Integration and Antiderivatives
Integration: Finding Antiderivatives
Power Rule for Derivatives
The power rule states that the derivative of x^n is nx^{n-1}.
Power Rule for Integration
The power rule for finding antiderivatives is:
\int x^n dx = \frac{x^{n+1}}{n+1} + C
Where:
- n is a constant exponent.
- C is the constant of integration.
Examples
Example 1: Antiderivative of 3x^2
- Add 1 to the exponent: 2 + 1 = 3. The expression becomes 3x^3.
- Divide by the new exponent: \frac{3x^3}{3} = x^3.
- Add the constant of integration: x^3 + C.
Example 2: Antiderivative of x^4
\int x^4 dx = \frac{x^{4+1}}{4+1} + C = \frac{x^5}{5} + C = \frac{1}{5}x^5 + C
Example 3: Antiderivative of x^2
\int x^2 dx = \frac{x^{2+1}}{2+1} + C = \frac{x^3}{3} + C
Example 4: Antiderivative of x^7
\int x^7 dx = \frac{x^{7+1}}{7+1} + C = \frac{x^8}{8} + C
Example 5: Antiderivative of x
\int x dx = \frac{x^{1+1}}{1+1} + C = \frac{x^2}{2} + C
Antiderivative with Fractions
Example 1: Antiderivative of \frac{x}{4}\int \frac{x}{4} dx = \frac{1}{4} \int x dx = \frac{1}{4} \cdot \frac{x^2}{2} + C = \frac{1}{8}x^2 + C
Example 2: Antiderivative of 8x^3
\int 8x^3 dx = 8 \int x^3 dx = 8 \cdot \frac{x^{3+1}}{3+1} + C = 8 \cdot \frac{x^4}{4} + C = 2x^4 + C
Antiderivative of a Constant
To find the antiderivative of a constant, add the variable of integration to the constant:
\int 4 dx = 4x + C
\int 5 dy = 5y + C
Explanation
We can rewrite 4 as 4x^0, because any number to the power of 0 is 1, so 4 \cdot 1 = 4.
Then, following the power rule, we have \frac{4x^{0+1}}{0+1} = \frac{4x^1}{1} = 4x
Example
\int -7 dz = -7z + C
Antiderivative of a Binomial
To find the antiderivative of a binomial, integrate each term separately:
\int (7x - 6) dx = \int 7x dx - \int 6 dx = \frac{7x^2}{2} - 6x + C
Antiderivative of a Trinomial
\int (6x^2 + 4x - 7) dx = \int 6x^2 dx + \int 4x dx - \int 7 dx = \frac{6x^3}{3} + \frac{4x^2}{2} - 7x + C = 2x^3 + 2x^2 - 7x + C
Antiderivatives of Radical Functions
Example 1: Antiderivative of \sqrt{x}
- Rewrite the radical as a rational exponent: \sqrt{x} = x^{\frac{1}{2}}.
- Apply the power rule:
\int x^{\frac{1}{2}} dx = \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{2}{3} x^{\frac{3}{2}} + C - Rewrite back in radical form: \frac{2}{3} \sqrt{x^3} + C
Example 2: Antiderivative of \sqrt[3]{x^4}
- Rewrite the radical as a rational exponent: \sqrt[3]{x^4} = x^{\frac{4}{3}}.
- Apply the power rule:
\int x^{\frac{4}{3}} dx = \frac{x^{\frac{4}{3} + 1}}{\frac{4}{3} + 1} + C = \frac{x^{\frac{7}{3}}}{\frac{7}{3}} + C = \frac{3}{7} x^{\frac{7}{3}} + C
- Rewrite back in radical form: \frac{3}{7} \sqrt[3]{x^7} + C
Example 3: Antiderivative of \sqrt[4]{x^7}
- Rewrite the radical as a rational exponent: \sqrt[4]{x^7} = x^{\frac{7}{4}}.
- Apply the power rule:
\int x^{\frac{7}{4}} dx = \frac{x^{\frac{7}{4} + 1}}{\frac{7}{4} + 1} + C = \frac{x^{\frac{11}{4}}}{\frac{11}{4}} + C = \frac{4}{11} x^{\frac{11}{4}} + C
- Rewrite back in radical form: \frac{4}{11} \sqrt[4]{x^{11}} + C
Integrals of Trigonometric Expressions
- \int \cos(x) dx = \sin(x) + C
- \int -\sin(x) dx = \cos(x) + C
- \int \sin(x) dx = -\cos(x) + C
- \int \sec^2(x) dx = \tan(x) + C
- \int -\csc^2(x) dx = \cot(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
\int (4 \sin(x) - 5 \cos(x) + 3 \sec^2(x)) dx = -4 \cos(x) - 5 \sin(x) + 3 \tan(x) + C
Indefinite vs. Definite Integrals
- Indefinite Integral: Results in an algebraic expression with a constant of integration (C). \int f(x) dx = F(x) + C
- Definite Integral: Results in a numerical value. It has a lower and upper limit of integration. \int_a^b f(x) dx = F(b) - F(a)
The constant of integration is not needed for definite integrals because it cancels out during evaluation.
