Calculus Examples

$\int_{0}^{6} [5 (2^{- \frac{x}{3}}) - \frac{x}{5}] d x$

Step 1

Remove parentheses.

$\int_{0}^{6} 5 (2^{- \frac{x}{3}}) - \frac{x}{5} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{0}^{6} 5 \cdot 2^{- \frac{x}{3}} d x + \int_{0}^{6} - \frac{x}{5} d x$

Step 3

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$5 \int_{0}^{6} 2^{- \frac{x}{3}} d x + \int_{0}^{6} - \frac{x}{5} d x$

Step 4

Let $u = - \frac{x}{3}$ . Then $d u = - \frac{1}{3} d x$ , so $- 3 d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

Let $u = - \frac{x}{3}$ . Find $\frac{d u}{d x}$ .

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Step 4.1.1

Differentiate $- \frac{x}{3}$ .

$\frac{d}{d x} [- \frac{x}{3}]$

Step 4.1.2

Since $- \frac{1}{3}$ is constant with respect to $x$ , the derivative of $- \frac{x}{3}$ with respect to $x$ is $- \frac{1}{3} \frac{d}{d x} [x]$ .

$- \frac{1}{3} \frac{d}{d x} [x]$

Step 4.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$- \frac{1}{3} \cdot 1$

Step 4.1.4

Multiply $- 1$ by $1$ .

$- \frac{1}{3}$

Step 4.2

Substitute the lower limit in for $x$ in $u = - \frac{x}{3}$ .

$u_{lower} = - \frac{0}{3}$

Step 4.3

Simplify.

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Step 4.3.1

Divide $0$ by $3$ .

$u_{lower} = - 0$

Step 4.3.2

Multiply $- 1$ by $0$ .

$u_{lower} = 0$

Step 4.4

Substitute the upper limit in for $x$ in $u = - \frac{x}{3}$ .

$u_{upper} = - \frac{6}{3}$

Step 4.5

Simplify.

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Step 4.5.1

Divide $6$ by $3$ .

$u_{upper} = - 1 \cdot 2$

Step 4.5.2

Multiply $- 1$ by $2$ .

$u_{upper} = - 2$

Step 4.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0$

$u_{upper} = - 2$

Step 4.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$5 \int_{0}^{- 2} - 2^{u} \cdot \frac{- 1}{- \frac{1}{3}} d u + \int_{0}^{6} - \frac{x}{5} d x$

Step 5

Simplify.

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Step 5.1

Dividing two negative values results in a positive value.

$5 \int_{0}^{- 2} - 2^{u} \cdot \frac{1}{\frac{1}{3}} d u + \int_{0}^{6} - \frac{x}{5} d x$

Step 5.2

Multiply by the reciprocal of the fraction to divide by $\frac{1}{3}$ .

$5 \int_{0}^{- 2} - 2^{u} \cdot (1 \cdot 3) d u + \int_{0}^{6} - \frac{x}{5} d x$

Step 5.3

Multiply $3$ by $1$ .

$5 \int_{0}^{- 2} - 2^{u} \cdot 3 d u + \int_{0}^{6} - \frac{x}{5} d x$

Step 5.4

Multiply $3$ by $- 1$ .

$5 \int_{0}^{- 2} - 3 \cdot 2^{u} d u + \int_{0}^{6} - \frac{x}{5} d x$

Step 6

Since $- 3$ is constant with respect to $u$ , move $- 3$ out of the integral.

$5 (- 3 \int_{0}^{- 2} 2^{u} d u) + \int_{0}^{6} - \frac{x}{5} d x$

Step 7

Multiply $- 3$ by $5$ .

$- 15 \int_{0}^{- 2} 2^{u} d u + \int_{0}^{6} - \frac{x}{5} d x$

Step 8

The integral of $2^{u}$ with respect to $u$ is $\frac{2^{u}}{\ln (2)}$ .

$- 15 (\frac{2^{u}}{\ln (2)}]_{0}^{- 2}) + \int_{0}^{6} - \frac{x}{5} d x$

Step 9

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

$- 15 (\frac{2^{u}}{\ln (2)}]_{0}^{- 2}) - \int_{0}^{6} \frac{x}{5} d x$

Step 10

Since $\frac{1}{5}$ is constant with respect to $x$ , move $\frac{1}{5}$ out of the integral.

$- 15 (\frac{2^{u}}{\ln (2)}]_{0}^{- 2}) - (\frac{1}{5} \int_{0}^{6} x d x)$

Step 11

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

$- 15 (\frac{2^{u}}{\ln (2)}]_{0}^{- 2}) - \frac{1}{5} \frac{1}{2} x^{2}]_{0}^{6}$

Step 12

Substitute and simplify.

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Step 12.1

Evaluate $\frac{2^{u}}{\ln (2)}$ at $- 2$ and at $0$ .

$- 15 ((\frac{2^{- 2}}{\ln (2)}) - \frac{2^{0}}{\ln (2)}) - \frac{1}{5} \frac{1}{2} x^{2}]_{0}^{6}$

Step 12.2

Evaluate $\frac{1}{2} x^{2}$ at $6$ and at $0$ .

