4.2 Properties of Indefinite Integrals

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In this section we examine several properties of the indefinite integral.

1 Properties of Indefinite Integrals

Constant Multiple Rule $\int cf(x) \ dx = c\int f(x) \ dx$ where c is a constant.

example 1 $\int 5\cos (x) \ dx = 5 \int \cos (x) \ dx = 5\sin (x) +C.$

(problem 1a) Compute

$\frac {d}{dx} \sin (x) = \cos (x)$

Do not add the +C to your answer

$\int 8\cos (x) \ dx = \answer {8\sin (x)} \ + C$

(problem 1b) Compute

$\frac {d}{dx} \cos (x) = -\sin (x)$

Do not add the +C to your answer

$\int 4\sin (x) \ dx = \answer {-4\cos (x)} \ + C$

example 2 $\int 3\sec ^2(x) \ dx = 3 \int \sec ^2(x) \ dx = 3\tan (x) +C.$

(problem 2a) Compute

$\frac {d}{dx} \tan (x) = \sec ^2(x)$

Do not add the +C to your answer

$\int 5\sec ^2(x) \ dx = \answer {5\tan (x)} \ + C$

(problem 2b) Compute

$\frac {d}{dx} \cot (x) = -\csc ^2(x)$

Do not add the +C to your answer

$\int 2\csc ^2(x) \ dx = \answer {-2\cot (x)} \ + C$

example 3 $\int 2e^x \ dx = 2 \int e^x \ dx = 2e^x +C.$

(problem 3a) Compute

$\frac {d}{dx} e^x = e^x$

Do not add the +C to your answer

$\int 7e^x \ dx = \answer {7e^x} \ + C$

(problem 3b) Compute

$\frac {d}{dx} e^x = e^x$

Do not add the +C to your answer

$\int \frac {e^x}{3} \ dx = \answer {e^x /3} \ + C$

example 4 $\int \frac {4}{1+x^2} \ dx = 4 \int \frac {1}{1+x^2} \ dx = 4\tan ^{-1}(x) +C.$

(problem 4a) Compute

$\frac {d}{dx} \tan ^{-1}(x) = \frac {1}{1+x^2}$

Do not add the +C to your answer

$\int \frac {9}{1+x^2} \ dx = \answer {9\tan ^{-1}(x)} \ + C$

(problem 4b) Compute

$\frac {d}{dx} \sin ^{-1}(x) = \frac {1}{\sqrt {1-x^2}}$

Do not add the +C to your answer

$\int \frac {6}{\sqrt {1-x^2}} \ dx = \answer {6\sin ^{-1}(x)} \ + C$

example 5 $\int \frac {7}{12x} \ dx = \frac {7}{12} \int \frac {1}{x} \ dx = \frac {7}{12} \ln |x| +C.$

(problem 5a) Compute

$\frac {d}{dx} \ln |x| = \frac {1}{x}$

Do not add the +C to your answer

$\int \frac {5}{x} \ dx = \answer {5\ln |x|} \ + C$

(problem 5b) Compute

$\frac {d}{dx} \ln |x| = \frac {1}{x}$

Do not add the +C to your answer

$\int \frac {1}{2x} \ dx = \answer {\ln |x| /2} \ + C$

example 6 $\int 4x^7 \ dx = 4 \int x^7 \ dx = 4\cdot \frac {x^8}{8} +C = \frac {x^8}{2}.$

(problem 6a) Compute

Use the power rule with $n=4$

The Power Rule says $\int x^n \ dx = \frac {x^{n+1}}{n+1} +$ C

Do not add the +C to your answer

$\int 6x^4 \ dx = \answer {6x^5 /5} \ + C$

(problem 6b) Compute

Use the power rule with $n=3$

The Power Rule says $\int x^n \ dx = \frac {x^{n+1}}{n+1} +$ C

Do not add the +C to your answer

$\int 8x^3 \ dx = \answer {2x^4} \ + C$

example 7 $\begin {aligned}[t] \int \frac {3}{5x^2} \ dx &= \frac {3}{5} \int \frac {1}{x^2} \ dx \\[3pt] &=\frac {3}{5} \int x^{-2} \ dx \\[3pt] &= \frac {3}{5} \cdot \frac {x^{-1}}{-1} +C \\[5pt] &= -\frac {3}{5} \cdot \frac {1}{x} +C \\[5pt] &= -\frac {3}{5x} +C. \end {aligned}$

(problem 7a) Compute

Negative exponents: $\frac {3}{x^2} = 3x^{-2}$

Use the power rule with $n=-2$

The Power Rule says $\int x^n \ dx = \frac {x^{n+1}}{n+1} +$ C

Do not add the +C to your answer

$\int \frac {3}{x^2} \ dx = \answer {-3x^{-1}} \ + C$

(problem 7b) Compute

Negative exponents: $\frac {4}{5x^6} = \frac 45 x^{-6}$

Use the power rule with $n=-6$

The Power Rule says $\int x^n \ dx = \frac {x^{n+1}}{n+1} +$ C

Do not add the +C to your answer

$\int \frac {4}{5x^6} \ dx = \answer {-4/25 x^{-5}} \ + C$

example 8 $\begin {aligned}[t] \int 4\sqrt [3]x \ dx &= 4 \int \sqrt [3] x \ dx \\ &= 4 \int x^{1/3} \ dx \\ &= 4 \cdot \frac {x^{4/3}}{4/3} +C \\[3pt] &= 4 \cdot \frac {3}{4} \cdot x^{4/3} +C \\[3pt] &= 3x^{4/3} +C \\ &= 3\sqrt [3] {x^4} +C \\ &= 3x\sqrt [3] x +C. \end {aligned}$

