rewrite using exponents
$$f(x) = x^{1/2} + x^{-2}$$
👉 Step 2 — Apply the Power Rule to each term
$$\frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2}$$
$$\frac{d}{dx}[x^{-2}] = -2x^{-3}$$
👉 Step 3 — Rewrite in simplified form
$$f'(x) = \frac{1}{2\sqrt{x}} – \frac{2}{x^3}$$
✅ Final Answer: \( \frac{1}{2\sqrt{x}} – \frac{2}{x^3} \)
Problem 3. Find dy/dx if \( y = (5x^3 – 2)^4 \)
👉 Step 1 — Identify the rule
Function inside a function → Chain Rule: \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)
Outer function: \( u^4 \) | Inner function: \( 5x^3 – 2 \)
👉 Step 2 — Differentiate the outer function
$$\frac{d}{du}[u^4] = 4u^3 = 4(5x^3 – 2)^3$$
👉 Step 3 — Differentiate the inner function
$$\frac{d}{dx}[5x^3 – 2] = 15x^2$$
👉 Step 4 — Multiply together
$$\frac{dy}{dx} = 4(5x^3 – 2)^3 \cdot 15x^2 = 60x^2(5x^3 – 2)^3$$
✅ Final Answer: \( 60x^2(5x^3 – 2)^3 \)
Problem 4. Find f'(x) if \( f(x) = x^3\cos(x) \)
👉 Step 1 — Identify the rule
Two functions multiplied → Product Rule: \( (fg)’ = f’g + fg’ \)
Let \( f = x^3 \) and \( g = \cos(x) \)
👉 Step 2 — Find f’ and g’
$$f'(x) = 3x^2 \qquad g'(x) = -\sin(x)$$
👉 Step 3 — Apply the Product Rule
$$f'(x) = 3x^2\cos(x) + x^3(-\sin(x))$$
👉 Step 4 — Simplify
$$f'(x) = 3x^2\cos(x) – x^3\sin(x)$$
✅ Final Answer: \( 3x^2\cos(x) – x^3\sin(x) \)
Problem 5. Find dy/dx if \( y = \frac{3x^2 + 1}{x – 2} \)
👉 Step 1 — Identify the rule
One function divided by another → Quotient Rule: \( \left(\frac{f}{g}\right)’ = \frac{f’g – fg’}{g^2} \)
Let \( f = 3x^2 + 1 \) and \( g = x – 2 \)
👉 Step 2 — Find f’ and g’
$$f'(x) = 6x \qquad g'(x) = 1$$
👉 Step 3 — Apply the Quotient Rule
$$\frac{dy}{dx} = \frac{6x(x-2) – (3x^2+1)(1)}{(x-2)^2}$$
👉 Step 4 — Expand the numerator
$$6x^2 – 12x – 3x^2 – 1 = 3x^2 – 12x – 1$$
👉 Step 5 — Write the final answer
$$\frac{dy}{dx} = \frac{3x^2 – 12x – 1}{(x-2)^2}$$
✅ Final Answer: \( \frac{3x^2 – 12x – 1}{(x-2)^2} \)
Problem 6. Find f'(x) if \( f(x) = e^{x^2} \)
👉 Step 1 — Identify the rule
Exponential with a function inside → Chain Rule
Outer: \( e^u \) | Inner: \( x^2 \)
👉 Step 2 — Differentiate the outer function
$$\frac{d}{du}[e^u] = e^u = e^{x^2}$$
👉 Step 3 — Differentiate the inner function
$$\frac{d}{dx}[x^2] = 2x$$
👉 Step 4 — Multiply together
$$f'(x) = e^{x^2} \cdot 2x = 2xe^{x^2}$$
✅ Final Answer: \( 2xe^{x^2} \)
Problem 7. Find f'(x) if \( f(x) = \ln(x^2 + 4) \)
👉 Step 1 — Identify the rule
Natural log with a function inside → Chain Rule
Outer: \( \ln(u) \) | Inner: \( x^2 + 4 \)
👉 Step 2 — Differentiate the outer function