$- 15 ((\frac{2^{- 2}}{\ln (2)}) - \frac{2^{0}}{\ln (2)}) - \frac{1}{5} ((\frac{1}{2} \cdot 6^{2}) - \frac{1}{2} \cdot 0^{2})$

Step 12.3

Simplify.

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Step 12.3.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- 15 (\frac{\frac{1}{2^{2}}}{\ln (2)} - \frac{2^{0}}{\ln (2)}) - \frac{1}{5} ((\frac{1}{2} \cdot 6^{2}) - \frac{1}{2} \cdot 0^{2})$

Step 12.3.2

Raise $2$ to the power of $2$ .

$- 15 (\frac{\frac{1}{4}}{\ln (2)} - \frac{2^{0}}{\ln (2)}) - \frac{1}{5} ((\frac{1}{2} \cdot 6^{2}) - \frac{1}{2} \cdot 0^{2})$

Step 12.3.3

Rewrite $\frac{\frac{1}{4}}{\ln (2)}$ as a product.

$- 15 (\frac{1}{4} \cdot \frac{1}{\ln (2)} - \frac{2^{0}}{\ln (2)}) - \frac{1}{5} ((\frac{1}{2} \cdot 6^{2}) - \frac{1}{2} \cdot 0^{2})$

Step 12.3.4

Multiply $\frac{1}{4}$ by $\frac{1}{\ln (2)}$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{2^{0}}{\ln (2)}) - \frac{1}{5} ((\frac{1}{2} \cdot 6^{2}) - \frac{1}{2} \cdot 0^{2})$

Step 12.3.5

Anything raised to $0$ is $1$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} ((\frac{1}{2} \cdot 6^{2}) - \frac{1}{2} \cdot 0^{2})$

Step 12.3.6

Raise $6$ to the power of $2$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} (\frac{1}{2} \cdot 36 - \frac{1}{2} \cdot 0^{2})$

Step 12.3.7

Combine $\frac{1}{2}$ and $36$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} (\frac{36}{2} - \frac{1}{2} \cdot 0^{2})$

Step 12.3.8

Cancel the common factor of $36$ and $2$ .

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Step 12.3.8.1

Factor $2$ out of $36$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} (\frac{2 \cdot 18}{2} - \frac{1}{2} \cdot 0^{2})$

Step 12.3.8.2

Cancel the common factors.

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Step 12.3.8.2.1

Factor $2$ out of $2$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} (\frac{2 \cdot 18}{2 (1)} - \frac{1}{2} \cdot 0^{2})$

Step 12.3.8.2.2

Cancel the common factor.

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} (\frac{2 \cdot 18}{2 \cdot 1} - \frac{1}{2} \cdot 0^{2})$

Step 12.3.8.2.3

Rewrite the expression.

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} (\frac{18}{1} - \frac{1}{2} \cdot 0^{2})$

Step 12.3.8.2.4

Divide $18$ by $1$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} (18 - \frac{1}{2} \cdot 0^{2})$

Step 12.3.9

Raising $0$ to any positive power yields $0$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} (18 - \frac{1}{2} \cdot 0)$

Step 12.3.10

Multiply $0$ by $- 1$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} (18 + 0 (\frac{1}{2}))$

Step 12.3.11

Multiply $0$ by $\frac{1}{2}$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} (18 + 0)$

Step 12.3.12

Add $18$ and $0$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{1}{5} \cdot 18$

Step 12.3.13

Multiply $18$ by $- 1$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - 18 (\frac{1}{5})$

Step 12.3.14

Combine $- 18$ and $\frac{1}{5}$ .

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) + \frac{- 18}{5}$

Step 12.3.15

Move the negative in front of the fraction.

$- 15 (\frac{1}{4 \ln (2)} - \frac{1}{\ln (2)}) - \frac{18}{5}$

Step 13

Simplify each term.

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Step 13.1

Apply the distributive property.

$- 15 \frac{1}{4 \ln (2)} - 15 (- \frac{1}{\ln (2)}) - \frac{18}{5}$

Step 13.2

Combine $- 15$ and $\frac{1}{4 \ln (2)}$ .

$\frac{- 15}{4 \ln (2)} - 15 (- \frac{1}{\ln (2)}) - \frac{18}{5}$

Step 13.3

Multiply $- 15 (- \frac{1}{\ln (2)})$ .

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Step 13.3.1

Multiply $- 1$ by $- 15$ .

$\frac{- 15}{4 \ln (2)} + 15 \frac{1}{\ln (2)} - \frac{18}{5}$

Step 13.3.2

Combine $15$ and $\frac{1}{\ln (2)}$ .

$\frac{- 15}{4 \ln (2)} + \frac{15}{\ln (2)} - \frac{18}{5}$

Step 13.4

Move the negative in front of the fraction.

$- \frac{15}{4 \ln (2)} + \frac{15}{\ln (2)} - \frac{18}{5}$

Step 14

The result can be shown in multiple forms.

Exact Form:

$- \frac{15}{4 \ln (2)} + \frac{15}{\ln (2)} - \frac{18}{5}$

Decimal Form:

$12.63031921 \dots$

Step 15