Calculus Examples

$f (x) = 2 e^{x} - 1$ ; $[- 2, 4]$

Step 1

Solve by substitution to find the intersection between the curves.

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Step 1.1

Eliminate the equal sides of each equation and combine.

$2 e^{x} - 1 = 0$

Step 1.2

Solve $2 e^{x} - 1 = 0$ for $x$ .

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Step 1.2.1

Add $1$ to both sides of the equation.

$2 e^{x} = 1$

Step 1.2.2

Divide each term in $2 e^{x} = 1$ by $2$ and simplify.

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Step 1.2.2.1

Divide each term in $2 e^{x} = 1$ by $2$ .

$\frac{2 e^{x}}{2} = \frac{1}{2}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{2 e^{x}}{2} = \frac{1}{2}$

Step 1.2.2.2.1.2

Divide $e^{x}$ by $1$ .

$e^{x} = \frac{1}{2}$

Step 1.2.3

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{x}) = \ln (\frac{1}{2})$

Step 1.2.4

Expand the left side.

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Step 1.2.4.1

Expand $\ln (e^{x})$ by moving $x$ outside the logarithm.

$x \ln (e) = \ln (\frac{1}{2})$

Step 1.2.4.2

The natural logarithm of $e$ is $1$ .

$x \cdot 1 = \ln (\frac{1}{2})$

Step 1.2.4.3

Multiply $x$ by $1$ .

$x = \ln (\frac{1}{2})$

Step 1.3

Substitute $\ln (\frac{1}{2})$ for $x$ .

$y = 0$

Step 1.4

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(\ln (\frac{1}{2}), 0)$

Step 2

The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region. The regions are determined by the intersection points of the curves. This can be done algebraically or graphically.

$A r e a = \int_{- 2}^{\ln (\frac{1}{2})} 0 d x - \int_{- 2}^{\ln (\frac{1}{2})} 2 e^{x} - 1 d x$

Step 3

Integrate to find the area between $- 2$ and $\ln (\frac{1}{2})$ .

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Step 3.1

Combine the integrals into a single integral.

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

Step 3.2

Subtract $2 e^{x} - 1$ from $0$ .

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

Step 3.3

Apply the distributive property.

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

Step 3.4

Multiply.

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Step 3.4.1

Multiply $2$ by $- 1$ .

$- 2 e^{x} - - 1$

Step 3.4.2

Multiply $- 1$ by $- 1$ .

$- 2 e^{x} + 1$

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

Step 3.5

Split the single integral into multiple integrals.

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

Step 3.6

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

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

Step 3.7

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

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

Step 3.8

Apply the constant rule.

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

Step 3.9

Substitute and simplify.

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Step 3.9.1

Evaluate $e^{x}$ at $\ln (\frac{1}{2})$ and at $- 2$ .

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

Step 3.9.2

Evaluate $x$ at $\ln (\frac{1}{2})$ and at $- 2$ .

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

Step 3.9.3

Exponentiation and log are inverse functions.

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

Step 3.10

Simplify.

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Step 3.10.1

Simplify each term.

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Step 3.10.1.1

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

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

Step 3.10.1.2

Apply the distributive property.

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

Step 3.10.1.3

Cancel the common factor of $2$ .

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Step 3.10.1.3.1

Factor $2$ out of $- 2$ .

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

Step 3.10.1.3.2

Cancel the common factor.

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

Step 3.10.1.3.3

Rewrite the expression.

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

Step 3.10.1.4

Multiply $- 2 (- \frac{1}{e^{2}})$ .

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Step 3.10.1.4.1

Multiply $- 1$ by $- 2$ .

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

Step 3.10.1.4.2

Combine $2$ and $\frac{1}{e^{2}}$ .

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

Step 3.10.2

Add $- 1$ and $2$ .

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

Step 4

$A r e a = \int_{\ln (\frac{1}{2})}^{4} 2 e^{x} - 1 d x - \int_{\ln (\frac{1}{2})}^{4} 0 d x$

Step 5

Integrate to find the area between $\ln (\frac{1}{2})$ and $4$ .

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Step 5.1

Integrate to find the area between $\ln (\frac{1}{2})$ and $4$ .

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Step 5.1.1

Combine the integrals into a single integral.

$\int_{\ln (\frac{1}{2})}^{4} 0 - (2 e^{x} - 1) d x$

Step 5.1.2

Subtract $2 e^{x} - 1$ from $0$ .

$\int_{\ln (\frac{1}{2})}^{4} - (2 e^{x} - 1) d x$

Step 5.1.3

Apply the distributive property.

