Calculus Examples

$\int_{1}^{6} \frac{9}{x} - e^{- x} d x$

Step 1

Split the single integral into multiple integrals.

$\int_{1}^{6} \frac{9}{x} d x + \int_{1}^{6} - e^{- x} d x$

Step 2

Since $9$ is constant with respect to $x$ , move $9$ out of the integral.

$9 \int_{1}^{6} \frac{1}{x} d x + \int_{1}^{6} - e^{- x} d x$

Step 3

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$9 (\ln (| x |)]_{1}^{6}) + \int_{1}^{6} - e^{- x} d x$

Step 4

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

$9 (\ln (| x |)]_{1}^{6}) - \int_{1}^{6} e^{- x} d x$

Step 5

Let $u = - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

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

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Step 5.1.1

Differentiate $- x$ .

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

Step 5.1.2

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

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

Step 5.1.3

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

$- 1 \cdot 1$

Step 5.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 5.2

Substitute the lower limit in for $x$ in $u = - x$ .

$u_{lower} = - 1 \cdot 1$

Step 5.3

Multiply $- 1$ by $1$ .

$u_{lower} = - 1$

Step 5.4

Substitute the upper limit in for $x$ in $u = - x$ .

$u_{upper} = - 1 \cdot 6$

Step 5.5

Multiply $- 1$ by $6$ .

$u_{upper} = - 6$

Step 5.6

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

$u_{lower} = - 1$

$u_{upper} = - 6$

Step 5.7

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

$9 (\ln (| x |)]_{1}^{6}) - \int_{- 1}^{- 6} - e^{u} d u$

Step 6

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

$9 (\ln (| x |)]_{1}^{6}) - - \int_{- 1}^{- 6} e^{u} d u$

Step 7

Simplify.

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Step 7.1

Multiply $- 1$ by $- 1$ .

$9 (\ln (| x |)]_{1}^{6}) + 1 \int_{- 1}^{- 6} e^{u} d u$

Step 7.2

Multiply $\int_{- 1}^{- 6} e^{u} d u$ by $1$ .

$9 (\ln (| x |)]_{1}^{6}) + \int_{- 1}^{- 6} e^{u} d u$

Step 8

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$9 (\ln (| x |)]_{1}^{6}) + e^{u}]_{- 1}^{- 6}$

Step 9

Substitute and simplify.

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Step 9.1

Evaluate $\ln (| x |)$ at $6$ and at $1$ .

$9 ((\ln (| 6 |)) - \ln (| 1 |)) + e^{u}]_{- 1}^{- 6}$

Step 9.2

Evaluate $e^{u}$ at $- 6$ and at $- 1$ .

$9 ((\ln (| 6 |)) - \ln (| 1 |)) + (e^{- 6}) - e^{- 1}$

Step 9.3

Remove unnecessary parentheses.

$9 (\ln (| 6 |) - \ln (| 1 |)) + e^{- 6} - e^{- 1}$

Step 10

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$9 \ln (\frac{| 6 |}{| 1 |}) + e^{- 6} - e^{- 1}$

Step 11

Simplify.

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Step 11.1

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$9 \ln (\frac{6}{| 1 |}) + e^{- 6} - e^{- 1}$

Step 11.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$9 \ln (\frac{6}{1}) + e^{- 6} - e^{- 1}$

Step 11.3

Divide $6$ by $1$ .

$9 \ln (6) + e^{- 6} - e^{- 1}$

Step 11.4

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

$9 \ln (6) + e^{- 6} - \frac{1}{e}$

Step 12

The result can be shown in multiple forms.

Exact Form:

$9 \ln (6) + e^{- 6} - \frac{1}{e}$

Decimal Form:

$15.76043453 \dots$

Step 13