Integration Basics – Lessons with Examples

Integration is the second major pillar of Calculus 1 — and once it clicks, it’s incredibly satisfying. Where derivatives measure how fast things change, integrals measure accumulation: area under a curve, total distance traveled, net change over time. This lesson walks through every integration technique you need for your Calc 1 final, with clear explanations and worked examples throughout.

📚 Table of Contents

1. What is an Integral?
2. Antiderivatives & Indefinite Integrals
3. Basic Integration Rules
4. Definite Integrals
5. Fundamental Theorem of Calculus
6. U-Substitution
7. Integrals of Trig Functions
8. Integrals of Exponential & Log Functions
9. Area Between Curves
10. Common Mistakes to Avoid
11. Practice Problems

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An integral answers the question: “What function, when differentiated, gives me this?” It also has a powerful geometric meaning — the area under a curve between two points.

There are two types of integrals you need to know:

Type	Notation	Result
Indefinite Integral	$$\int f(x)\,dx$$	A family of functions + C
Definite Integral	$$\int_a^b f(x)\,dx$$	A number (the net area)

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An antiderivative of \(f(x)\) is any function \(F(x)\) such that \(F'(x) = f(x)\). The indefinite integral finds all antiderivatives at once — that’s why we always add + C.

The constant C represents the fact that infinitely many functions share the same derivative. For example:

$$\frac{d}{dx}[x^2] = 2x \qquad \frac{d}{dx}[x^2 + 5] = 2x \qquad \frac{d}{dx}[x^2 – 100] = 2x$$

So \(\int 2x\,dx = x^2 + C\) captures all of them at once.

⚠️ Never forget + C on an indefinite integral. It costs points every single time.

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These are the core rules you need to memorize. Notice they are the reverse of the derivative rules.

Rule	Formula
Power Rule	$$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$$
Constant Rule	$$\int k\,dx = kx + C$$
Constant Multiple Rule	$$\int k\cdot f(x)\,dx = k\int f(x)\,dx$$
Sum/Difference Rule	$$\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx$$

✏️ Example 1 — Power Rule

Problem: Find \(\int x^4\,dx\)

$$\int x^4\,dx = \frac{x^5}{5} + C$$

✏️ Example 2 — Combining Rules

Problem: Find \(\int (3x^2 – 5x + 2)\,dx\)

$$= x^3 – \frac{5x^2}{2} + 2x + C$$

✏️ Example 3 — Negative & Fractional Exponents

Problem: Find \(\int \frac{1}{x^3}\,dx\)

Rewrite as \(x^{-3}\), then apply Power Rule:

$$\int x^{-3}\,dx = \frac{x^{-2}}{-2} + C = \frac{-1}{2x^2} + C$$

✏️ Example 4 — Square Root

Problem: Find \(\int \sqrt{x}\,dx\)

Rewrite as \(x^{1/2}\):

$$\int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$$

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A definite integral has limits of integration — a lower bound \(a\) and upper bound \(b\). The result is a number, not a function, and there is no + C.

Key Properties

Property	Formula
Reverse limits	$$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$$
Same limits	$$\int_a^a f(x)\,dx = 0$$
Additive interval	$$\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$$
Constant multiple	$$\int_a^b k\cdot f(x)\,dx = k\int_a^b f(x)\,dx$$

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This is the most important theorem in all of Calculus 1. It connects derivatives and integrals — two ideas that seem completely different — into one unified framework.

Part 1 — Differentiation of an Integral

$$\frac{d}{dx}\left[\int_a^x f(t)\,dt\right] = f(x)$$

In plain English: if you differentiate an integral, you get the integrand back.

Part 2 — Evaluating Definite Integrals

$$\int_a^b f(x)\,dx = F(b) – F(a)$$

Where \(F'(x) = f(x)\). Find the antiderivative, plug in the top limit, subtract the bottom limit.

✏️ Example 5

Problem: Evaluate \(\int_1^4 (2x + 3)\,dx\)

Step 1: Find the antiderivative: \(F(x) = x^2 + 3x\)

Step 2: Apply FTC Part 2:

$$F(4) – F(1) = (16 + 12) – (1 + 3) = 28 – 4 = 24$$

✅ Answer: 24

✏️ Example 6

Problem: Evaluate \(\int_0^3 (x^2 – 2x + 1)\,dx\)

$$F(x) = \frac{x^3}{3} – x^2 + x$$ $$F(3) – F(0) = (9 – 9 + 3) – 0 = 3$$

✅ Answer: 3

✏️ Example 7 — FTC Part 1

Problem: Find \(\frac{d}{dx}\!\left[\int_1^x \sin(t^2)\,dt\right]\)

By FTC Part 1, the answer is simply the integrand with \(t\) replaced by \(x\):

$$\sin(x^2)$$

✅ Answer: \(\sin(x^2)\)

✏️ Example 8 — FTC Part 1 with Chain Rule

Problem: Find \(\frac{d}{dx}\!\left[\int_1^{x^3} \sin(t^2)\,dt\right]\)

Upper limit is \(x^3\), so apply Chain Rule:

$$\sin((x^3)^2) \cdot 3x^2 = 3x^2\sin(x^6)$$

✅ Answer: \(3x^2\sin(x^6)\)

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U-substitution is the integration version of the Chain Rule. Use it when you see a composite function — especially when the derivative of the inside function is also present.

