The Anti-Derivative
Learning Objectives
By the end of this lesson, you will be able to:
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Understand what an anti-derivative (indefinite integral) is.
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Compute anti-derivatives of basic functions.
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Apply power rule for integration.
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Recognize and handle the constant of integration.
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Interpret anti-derivatives as the reverse process of differentiation.
Introduction: What Is an Anti-Derivative?
The anti-derivative is the reverse of differentiation.
If you know $$f^\prime(x)$$, the anti-derivative finds the original function f(x).
Formally:
If $$F^\prime(x) = f(x)$$, then F(x) is an anti-derivative of f(x).
Since differentiation removes constants, there can be many functions whose derivative is the same.
So we add a constant of integration, C.
Basic Idea
Differentiation:
$$\frac{d}{dx}(x^{2}) = 2x$$
Anti-differentiation (reverse):
$$\int 2x dx = x^{2} + C $$
That’s why integration is sometimes called “undoing the derivative.”
Notation
The symbol $$\int$$(integral sign) represents the process of finding the anti-derivative,
and indicates the variable of integration.
$$\int f(x) dx = F(x) + C$$
Power Rule for Anti-Derivatives
If $$n \neq -1 $$, then
Examples:
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$$\int x^{3} dx = \frac{1}{3+1} x^{3+1} +C = \frac{x^{4}}{4} + C$$
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$$\int f^{-2} dx = \frac{1}{-2+1} x^{-2+1} +C = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C $$
natural logarithm integration
Derivative natural logarithm is $$\frac{d}{dx}ln(x) = \frac{1}{x}$$
Anti-Derivative is
$$\int\frac{1}{x}dx=ln|x|+C$$
this is very important technique for integration Trigonometry
Addition and Subtraction
$$\int f(x)+g(x)+h(x) dx = \int f(x) dx + \int g(x) dx + \int h(x) dx$$
example:
$$\int 2x^{3} -7x^{2} + 3x – 9 dx$$
$$= \int 2x^{3} dx – \int 7x^{2} dx + \int 3x dx – \int 9 dx$$
$$= 2\int x^{3} dx -7 \int x^{2} dx + 3 \int x dx – \int 9 dx$$
$$= 2\cdot\frac{1}{4} x^{4} +C -7\cdot\frac{1}{3} x^{3} + C + 3\cdot\frac{1}{2} x^{2} +C – 9x +C $$
$$= \frac{1}{2} x^{4} -\frac{7}{3} x^{3} +\frac{3}{2}x^{2}-9x+C$$
Find the equation (known initial value , find out integration constant C)
example:
$$\frac{dy}{dx} =x^{2}+4x-7$$ and the graph pass through (0,9) , find the equation.
first, Let integral $$\frac{dy}{dx}$$ :
$$f(x)=\int x^{2}+4x-7dx = \frac{1}{3}x^{3}+4\cdot\frac{1}{2}x^{2}-7x+C = \frac{1}{3}x^{3}+2x^{2}-7x+C$$
we know $$f(x)$$ pass through (0,9). then
$$9=\frac{1}{3}\cdot0^{3}+2\cdot0^{2}-7\cdot0+C$$ therefore $$C = 9$$
$$f(x)=\frac{1}{3}x^{3}+2x^{2}-7x+9$$