Calculus Examples

$C'' (x) = \frac{36000}{x^{3}}$

Step 1

The function $C' (x)$ can be found by evaluating the indefinite integral of the derivative $C'' (x)$ .

$C' (x) = \int C'' (x) d x$

Step 2

Since $36000$ is constant with respect to $x$ , move $36000$ out of the integral.

$36000 \int \frac{1}{x^{3}} d x$

Step 3

Apply basic rules of exponents.

Tap for more steps...

Step 3.1

Move $x^{3}$ out of the denominator by raising it to the $- 1$ power.

$36000 \int {(x^{3})}^{- 1} d x$

Step 3.2

Multiply the exponents in ${(x^{3})}^{- 1}$ .

Tap for more steps...

Step 3.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$36000 \int x^{3 \cdot - 1} d x$

Step 3.2.2

Multiply $3$ by $- 1$ .

$36000 \int x^{- 3} d x$

Step 4

By the Power Rule, the integral of $x^{- 3}$ with respect to $x$ is $- \frac{1}{2} x^{- 2}$ .

$36000 (- \frac{1}{2} x^{- 2} + C)$

Step 5

Simplify the answer.

Tap for more steps...

Step 5.1

Simplify.

Tap for more steps...

Step 5.1.1

Combine $x^{- 2}$ and $\frac{1}{2}$ .

$36000 (- \frac{x^{- 2}}{2} + C)$

Step 5.1.2

Move $x^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$36000 (- \frac{1}{2 x^{2}} + C)$

Step 5.2

Simplify.

$36000 (- \frac{1}{2 x^{2}}) + C$

Step 5.3

Simplify.

Tap for more steps...

Step 5.3.1

Multiply $- 1$ by $36000$ .

$- 36000 \frac{1}{2 x^{2}} + C$

Step 5.3.2

Combine $- 36000$ and $\frac{1}{2 x^{2}}$ .

$\frac{- 36000}{2 x^{2}} + C$

Step 5.3.3

Cancel the common factor of $- 36000$ and $2$ .

Tap for more steps...

Step 5.3.3.1

Factor $2$ out of $- 36000$ .

$\frac{2 \cdot - 18000}{2 x^{2}} + C$

Step 5.3.3.2

Cancel the common factors.

Tap for more steps...

Step 5.3.3.2.1

Factor $2$ out of $2 x^{2}$ .

$\frac{2 \cdot - 18000}{2 (x^{2})} + C$

Step 5.3.3.2.2

Cancel the common factor.

$\frac{2 \cdot - 18000}{2 x^{2}} + C$

Step 5.3.3.2.3

Rewrite the expression.

$\frac{- 18000}{x^{2}} + C$

Step 5.3.4

Move the negative in front of the fraction.

$- \frac{18000}{x^{2}} + C$

Step 6

The function $C'$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$C' (x) = - \frac{18000}{x^{2}} + C$

Step 7

The function $C (x)$ can be found by evaluating the indefinite integral of the derivative $C' (x)$ .

$C (x) = \int C' (x) d x$

Step 8

Split the single integral into multiple integrals.

$\int - \frac{18000}{x^{2}} d x + \int C d x$

Step 9

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$- \int \frac{18000}{x^{2}} d x + \int C d x$

Step 10

Since $18000$ is constant with respect to $x$ , move $18000$ out of the integral.

$- (18000 \int \frac{1}{x^{2}} d x) + \int C d x$

Step 11

Simplify the expression.

Tap for more steps...

Step 11.1

Multiply $18000$ by $- 1$ .

$- 18000 \int \frac{1}{x^{2}} d x + \int C d x$

Step 11.2

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

$- 18000 \int {(x^{2})}^{- 1} d x + \int C d x$

Step 11.3

Multiply the exponents in ${(x^{2})}^{- 1}$ .

Tap for more steps...

Step 11.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- 18000 \int x^{2 \cdot - 1} d x + \int C d x$

Step 11.3.2

Multiply $2$ by $- 1$ .

$- 18000 \int x^{- 2} d x + \int C d x$

Step 12

By the Power Rule, the integral of $x^{- 2}$ with respect to $x$ is $- x^{- 1}$ .

$- 18000 (- x^{- 1} + C) + \int C d x$

Step 13

Apply the constant rule.

$- 18000 (- x^{- 1} + C) + C x + C$

Step 14

Simplify.

Tap for more steps...

Step 14.1

Simplify.

$- 18000 (- \frac{1}{x}) + C x + C$

Step 14.2

Simplify.

Tap for more steps...

Step 14.2.1

Multiply $- 1$ by $- 18000$ .

$18000 \frac{1}{x} + C x + C$

Step 14.2.2

Combine $18000$ and $\frac{1}{x}$ .

$\frac{18000}{x} + C x + C$

Step 15

The function $C$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$C (x) = \frac{18000}{x} + C x + C$