(problem 8a) Compute

Rational exponents: $\sqrt x = x^{1/2}$

Use the power rule with $n=1/2$

The Power Rule says $\int x^n \ dx = \frac {x^{n+1}}{n+1} +$ C

Do not add the +C to your answer

$\int 3\sqrt x \ dx = \answer {2x^{3/2}} \ + C$

(problem 8b) Compute

Rational exponents: $\sqrt [4]x = x^{1/4}$

Use the power rule with $n=1/2$

The Power Rule says $\int x^n \ dx = \frac {x^{n+1}}{n+1} +$ C

Do not add the +C to your answer

$\int 5\sqrt [4]x \ dx = \answer {4x^{5/4}} \ + C$

Sum And Difference Rules The integral of a sum or difference is the sum or difference of the integrals: $\int \left [f(x)\pm g(x)\right ] \ dx = \int f(x) \; dx \pm \int g(x) \; dx.$

example 9 $\int \left (\cos (x) + e^x\right ) \ dx = \sin (x) + e^x + C.$

(problem 9a) Compute $\int \big (1 + \cos (x) \big ) \ dx.$

$\frac {d}{dx} x = 1$

$\frac {d}{dx} \sin (x) = \cos (x)$

Do not add the +C to your answer

$\int \left [1 + \cos (x) \right ] \ dx = \answer {x + \sin (x)} \ + C$

(problem 9b) Compute

$\frac {d}{dx} e^x = e^x$

$\frac {d}{dx} \cos (x) = -\sin (x)$

Do not add the +C to your answer

$\int \left (e^x + \sin (x) \right ) \ dx = \answer {e^x -\cos (x)} \ + C$

example 10 $\int \left (1 - e^x\right ) \ dx = x - e^x + C.$

(problem 10) Compute $\int (3e^x - 4x) \; dx = \answer {3e^x - 2x^2} + C.$

example 11 $\int \left (x^2 - x\right ) \ dx = \dfrac {x^3}{3} - \dfrac {x^2}{2} + C.$

(problem 11) Compute $\int (x^5 - x^3) \; dx = \answer {1/6 x^6 - 1/4 x^4} + C.$

example 12 $\int \left (x + \sec ^2(x)\right ) \ dx = \frac {x^2}{2} + \tan (x) + C.$

(problem 12a) Compute

$\frac {d}{dx} x = 1$

$\frac {d}{dx} \sec (x) = \sec (x)\tan (x)$

Do not add the +C to your answer

$\int \big (1 + \sec (x)\tan (x) \big ) \ dx = \answer {x + \sec (x)} \ + C$

(problem 12b) Compute

$\frac {d}{dx} x^2 = 2x$

$\frac {d}{dx} \cot (x) = -\csc ^2(x)$

Do not add the +C to your answer

$\int \left (x + \csc ^2(x) \right ) \ dx = \answer {x^2 /2 - \cot (x)} \ + C$

example 13 $\int \left [3\cos (x) - 2\sin (x)\right ] \ dx = 3\sin (x) + 2\cos (x) + C.$

(problem 13) Compute $\int \left [4\sin (t) - 6\cos (t)\right ] \; dt = \answer {-4\cos (t) - 6\sin (t)} + C.$

example 14 $\int 4\csc (x)\big [\csc (x) - \cot (x)\big ] \ dx = \int \big (4\csc ^2(x) - 4\csc (x)\cot (x)\big ) \ dx$ $= -4\cot (x) + 4\csc (x) + C = 4\csc (x) - 4 \cot (x) + C.$

(problem 14a) Compute

Distribute $\sec (x)$

$\frac {d}{dx} \tan (x) = \sec ^2(x)$

$\frac {d}{dx} \sec (x) = \sec (x)\tan (x)$

Do not add the +C to your answer

$\int \sec (x)\big [\sec (x) + \tan (x) \big ] \ dx = \answer {\tan (x) + \sec (x)} \ + C$

(problem 14b) Compute $\int 3\cot (x)\left [2\csc (x) - \tan (x)\right ]\; dx = \answer {-6\csc (x) - 3x} + C.$

example 15 $\int x^2 + \dfrac {1}{x^2} \ dx = \int x^2 + x^{-2} \ dx = \dfrac {x^3}{3} + \dfrac {x^{-1}}{-1} + C = \dfrac {x^3}{3} - \dfrac {1}{x} + C.$

example 16 $\int \dfrac {x+2}{x} \ dx = \int \dfrac {x}{x} + \dfrac {2}{x} \ dx = \int 1 + \dfrac {2}{x} \ dx = x + 2\ln |x| +C.$