$$\frac{d}{du}[\ln(u)] = \frac{1}{u} = \frac{1}{x^2 + 4}$$
👉 Step 3 — Differentiate the inner function
$$\frac{d}{dx}[x^2 + 4] = 2x$$
👉 Step 4 — Multiply together
$$f'(x) = \frac{1}{x^2 + 4} \cdot 2x = \frac{2x}{x^2 + 4}$$
✅ Final Answer: \( \frac{2x}{x^2 + 4} \)
Problem 8. Find f'(x) if \( f(x) = \sin^3(x) \)
👉 Step 1 — Rewrite clearly
$$f(x) = [\sin(x)]^3$$
Outer: \( u^3 \) | Inner: \( \sin(x) \)
👉 Step 2 — Differentiate the outer function
$$\frac{d}{du}[u^3] = 3u^2 = 3\sin^2(x)$$
👉 Step 3 — Differentiate the inner function
$$\frac{d}{dx}[\sin(x)] = \cos(x)$$
👉 Step 4 — Multiply together
$$f'(x) = 3\sin^2(x) \cdot \cos(x)$$
✅ Final Answer: \( 3\sin^2(x)\cos(x) \)
Problem 9. Find f'(x) if \( f(x) = x^2 e^x \)
👉 Step 1 — Identify the rule
Two functions multiplied → Product Rule
Let \( f = x^2 \) and \( g = e^x \)
👉 Step 2 — Find f’ and g’
$$f'(x) = 2x \qquad g'(x) = e^x$$
👉 Step 3 — Apply the Product Rule
$$f'(x) = 2x \cdot e^x + x^2 \cdot e^x$$
👉 Step 4 — Factor and simplify
$$f'(x) = xe^x(2 + x)$$
✅ Final Answer: \( xe^x(2 + x) \)
Problem 10. Find dy/dx if \( y = \tan(4x) \)
👉 Step 1 — Identify the rule
Trig function with something inside → Chain Rule
Outer: \( \tan(u) \) | Inner: \( 4x \)
👉 Step 2 — Differentiate the outer function
$$\frac{d}{du}[\tan(u)] = \sec^2(u) = \sec^2(4x)$$
👉 Step 3 — Differentiate the inner function
$$\frac{d}{dx}[4x] = 4$$
👉 Step 4 — Multiply together
$$\frac{dy}{dx} = \sec^2(4x) \cdot 4 = 4\sec^2(4x)$$
✅ Final Answer: \( 4\sec^2(4x) \)
Problem 11. Find f'(x) if \( f(x) = \frac{\sin(x)}{x^2} \)
👉 Step 1 — Identify the rule
One function divided by another → Quotient Rule
Let \( f = \sin(x) \) and \( g = x^2 \)
👉 Step 2 — Find f’ and g’
$$f'(x) = \cos(x) \qquad g'(x) = 2x$$
👉 Step 3 — Apply the Quotient Rule
$$f'(x) = \frac{\cos(x) \cdot x^2 – \sin(x) \cdot 2x}{x^4}$$
👉 Step 4 — Factor and simplify
$$f'(x) = \frac{x[x\cos(x) – 2\sin(x)]}{x^4} = \frac{x\cos(x) – 2\sin(x)}{x^3}$$
✅ Final Answer: \( \frac{x\cos(x) – 2\sin(x)}{x^3} \)
3. 📌 Applications of Derivatives — Extra Practice
Critical points, increasing/decreasing intervals, concavity, and optimization — step through each one carefully!
Problem 1. Find the critical points of \( f(x) = x^3 – 3x^2 – 9x + 5 \)
👉 Step 1 — Find f'(x)
$$f'(x) = 3x^2 – 6x – 9$$
👉 Step 2 — Factor f'(x)
$$f'(x) = 3(x^2 – 2x – 3) = 3(x-3)(x+1)$$
👉 Step 3 — Set f'(x) = 0 and solve
$$3(x-3)(x+1) = 0 \Rightarrow x = 3 \quad \text{and} \quad x = -1$$
✅ Final Answer: Critical points at x = −1 and x = 3
Problem 2. Determine where \( f(x) = x^3 – 3x^2 – 9x + 5 \) is increasing and decreasing.