Example of Indefinite Integral
\int 6x^2 dx = 2x^3 + C
Example of Definite Integral
\int1^2 6x^2 dx = [2x^3]1^2 = 2(2^3) - 2(1^3) = 16 - 2 = 14
Example 2
\int2^3 (8x - 3) dx = [4x^2 - 3x]2^3 = (4(3^2) - 3(3)) - (4(2^2) - 3(2)) = (36 - 9) - (16 - 6) = 27 - 10 = 17
Fundamental Theorem of Calculus
The fundamental theorem of calculus states that if f(x) is a continuous function on the closed interval [a, b], then:
\int_a^b f(x) dx = F(b) - F(a)
Where F(x) is the antiderivative of f(x), i.e., F'(x) = f(x).
Also,
\int f(x) dx = F(x) + C
Exponential Functions
The derivative of e^u is e^u \cdot u'. For antiderivatives, if u is a linear function ax + b, then:
\int e^u dx = \frac{e^u}{u'} + C
Where u' is the derivative of u with respect to x.
Example 1
\int e^{3x} dx = \frac{e^{3x}}{3} + C
Example 2
\int e^{5x} dx = \frac{e^{5x}}{5} + C
Example 3
\int e^{-7x} dx = \frac{e^{-7x}}{-7} + C
Example 4
\int e^{3x - 5} dx = \frac{e^{3x - 5}}{3} + C
U-Substitution
U-substitution is a technique used to find the antiderivative of composite functions.
Example 1
Find the antiderivative of e^{8x}.
- Let u = 8x.
- Find \frac{du}{dx} = 8, so du = 8 dx.
- Solve for dx: dx = \frac{du}{8}.
- Substitute: \int e^{8x} dx = \int e^u \frac{du}{8} = \frac{1}{8} \int e^u du.
- Integrate: \frac{1}{8} e^u + C.
- Substitute back: \frac{1}{8} e^{8x} + C.
Example 2
Find the antiderivative of 4x e^{x^2}.
- Let u = x^2.
- Find \frac{du}{dx} = 2x, so du = 2x dx.
- Solve for dx: dx = \frac{du}{2x}.
- Substitute: \int 4x e^{x^2} dx = \int 4x e^u \frac{du}{2x} = 2 \int e^u du.
- Integrate: 2 e^u + C.
- Substitute back: 2 e^{x^2} + C.
Antiderivatives of Rational Functions
Example 1
Find the antiderivative of \frac{1}{x^2}.
- Rewrite: \frac{1}{x^2} = x^{-2}.
- Apply the power rule: \int x^{-2} dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C
Example 2
Find the antiderivative of \frac{1}{x^3}.
- Rewrite: \frac{1}{x^3} = x^{-3}.
- Apply the power rule: \int x^{-3} dx = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C
Example 3
Find the antiderivative of \frac{8}{x^4}.
- Rewrite: \frac{8}{x^4} = 8x^{-4}.
- Apply the power rule: \int 8x^{-4} dx = 8 \frac{x^{-3}}{-3} + C = -\frac{8}{3x^3} + C
Example 4
What is the antiderivative of \frac{1}{(4x - 3)^2}?
- Let u = 4x - 3.
- Then \frac{du}{dx} = 4, so dx = \frac{du}{4}.
- Substituting gives: \int \frac{1}{u^2} \frac{du}{4} = \frac{1}{4} \int u^{-2} du.
- Integrating: \frac{1}{4} \frac{u^{-1}}{-1} + C = -\frac{1}{4u} + C.
- Substituting back gives: \frac{-1}{4(4x - 3)} + C
Example 5
What is the antiderivative of \frac{7}{(5x - 3)^4}?
- Let u = 5x - 3.
- Then \frac{du}{dx} = 5, so dx = \frac{du}{5}.
- Substituting gives: \int \frac{7}{u^4} \frac{du}{5} = \frac{7}{5} \int u^{-4} du.
- Integrating: \frac{7}{5} \frac{u^{-3}}{-3} + C = -\frac{7}{15u^3} + C.
- Substituting back gives: \frac{-7}{15(5x - 3)^3} + C
Antiderivative of \frac{1}{x}
The antiderivative of \frac{1}{x} is \ln(x).
\int \frac{1}{x} dx = \ln(x) + C
Rationale
Since the derivative of \ln(u) is \frac{u'}{u}, the derivative of \ln(x) is \frac{1}{x}.
Example 1
Find the antiderivative of \frac{7}{x}:
\int \frac{7}{x} dx = 7 \int \frac{1}{x} dx = 7 \ln(x) + C
Example 2
Find the antiderivative of \frac{1}{x + 5}:
\int \frac{1}{x + 5} dx = \ln(x + 5) + C
Using u-substitution:
- Let u = x + 5.
- Then du = dx.
- So, \int \frac{1}{u} du = \ln(u) + C = \ln(x + 5) + C