Integral calculus is a branch of mathematics that focuses on finding areas, volumes, and totals. It is the opposite of differential calculus, which focuses on rates of change.
Integral calculus helps us solve problems where we need to calculate the total value of something over a range, like the area under a curve. It is widely used in physics, engineering, and economics.
Fig: Integral Calculus Example
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There are two main types of integrals:
Indefinite Integrals
An indefinite integral finds the general antiderivative of a function. It is expressed as:
∫ f(x) dx = F(x) + C
Here, F(x) is the antiderivative, and C is the constant of integration.
Example:
∫ x² dx = (x³ / 3) + C
Definite Integrals
A definite integral calculates the exact area under a curve between limits a and b. It is expressed as:
∫[a to b] f(x) dx = F(b) - F(a)
Example:
∫[0 to 2] x² dx = [(x³ / 3)] from 0 to 2 = (8/3) - (0) = 8/3
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Explanation: If you have two functions being added together, you can integrate them separately, multiply by the constants, and then add the results.
Explanation: If you're finding the area under the curve from the same point to the same point, there's no area, so the result is zero.
Explanation: You can combine integrals over different parts of the interval into one integral over the entire interval.
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Here are some fundamental formulas of Integral Calculus:
Category | Integral Formula |
Basic Integration Formulas | ∫ dx = x + C |
∫ a dx = ax + C | |
∫ xⁿ dx = (xⁿ⁺¹ / (n+1)) + C, (n ≠ -1) | |
∫ eˣ dx = eˣ + C | |
∫ aˣ dx = (aˣ / ln(a)) + C, (a > 0, a ≠ 1) | |
∫ 1/x dx = ln | |
Trigonometric Integration | ∫ sin(x) dx = -cos(x) + C |
∫ cos(x) dx = sin(x) + C | |
∫ sec²(x) dx = tan(x) + C | |
∫ csc²(x) dx = -cot(x) + C | |
∫ sec(x)tan(x) dx = sec(x) + C | |
∫ csc(x)cot(x) dx = -csc(x) + C | |
Inverse Trigonometric | ∫ 1/√(1 - x²) dx = sin⁻¹(x) + C |
∫ -1/√(1 - x²) dx = cos⁻¹(x) + C | |
∫ 1/(1 + x²) dx = tan⁻¹(x) + C | |
∫ -1/(1 + x²) dx = cot⁻¹(x) + C | |
∫ 1/( | |
∫ -1/( | |
Exponential and Logarithmic | ∫ eˣ dx = eˣ + C |
∫ e⁻ˣ dx = -e⁻ˣ + C | |
∫ ln(x) dx = x ln(x) - x + C | |
Special Integration | ∫ 1/(x² + a²) dx = (1/a) tan⁻¹(x/a) + C |
∫ 1/(x² - a²) dx = (1/2a) ln | |
∫ 1/√(x² + a²) dx = ln | |
∫ 1/√(x² - a²) dx = ln |
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1. Substitution Method
2. Integration by Parts
3. Partial Fractions
4. Trigonometric Substitution
Example 1: Substitution Method
Evaluate ∫ x cos(x²) dx.
Let u = x², so du = 2x dx. Rewrite the integral:
(1/2) ∫ cos(u) du = (1/2) sin(u) + C = (1/2) sin(x²) + C
Example 2: Definite Integral
Find the area under f(x) = 3x from x = 1 to x = 4.
∫[1 to 4] 3x dx = [(3x² / 2)] from 1 to 4
= [(3(4²) / 2) - (3(1²) / 2)]
= (24 - 1.5) = 22.5
Example 3: Integration by Parts
Evaluate ∫ x e^x dx.
Use the formula: ∫ u dv = uv - ∫ v du.
Let u = x and dv = e^x dx. Then, du = dx and v = e^x.
Now, apply the formula:
∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C
So, the result is: x e^x - e^x + C.
Example 4: Trigonometric Substitution
Evaluate ∫ (1 / √(1 - x²)) dx.
This is a standard integral that can be solved using trigonometric substitution:
Let x = sin(θ), so dx = cos(θ) dθ.
Substitute and simplify:
∫ (1 / √(1 - sin²(θ))) cos(θ) dθ = ∫ dθ = θ + C
Finally, substitute back: θ = arcsin(x), so the result is:
arcsin(x) + C.
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