👉 Step 1 — Use the critical points from Problem 1
Critical points divide the number line into three intervals:
\( (-\infty, -1) \), \( (-1, 3) \), \( (3, \infty) \)
👉 Step 2 — Test a value in each interval
Use \( f'(x) = 3(x-3)(x+1) \):
- Test x = −2: \( 3(-5)(-1) = 15 > 0 \) → positive
- Test x = 0: \( 3(-3)(1) = -9 < 0 \) → negative
- Test x = 4: \( 3(1)(5) = 15 > 0 \) → positive
👉 Step 3 — State the intervals
- f'(x) > 0 on (−∞, −1) → increasing
- f'(x) < 0 on (−1, 3) → decreasing
- f'(x) > 0 on (3, ∞) → increasing
✅ Final Answer: Increasing on (−∞, −1) ∪ (3, ∞); Decreasing on (−1, 3)
Problem 3. Find the local max and min of \( f(x) = x^3 – 3x^2 – 9x + 5 \)
👉 Step 1 — Use the sign changes from Problem 2
- At x = −1: f’ changes from + to − → local maximum
- At x = 3: f’ changes from − to + → local minimum
👉 Step 2 — Calculate f(−1)
$$f(-1) = (-1)^3 – 3(-1)^2 – 9(-1) + 5 = -1 – 3 + 9 + 5 = 10$$
👉 Step 3 — Calculate f(3)
$$f(3) = 27 – 27 – 27 + 5 = -22$$
👉 Step 4 — State the answer
✅ Final Answer: Local max at (−1, 10); Local min at (3, −22)
Problem 4. Find the inflection point and intervals of concavity for \( f(x) = x^3 – 3x^2 – 9x + 5 \)
👉 Step 1 — Find f”(x)
$$f'(x) = 3x^2 – 6x – 9 \Rightarrow f”(x) = 6x – 6$$
👉 Step 2 — Set f”(x) = 0
$$6x – 6 = 0 \Rightarrow x = 1$$
👉 Step 3 — Test each interval
- x < 1: f”(0) = −6 < 0 → concave down
- x > 1: f”(2) = 6 > 0 → concave up
👉 Step 4 — Find the inflection point coordinates
$$f(1) = 1 – 3 – 9 + 5 = -6$$
✅ Final Answer: Inflection point at (1, −6); Concave down on (−∞, 1); Concave up on (1, ∞)
Problem 5. Find the absolute max and min of \( f(x) = x^2 – 4x + 1 \) on [0, 3].
👉 Step 1 — Find f'(x) and critical points
$$f'(x) = 2x – 4 = 0 \Rightarrow x = 2$$
x = 2 is inside [0, 3] ✓
👉 Step 2 — Evaluate f at the critical point
$$f(2) = 4 – 8 + 1 = -3$$
👉 Step 3 — Evaluate f at both endpoints
$$f(0) = 0 – 0 + 1 = 1$$
$$f(3) = 9 – 12 + 1 = -2$$
👉 Step 4 — Compare all values
Values: f(0) = 1, f(2) = −3, f(3) = −2
✅ Final Answer: Absolute max = 1 at x = 0; Absolute min = −3 at x = 2
Problem 6. A farmer has 200 meters of fencing to enclose a rectangle. What dimensions maximize the area?
👉 Step 1 — Write the constraint equation
$$2x + 2y = 200 \Rightarrow y = 100 – x$$
👉 Step 2 — Write the area function
$$A = xy = x(100 – x) = 100x – x^2$$
👉 Step 3 — Take the derivative and set it equal to zero
$$A’ = 100 – 2x = 0 \Rightarrow x = 50$$
👉 Step 4 — Find y and state the answer
$$y = 100 – 50 = 50$$
✅ Final Answer: 50 m × 50 m — a square always maximizes area with a fixed perimeter
Problem 7. A box with a square base and no top must hold 32 cubic feet. Find the dimensions that minimize surface area.
👉 Step 1 — Define variables and write the volume constraint
Let x = side length of base, h = height:
$$x^2 h = 32 \Rightarrow h = \frac{32}{x^2}$$
👉 Step 2 — Write the surface area function
Bottom + 4 sides (no top):
$$S = x^2 + 4xh = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x}$$
👉 Step 3 — Take the derivative
$$S’ = 2x – \frac{128}{x^2}$$
👉 Step 4 — Set S’ = 0 and solve
$$2x = \frac{128}{x^2} \Rightarrow 2x^3 = 128 \Rightarrow x^3 = 64 \Rightarrow x = 4$$
👉 Step 5 — Find h and state the answer
$$h = \frac{32}{16} = 2$$
✅ Final Answer: Base = 4 ft × 4 ft, Height = 2 ft
Problem 8. A ladder 10 feet long leans against a wall. The bottom slides away at 2 ft/sec. How fast is the top sliding down when the bottom is 6 feet from the wall?