Steps

1. Choose \(u\) — usually the “inside” function
2. Find \(du = u’\,dx\) and solve for \(dx\)
3. Substitute everything in terms of \(u\)
4. Integrate with respect to \(u\)
5. Substitute back in terms of \(x\)

✏️ Example 9 — Basic U-Sub

Problem: Find \(\int 2x(x^2 + 1)^5\,dx\)

Let \(u = x^2 + 1\), then \(du = 2x\,dx\)

$$\int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2+1)^6}{6} + C$$

✅ Answer: \(\frac{(x^2+1)^6}{6} + C\)

✏️ Example 10 — Adjusting the Constant

Problem: Find \(\int x^2(x^3 + 4)^7\,dx\)

Let \(u = x^3 + 4\), then \(du = 3x^2\,dx\), so \(x^2\,dx = \frac{du}{3}\)

$$\int u^7 \cdot \frac{du}{3} = \frac{1}{3} \cdot \frac{u^8}{8} + C = \frac{(x^3+4)^8}{24} + C$$

✅ Answer: \(\frac{(x^3+4)^8}{24} + C\)

✏️ Example 11 — U-Sub with Trig

Problem: Find \(\int \sin(5x)\,dx\)

Let \(u = 5x\), then \(du = 5\,dx\), so \(dx = \frac{du}{5}\)

$$\int \sin(u) \cdot \frac{du}{5} = -\frac{\cos(u)}{5} + C = -\frac{\cos(5x)}{5} + C$$

✅ Answer: \(-\frac{\cos(5x)}{5} + C\)

✏️ Example 12 — U-Sub with Exponential

Problem: Find \(\int e^{4x}\,dx\)

Let \(u = 4x\), then \(du = 4\,dx\), so \(dx = \frac{du}{4}\)

$$\int e^u \cdot \frac{du}{4} = \frac{e^{4x}}{4} + C$$

✅ Answer: \(\frac{e^{4x}}{4} + C\)

✏️ Example 13 — Definite Integral with U-Sub

Problem: Evaluate \(\int_0^1 2x(x^2+1)^3\,dx\)

Let \(u = x^2 + 1\), \(du = 2x\,dx\)

Change the limits: when \(x = 0\): \(u = 1\); when \(x = 1\): \(u = 2\)

$$\int_1^2 u^3\,du = \left[\frac{u^4}{4}\right]_1^2 = \frac{16}{4} – \frac{1}{4} = \frac{15}{4}$$

✅ Answer: \(\frac{15}{4}\)

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These six must be memorized — they are the direct reverse of the trig derivatives.

Integral	Result
$$\int \sin(x)\,dx$$	$$-\cos(x) + C$$
$$\int \cos(x)\,dx$$	$$\sin(x) + C$$
$$\int \sec^2(x)\,dx$$	$$\tan(x) + C$$
$$\int \csc^2(x)\,dx$$	$$-\cot(x) + C$$
$$\int \sec(x)\tan(x)\,dx$$	$$\sec(x) + C$$
$$\int \csc(x)\cot(x)\,dx$$	$$-\csc(x) + C$$

✏️ Example 14

Problem: Find \(\int (3\sin(x) + 2\cos(x))\,dx\)

$$= -3\cos(x) + 2\sin(x) + C$$

✏️ Example 15 — U-Sub with Trig

Problem: Find \(\int \cos(3x^2) \cdot 6x\,dx\)

Let \(u = 3x^2\), \(du = 6x\,dx\)

$$\int \cos(u)\,du = \sin(u) + C = \sin(3x^2) + C$$

✅ Answer: \(\sin(3x^2) + C\)

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Integral	Result
$$\int e^x\,dx$$	$$e^x + C$$
$$\int a^x\,dx$$	$$\frac{a^x}{\ln(a)} + C$$
$$\int \frac{1}{x}\,dx$$	$$\ln|x| + C$$

⚠️ The absolute value in \(\ln|x|\) matters — don’t drop it!

✏️ Example 16

Problem: Find \(\int \frac{2x}{x^2 + 1}\,dx\)

Let \(u = x^2 + 1\), \(du = 2x\,dx\)

$$\int \frac{du}{u} = \ln|u| + C = \ln(x^2 + 1) + C$$

(No absolute value needed since \(x^2 + 1 > 0\) always)

✅ Answer: \(\ln(x^2 + 1) + C\)

✏️ Example 17

Problem: Find \(\int 3e^{2x}\,dx\)

Let \(u = 2x\), \(du = 2\,dx\):

$$3 \cdot \frac{e^{2x}}{2} + C = \frac{3e^{2x}}{2} + C$$

✅ Answer: \(\frac{3e^{2x}}{2} + C\)

