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Integral Calculator Examples

The Integral Calculator computes both definite and indefinite integrals with full step-by-step solutions. Below are practical examples covering common integration techniques used in calculus courses and professional work.

Basic Indefinite Integrals (Power Rule)

The power rule is the most fundamental integration technique. For any function x^n, the integral is x^(n+1)/(n+1) + C.

∫ x^3 dx

Step 1: Apply the power rule: ∫ x^n dx = x^(n+1)/(n+1) + C
Step 2: n = 3, so n+1 = 4
Step 3: Result = x^4/4 + C

Answer: x^4/4 + C
∫ (3x^2 + 2x - 5) dx

Step 1: Integrate term by term
Step 2: ∫ 3x^2 dx = 3 · x^3/3 = x^3
Step 3: ∫ 2x dx = 2 · x^2/2 = x^2
Step 4: ∫ -5 dx = -5x

Answer: x^3 + x^2 - 5x + C

Definite Integral Example

Definite integrals compute the area under a curve between two bounds. The calculator evaluates the antiderivative at both bounds and subtracts.

∫₀² (x^2 + 1) dx

Step 1: Find antiderivative: F(x) = x^3/3 + x
Step 2: Evaluate at upper bound: F(2) = 8/3 + 2 = 14/3
Step 3: Evaluate at lower bound: F(0) = 0/3 + 0 = 0
Step 4: Subtract: 14/3 - 0 = 14/3

Answer: 14/3 ≈ 4.667

U-Substitution

U-substitution handles composite functions by substituting a new variable to simplify the integral.

∫ 2x · (x^2 + 1)^5 dx

Step 1: Let u = x^2 + 1
Step 2: du/dx = 2x → du = 2x dx
Step 3: Substitute: ∫ u^5 du
Step 4: Apply power rule: u^6/6 + C
Step 5: Back-substitute: (x^2 + 1)^6 / 6 + C

Answer: (x^2 + 1)^6 / 6 + C
∫ cos(3x) dx

Step 1: Let u = 3x
Step 2: du = 3 dx → dx = du/3
Step 3: Substitute: ∫ cos(u) · (1/3) du
Step 4: = (1/3) sin(u) + C
Step 5: Back-substitute: (1/3) sin(3x) + C

Answer: sin(3x)/3 + C

Integration by Parts

Integration by parts handles products of functions using the formula: ∫ u dv = uv - ∫ v du.

∫ x · e^x dx

Step 1: Choose u = x, dv = e^x dx
Step 2: Then du = dx, v = e^x
Step 3: Apply formula: uv - ∫ v du = x·e^x - ∫ e^x dx
Step 4: = x·e^x - e^x + C
Step 5: Factor: e^x(x - 1) + C

Answer: e^x(x - 1) + C
∫ x · ln(x) dx

Step 1: Choose u = ln(x), dv = x dx
Step 2: Then du = (1/x) dx, v = x^2/2
Step 3: Apply formula: (x^2/2)·ln(x) - ∫ (x^2/2)·(1/x) dx
Step 4: = (x^2/2)·ln(x) - ∫ x/2 dx
Step 5: = (x^2/2)·ln(x) - x^2/4 + C

Answer: (x^2/2)·ln(x) - x^2/4 + C

Trigonometric Integrals

∫ sin^2(x) dx

Step 1: Use identity: sin^2(x) = (1 - cos(2x))/2
Step 2: ∫ (1 - cos(2x))/2 dx
Step 3: = (1/2)∫ 1 dx - (1/2)∫ cos(2x) dx
Step 4: = x/2 - (1/2)·sin(2x)/2 + C

Answer: x/2 - sin(2x)/4 + C

Partial Fractions

Partial fractions decompose rational functions into simpler fractions that can be integrated individually.

∫ 1/(x^2 - 1) dx

Step 1: Factor denominator: (x-1)(x+1)
Step 2: Decompose: 1/((x-1)(x+1)) = A/(x-1) + B/(x+1)
Step 3: Solve: A = 1/2, B = -1/2
Step 4: ∫ [1/(2(x-1)) - 1/(2(x+1))] dx
Step 5: = (1/2)ln|x-1| - (1/2)ln|x+1| + C

Answer: (1/2)ln|(x-1)/(x+1)| + C

Improper Integrals

∫₁^∞ 1/x^2 dx

Step 1: Replace ∞ with limit: lim(b→∞) ∫₁^b x^(-2) dx
Step 2: Antiderivative: -x^(-1) = -1/x
Step 3: Evaluate: lim(b→∞) [-1/b - (-1/1)]
Step 4: = lim(b→∞) [-1/b + 1]
Step 5: As b→∞, -1/b → 0

Answer: 1 (converges)

Common Integration Formulas Reference

Real-World Application: Area Between Curves

Find the area between y = x^2 and y = x from x=0 to x=1

Step 1: Area = ∫₀¹ (x - x^2) dx  [top minus bottom]
Step 2: Antiderivative: x^2/2 - x^3/3
Step 3: Evaluate at 1: 1/2 - 1/3 = 3/6 - 2/6 = 1/6
Step 4: Evaluate at 0: 0

Answer: 1/6 square units

The Integral Calculator handles all these techniques automatically, showing each step so you can follow the reasoning, verify your own work, or learn the method for future problems.

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