👉 Step 1 — Draw and label the diagram
Let x = distance from wall to base of ladder, y = height of top of ladder on wall.
Given: ladder length = 10, \( \frac{dx}{dt} = 2 \) ft/sec, find \( \frac{dy}{dt} \) when x = 6.
👉 Step 2 — Write the equation connecting x and y
$$x^2 + y^2 = 100$$
👉 Step 3 — Differentiate both sides with respect to time
$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$
👉 Step 4 — Find y when x = 6
$$y = \sqrt{100 – 36} = \sqrt{64} = 8$$
👉 Step 5 — Substitute and solve
$$2(6)(2) + 2(8)\frac{dy}{dt} = 0 \Rightarrow 24 + 16\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{3}{2}$$
The negative sign means the top is sliding down.
✅ Final Answer: The top slides down at 3/2 ft/sec
4. 📌 Integration Basics — Extra Practice
Indefinite integrals, definite integrals, and u-substitution — the three things you must nail. Step through each one!
Problem 1. Find \( \int (5x^4 – 3x^2 + 7)\, dx \)
👉 Step 1 — Identify the rule
Each term is a power of x → use the Power Rule for Integration: \( \int x^n\, dx = \frac{x^{n+1}}{n+1} + C \)
👉 Step 2 — Integrate each term
$$\int 5x^4\, dx = x^5$$
$$\int -3x^2\, dx = -x^3$$
$$\int 7\, dx = 7x$$
👉 Step 3 — Combine and add + C
$$x^5 – x^3 + 7x + C$$
✅ Final Answer: \( x^5 – x^3 + 7x + C \)
Problem 2. Find \( \int \left(\sqrt{x} + \frac{1}{x^3}\right) dx \)
👉 Step 1 — Rewrite using exponents
$$\int \left(x^{1/2} + x^{-3}\right) dx$$
👉 Step 2 — Integrate each term
$$\int x^{1/2}\, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$$
$$\int x^{-3}\, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$$
👉 Step 3 — Combine and add + C
$$\frac{2}{3}x^{3/2} – \frac{1}{2x^2} + C$$
✅ Final Answer: \( \frac{2}{3}x^{3/2} – \frac{1}{2x^2} + C \)
Problem 3. Evaluate \( \int_0^3 (x^2 – 2x + 4)\, dx \)
👉 Step 1 — Find the antiderivative
$$F(x) = \frac{x^3}{3} – x^2 + 4x$$
👉 Step 2 — Evaluate F(3)
$$F(3) = \frac{27}{3} – 9 + 12 = 9 – 9 + 12 = 12$$
👉 Step 3 — Evaluate F(0)
$$F(0) = 0 – 0 + 0 = 0$$
👉 Step 4 — Apply the Fundamental Theorem
$$F(3) – F(0) = 12 – 0 = 12$$
✅ Final Answer: 12
Problem 4. Find \( \int 4x(x^2 + 3)^5\, dx \)
👉 Step 1 — Identify u-substitution
The “inside” function is \( x^2 + 3 \) and its derivative \( 2x \) is sitting nearby → U-Substitution
$$u = x^2 + 3$$
👉 Step 2 — Find du
$$du = 2x\, dx \Rightarrow 4x\, dx = 2\, du$$
👉 Step 3 — Substitute everything
$$\int 2u^5\, du$$
👉 Step 4 — Integrate
$$\frac{2u^6}{6} + C = \frac{u^6}{3} + C$$
👉 Step 5 — Substitute back
$$\frac{(x^2+3)^6}{3} + C$$
✅ Final Answer: \( \frac{(x^2+3)^6}{3} + C \)
Problem 5. Find \( \int \cos(7x)\, dx \)
👉 Step 1 — Identify u-substitution
$$u = 7x$$
👉 Step 2 — Find du
$$du = 7\, dx \Rightarrow dx = \frac{du}{7}$$
👉 Step 3 — Substitute and integrate
$$\int \cos(u) \cdot \frac{du}{7} = \frac{1}{7}\sin(u) + C$$
👉 Step 4 — Substitute back
$$\frac{\sin(7x)}{7} + C$$
✅ Final Answer: \( \frac{\sin(7x)}{7} + C \)
Problem 6. Find \( \int \frac{3x^2}{x^3 + 1}\, dx \)
👉 Step 1 — Identify u-substitution
The numerator \( 3x^2 \) is the derivative of the denominator \( x^3 + 1 \) → perfect for u-sub!