(problem 16) Compute

Divide each term by $x^2$

Use the power rule where appropriate ( $n\neq -1$ )

Do not add the +C to your answer

$\int \dfrac {x^3 + 3x^2 - 4x + 5}{x^2} \ dx = \answer {x^2 /2 + 3x - 4\ln |x| -5/x} \ + C$

example 17 $\int \left (3x^2 - \dfrac {3}{x} - \dfrac {1}{1+x^2}\right ) \ dx = x^3 - 3\ln |x| - \tan ^{-1}(x) + C.$

(problem 17) Compute $\int \left (\frac {2}{x^3} - \frac {3}{x} \right ) \; dx = \answer {-x^{-2} -3\ln |x|} + C.$

example 18 $\begin {aligned}[t] \int \big (\sqrt x + \sqrt [3] x \big ) \ dx &= \int \big (x^{1/2} + x^{1/3}\big ) \ dx \\ &= \frac {x^{3/2}}{3/2} + \frac {x^{4/3}}{4/3} + C \\ &= \tfrac {2}{3} x^{3/2} + \tfrac {3}{4} x^{4/3} + C \\ &= \tfrac {2}{3} \sqrt {x^3} + \tfrac {3}{4} \sqrt [3] {x^4} + C \\ &= \tfrac {2}{3} x\sqrt x + \tfrac {3}{4} x \sqrt [3] x + C. \end {aligned}$

(problem 18) Compute

Distribute the x (add exponents)

Use the power rule where appropriate ( $n\neq -1$ )

Do not add the +C to your answer

$\int x\big (\sqrt x + \sqrt [3] x \big ) \ dx = \answer {2/5 x^{5/2} + 3/7 x^{7/3}} \ + C$

example 19 $\int x^2(x^2 - 3x + 4) \ dx = \int (x^4 - 3x^3 + 4x^2) \ dx = \frac 15 x^5 - \frac {3}{4}x^4 + \frac 43 x^3 + C.$

(problem 19) Compute $\int \left (x^4 - 2x^3 + 6x^2 - 8x + 7 \right ) \; dx = \answer { x^5/5 - x^4/2 + 2x^3 - 4x^2 + 7x} + C.$

2 Initial Value Problems

One reason we might be interested in computing an indefinite integral is to solve a differential equation. Consider the following initial value problem:
Given, $f'(x) = h(x)$ and $f(x_0) = y_0$ , find $f(x)$ .
The condition, $f(x_0) = y_0$ , is called an initial condition. The main step in slving this problem is to observe that the function $f(x)$ that we seek is an anti-derivative of the given function $h(x)$ . Hence, to find $f(x)$ we will compute the definite integral,

$f(x) = \int f'(x) \ dx = \int h(x) \ dx.$

Lets look at some examples.

example 20 Solve the initial value problem: $f'(x) = x - 3, \ f(2) = 4.$

First, $f(x) = \int f'(x) \ dx = \int (x-3) \ dx = \frac {x^2}{2} - 3x + C.$ Then, we use the initial condition, $f(2) = 4$ , to find the appropriate value of the constant $C$ . $f(2) = \frac {2^2}{2} - 3(2) + C = -4 + C = 4,$ which yields $C = 8$ . So, the final answer is $f(x) = \frac {x^2}{2} -3x + 8.$

example 21 Solve the initial value problem: $f'(x) = 2\cos (x) -3\sin (x), \ f(0) = 1.$

First, $f(x) = \int f'(x) \ dx = \int \big [2\cos (x) -3\sin (x)\big ] \ dx = 2\sin (x) + 3\cos (x) + C.$ Then, we use the initial condition, $f(0) = 1$ , to find the appropriate value of the constant $C$ . $f(0) = 2\sin (0) + 3\cos (0) + C = 0 + 3 + C = 1,$ which yields $C = -2$ . So, the final answer is $f(x) = 2\sin (x) + 3\cos (x)-2.$

(problem 21a) Solve the initial value problem: $f'(x) = x-4, \ f(2) = 0.$

anti-differentiate $f'(x)$

Solve for the constant of integration, $C$

$f(x) = \answer {\frac {x^2}{2} - 4x + 6}.$

(problem 21b) Solve the initial value problem: $f'(x) = 3\cos (x), \ f(0) = -2.$

anti-differentiate $f'(x)$

Solve for the constant of integration, $C$

$f(x) = \answer {3\sin (x) - 2}.$

(problem 21c) Solve the initial value problem: $f'(x) = \frac {1}{x}, \ f(-1) = 5.$

anti-differentiate $f'(x)$

Solve for the constant of integration, $C$

$f(x) = \answer {\ln |x| + 5}.$

(problem 21d) Solve the initial value problem: $f'(x) = e^{2x}, \ f(0) = 1.$

Anti-differentiate $f'(x)$

Solve for the constant of integration, $C$

$f(x) = \answer {\frac {e^{2x}}{2} + \frac 12}.$

2024-09-27 13:56:31