Calculus Examples

$\int_{0}^{\ln (2)} \frac{e^{x} + e^{- x}}{3} d x$

Step 1

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} \int_{0}^{\ln (2)} e^{x} + e^{- x} d x$

Step 2

Split the single integral into multiple integrals.

$\frac{1}{3} (\int_{0}^{\ln (2)} e^{x} d x + \int_{0}^{\ln (2)} e^{- x} d x)$

Step 3

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

$\frac{1}{3} (e^{x}]_{0}^{\ln (2)} + \int_{0}^{\ln (2)} e^{- x} d x)$

Step 4

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

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Step 4.1

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

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Step 4.1.1

Differentiate $- x$ .

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

Step 4.1.2

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 4.1.3

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

$- 1 \cdot 1$

Step 4.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 4.2

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

$u_{lower} = - 0$

Step 4.3

Multiply $- 1$ by $0$ .

$u_{lower} = 0$

Step 4.4

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

$u_{upper} = - \ln (2)$

Step 4.5

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

$u_{lower} = 0$

$u_{upper} = - \ln (2)$

Step 4.6

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

$\frac{1}{3} (e^{x}]_{0}^{\ln (2)} + \int_{0}^{- \ln (2)} - e^{u} d u)$

Step 5

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

$\frac{1}{3} (e^{x}]_{0}^{\ln (2)} - \int_{0}^{- \ln (2)} e^{u} d u)$

Step 6

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

$\frac{1}{3} (e^{x}]_{0}^{\ln (2)} - (e^{u}]_{0}^{- \ln (2)}))$

Step 7

Substitute and simplify.

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Step 7.1

Evaluate $e^{x}$ at $\ln (2)$ and at $0$ .

$\frac{1}{3} ((e^{\ln (2)}) - e^{0} - (e^{u}]_{0}^{- \ln (2)}))$

Step 7.2

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

$\frac{1}{3} ((e^{\ln (2)}) - e^{0} - ((e^{- \ln (2)}) - e^{0}))$

Step 7.3

Simplify.

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Step 7.3.1

Exponentiation and log are inverse functions.

$\frac{1}{3} (2 - e^{0} - ((e^{- \ln (2)}) - e^{0}))$

Step 7.3.2

Anything raised to $0$ is $1$ .

$\frac{1}{3} (2 - 1 \cdot 1 - ((e^{- \ln (2)}) - e^{0}))$

Step 7.3.3

Multiply $- 1$ by $1$ .

$\frac{1}{3} (2 - 1 - ((e^{- \ln (2)}) - e^{0}))$

Step 7.3.4

Subtract $1$ from $2$ .

$\frac{1}{3} (1 - ((e^{- \ln (2)}) - e^{0}))$

Step 7.3.5

Anything raised to $0$ is $1$ .

$\frac{1}{3} (1 - (e^{- \ln (2)} - 1 \cdot 1))$

Step 7.3.6

Multiply $- 1$ by $1$ .

$\frac{1}{3} (1 - (e^{- \ln (2)} - 1))$

Step 8

Simplify.

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Step 8.1

Simplify each term.

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Step 8.1.1

Simplify each term.

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Step 8.1.1.1

Simplify $- \ln (2)$ by moving $- 1$ inside the logarithm.

$\frac{1}{3} (1 - (e^{\ln (2^{- 1})} - 1))$

Step 8.1.1.2

Exponentiation and log are inverse functions.

$\frac{1}{3} (1 - (2^{- 1} - 1))$

Step 8.1.1.3

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

$\frac{1}{3} (1 - (\frac{1}{2} - 1))$

Step 8.1.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{3} (1 - (\frac{1}{2} - 1 \cdot \frac{2}{2}))$

Step 8.1.3

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

$\frac{1}{3} (1 - (\frac{1}{2} + \frac{- 1 \cdot 2}{2}))$

Step 8.1.4

Combine the numerators over the common denominator.

$\frac{1}{3} (1 - \frac{1 - 1 \cdot 2}{2})$

Step 8.1.5

Simplify the numerator.

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Step 8.1.5.1

Multiply $- 1$ by $2$ .

$\frac{1}{3} (1 - \frac{1 - 2}{2})$

Step 8.1.5.2

Subtract $2$ from $1$ .

$\frac{1}{3} (1 - \frac{- 1}{2})$

Step 8.1.6

Move the negative in front of the fraction.

$\frac{1}{3} (1 - - \frac{1}{2})$

Step 8.1.7

Multiply $- - \frac{1}{2}$ .

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Step 8.1.7.1

Multiply $- 1$ by $- 1$ .

$\frac{1}{3} (1 + 1 (\frac{1}{2}))$

Step 8.1.7.2

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

$\frac{1}{3} (1 + \frac{1}{2})$

Step 8.2

Write $1$ as a fraction with a common denominator.

$\frac{1}{3} (\frac{2}{2} + \frac{1}{2})$

Step 8.3

Combine the numerators over the common denominator.

$\frac{1}{3} \cdot \frac{2 + 1}{2}$

Step 8.4

Add $2$ and $1$ .

$\frac{1}{3} \cdot \frac{3}{2}$

Step 8.5

Cancel the common factor of $3$ .

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Step 8.5.1

Cancel the common factor.

$\frac{1}{3} \cdot \frac{3}{2}$

Step 8.5.2

Rewrite the expression.

$\frac{1}{2}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{2}$

Decimal Form:

$0.5$

Step 10