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When you need the area between two curves \(f(x)\) and \(g(x)\) on \([a, b]\), integrate the difference of the top curve minus the bottom curve:

$$A = \int_a^b [f(x) – g(x)]\,dx \quad \text{where } f(x) \geq g(x) \text{ on } [a, b]$$

Steps

1. Find the intersection points by setting \(f(x) = g(x)\) — these are usually your limits
2. Determine which function is on top
3. Integrate [top − bottom]

✏️ Example 18

Problem: Find the area between \(f(x) = x^2\) and \(g(x) = x\) on \([0, 1]\)

On \([0, 1]\): \(x \geq x^2\), so top = \(x\), bottom = \(x^2\)

$$A = \int_0^1 (x – x^2)\,dx = \left[\frac{x^2}{2} – \frac{x^3}{3}\right]_0^1 = \frac{1}{2} – \frac{1}{3} = \frac{1}{6}$$

✅ Answer: \(\frac{1}{6}\)

✏️ Example 19

Problem: Find the area enclosed by \(f(x) = 4 – x^2\) and \(g(x) = x + 2\)

Step 1: Find intersections:

$$4 – x^2 = x + 2 \Rightarrow x^2 + x – 2 = 0 \Rightarrow (x+2)(x-1) = 0$$

Intersect at \(x = -2\) and \(x = 1\)

Step 2: On \([-2, 1]\): \(4 – x^2 \geq x + 2\)

$$A = \int_{-2}^1 [(4-x^2) – (x+2)]\,dx = \int_{-2}^1 (2 – x – x^2)\,dx$$ $$= \left[2x – \frac{x^2}{2} – \frac{x^3}{3}\right]_{-2}^1$$ $$= \left(2 – \frac{1}{2} – \frac{1}{3}\right) – \left(-4 – 2 + \frac{8}{3}\right) = \frac{7}{6} + 6 – \frac{8}{3} = \frac{9}{2}$$

✅ Answer: \(\frac{9}{2}\)

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• ❌ Forgetting + C on every indefinite integral — this is the #1 point-loser
• ❌ Power Rule error: \(\int x^2\,dx = \frac{x^3}{3} + C\) — add 1 to the exponent AND divide by it
• ❌ Using Power Rule when n = −1: \(\int \frac{1}{x}\,dx = \ln|x| + C\), not \(\frac{x^0}{0}\)
• ❌ Forgetting to change limits in definite integral u-substitution
• ❌ Wrong sign on trig integrals: \(\int \sin(x)\,dx = -\cos(x) + C\) — don’t drop the negative
• ❌ Not checking which curve is on top in area between curves — always verify before integrating
• ❌ Dropping the absolute value: \(\int \frac{1}{x}\,dx = \ln|x| + C\), not \(\ln(x) + C\)

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Try each problem before clicking Show Solution!

1. \(\displaystyle\int (5x^4 – 3x^2 + 7)\,dx\)

👁️ Show Solution

$$x^5 – x^3 + 7x + C$$

2. \(\displaystyle\int \frac{1}{x^4}\,dx\)

👁️ Show Solution

Rewrite as \(x^{-4}\):

$$\frac{x^{-3}}{-3} + C = \frac{-1}{3x^3} + C$$

3. Evaluate \(\displaystyle\int_0^2 (x^3 + 1)\,dx\)

👁️ Show Solution

$$F(x) = \frac{x^4}{4} + x$$ $$F(2) – F(0) = (4 + 2) – 0 = 6$$

4. \(\displaystyle\int 3x^2(x^3 – 1)^5\,dx\)

👁️ Show Solution

Let \(u = x^3 – 1\), \(du = 3x^2\,dx\)

$$\int u^5\,du = \frac{u^6}{6} + C = \frac{(x^3-1)^6}{6} + C$$

5. \(\displaystyle\int \cos(7x)\,dx\)

👁️ Show Solution

Let \(u = 7x\), \(du = 7\,dx\):

$$\frac{\sin(7x)}{7} + C$$

6. \(\displaystyle\int \frac{4x}{x^2 + 3}\,dx\)

👁️ Show Solution

Let \(u = x^2 + 3\), \(du = 2x\,dx\), so \(4x\,dx = 2\,du\)

$$\int \frac{2\,du}{u} = 2\ln|u| + C = 2\ln(x^2+3) + C$$

7. Find \(\displaystyle\frac{d}{dx}\!\left[\int_2^x (t^3 + \sin t)\,dt\right]\)

👁️ Show Solution

By FTC Part 1:

$$x^3 + \sin(x)$$

8. Find the area between \(f(x) = x^2\) and \(g(x) = 2x\) on \([0, 2]\)

👁️ Show Solution

On \([0, 2]\): \(2x \geq x^2\)

$$A = \int_0^2 (2x – x^2)\,dx = \left[x^2 – \frac{x^3}{3}\right]_0^2 = 4 – \frac{8}{3} = \frac{4}{3}$$

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🔗 Continue Your Calculus 1 Journey

👉 Integration Extra Practice Problem Set — Coming soon!

👉 Integration by Parts Lesson — Coming soon!

👉 Back to Derivatives Applications Lesson — Coming soon!

👉 Full Calculus 1 Study Guide — Coming soon!