$$u = x^3 + 1$$
👉 Step 2 — Find du
$$du = 3x^2\, dx$$
👉 Step 3 — Substitute and integrate
$$\int \frac{du}{u} = \ln|u| + C$$
👉 Step 4 — Substitute back
$$\ln|x^3 + 1| + C$$
✅ Final Answer: \( \ln|x^3 + 1| + C \)
Problem 7. Evaluate \( \int_1^4 \frac{1}{\sqrt{x}}\, dx \)
👉 Step 1 — Rewrite using exponents
$$\int_1^4 x^{-1/2}\, dx$$
👉 Step 2 — Find the antiderivative
$$F(x) = \frac{x^{1/2}}{1/2} = 2\sqrt{x}$$
👉 Step 3 — Evaluate F(4) and F(1)
$$F(4) = 2\sqrt{4} = 4 \qquad F(1) = 2\sqrt{1} = 2$$
👉 Step 4 — Apply the Fundamental Theorem
$$F(4) – F(1) = 4 – 2 = 2$$
✅ Final Answer: 2
Problem 8. Find \( \int x \cdot e^{x^2}\, dx \)
👉 Step 1 — Identify u-substitution
The inside of the exponential is \( x^2 \), and its derivative \( 2x \) is nearby:
$$u = x^2$$
👉 Step 2 — Find du
$$du = 2x\, dx \Rightarrow x\, dx = \frac{du}{2}$$
👉 Step 3 — Substitute and integrate
$$\int e^u \cdot \frac{du}{2} = \frac{e^u}{2} + C$$
👉 Step 4 — Substitute back
$$\frac{e^{x^2}}{2} + C$$
✅ Final Answer: \( \frac{e^{x^2}}{2} + C \)
Problem 9. Evaluate \( \int_0^{\pi} \sin(x)\, dx \)
👉 Step 1 — Find the antiderivative
$$F(x) = -\cos(x)$$
👉 Step 2 — Evaluate F(π)
$$F(\pi) = -\cos(\pi) = -(-1) = 1$$
👉 Step 3 — Evaluate F(0)
$$F(0) = -\cos(0) = -(1) = -1$$
👉 Step 4 — Apply the Fundamental Theorem
$$F(\pi) – F(0) = 1 – (-1) = 2$$
✅ Final Answer: 2
Problem 10. Find \( \int \sin(x)\cos(x)\, dx \)
👉 Step 1 — Identify u-substitution
Let \( u = \sin(x) \) — its derivative \( \cos(x) \) is sitting right next to it:
$$u = \sin(x)$$
👉 Step 2 — Find du
$$du = \cos(x)\, dx$$
👉 Step 3 — Substitute and integrate
$$\int u\, du = \frac{u^2}{2} + C$$
👉 Step 4 — Substitute back
$$\frac{\sin^2(x)}{2} + C$$
✅ Final Answer: \( \frac{\sin^2(x)}{2} + C \)
Problem 11. Find \( \int \left(e^x + \cos(x) – \frac{1}{x}\right) dx \)
👉 Step 1 — Split into three separate integrals
$$\int e^x\, dx + \int \cos(x)\, dx – \int \frac{1}{x}\, dx$$
👉 Step 2 — Integrate each term using standard rules
$$\int e^x\, dx = e^x$$
$$\int \cos(x)\, dx = \sin(x)$$
$$\int \frac{1}{x}\, dx = \ln|x|$$
👉 Step 3 — Combine and add + C
$$e^x + \sin(x) – \ln|x| + C$$
✅ Final Answer: \( e^x + \sin(x) – \ln|x| + C \)
Keep going — every problem you solve is one less surprise on exam day. You’ve got this! 🎓
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