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