$\int_{\ln (\frac{1}{2})}^{4} - (2 e^{x}) + 1 d x$

Step 5.1.4

Multiply.

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Step 5.1.4.1

Multiply $2$ by $- 1$ .

$- 2 e^{x} - - 1$

Step 5.1.4.2

Multiply $- 1$ by $- 1$ .

$- 2 e^{x} + 1$

$\int_{\ln (\frac{1}{2})}^{4} - 2 e^{x} + 1 d x$

Step 5.1.5

Split the single integral into multiple integrals.

$\int_{\ln (\frac{1}{2})}^{4} - 2 e^{x} d x + \int_{\ln (\frac{1}{2})}^{4} d x$

Step 5.1.6

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

$- 2 \int_{\ln (\frac{1}{2})}^{4} e^{x} d x + \int_{\ln (\frac{1}{2})}^{4} d x$

Step 5.1.7

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

$- 2 (e^{x}]_{\ln (\frac{1}{2})}^{4}) + \int_{\ln (\frac{1}{2})}^{4} d x$

Step 5.1.8

Apply the constant rule.

$- 2 (e^{x}]_{\ln (\frac{1}{2})}^{4}) + x]_{\ln (\frac{1}{2})}^{4}$

Step 5.1.9

Substitute and simplify.

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Step 5.1.9.1

Evaluate $e^{x}$ at $4$ and at $\ln (\frac{1}{2})$ .

$- 2 ((e^{4}) - e^{\ln (\frac{1}{2})}) + x]_{\ln (\frac{1}{2})}^{4}$

Step 5.1.9.2

Evaluate $x$ at $4$ and at $\ln (\frac{1}{2})$ .

$- 2 (e^{4} - e^{\ln (\frac{1}{2})}) + 4 - \ln (\frac{1}{2})$

Step 5.1.9.3

Exponentiation and log are inverse functions.

$- 2 (e^{4} - \frac{1}{2}) + 4 - \ln (\frac{1}{2})$

Step 5.1.10

Simplify.

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Step 5.1.10.1

Simplify each term.

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Step 5.1.10.1.1

Apply the distributive property.

$- 2 e^{4} - 2 (- \frac{1}{2}) + 4 - \ln (\frac{1}{2})$

Step 5.1.10.1.2

Cancel the common factor of $2$ .

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Step 5.1.10.1.2.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$- 2 e^{4} - 2 (\frac{- 1}{2}) + 4 - \ln (\frac{1}{2})$

Step 5.1.10.1.2.2

Factor $2$ out of $- 2$ .

$- 2 e^{4} + 2 (- 1) \frac{- 1}{2} + 4 - \ln (\frac{1}{2})$

Step 5.1.10.1.2.3

Cancel the common factor.

$- 2 e^{4} + 2 \cdot - 1 \frac{- 1}{2} + 4 - \ln (\frac{1}{2})$

Step 5.1.10.1.2.4

Rewrite the expression.

$- 2 e^{4} - - 1 + 4 - \ln (\frac{1}{2})$

Step 5.1.10.1.3

Multiply $- 1$ by $- 1$ .

$- 2 e^{4} + 1 + 4 - \ln (\frac{1}{2})$

Step 5.1.10.2

Add $1$ and $4$ .

$- 2 e^{4} + 5 - \ln (\frac{1}{2})$

Step 5.2

Combine the integrals into a single integral.

$\int_{\ln (\frac{1}{2})}^{4} 2 e^{x} - 1 - (0) d x$

Step 5.3

Subtract $0$ from $2 e^{x} - 1$ .

$\int_{\ln (\frac{1}{2})}^{4} 2 e^{x} - 1 d x$

Step 5.4

Split the single integral into multiple integrals.

$\int_{\ln (\frac{1}{2})}^{4} 2 e^{x} d x + \int_{\ln (\frac{1}{2})}^{4} - 1 d x$

Step 5.5

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$2 \int_{\ln (\frac{1}{2})}^{4} e^{x} d x + \int_{\ln (\frac{1}{2})}^{4} - 1 d x$

Step 5.6

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

$2 (e^{x}]_{\ln (\frac{1}{2})}^{4}) + \int_{\ln (\frac{1}{2})}^{4} - 1 d x$

Step 5.7

Apply the constant rule.

$2 (e^{x}]_{\ln (\frac{1}{2})}^{4}) + - x]_{\ln (\frac{1}{2})}^{4}$

Step 5.8

Substitute and simplify.

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Step 5.8.1

Evaluate $e^{x}$ at $4$ and at $\ln (\frac{1}{2})$ .

$2 ((e^{4}) - e^{\ln (\frac{1}{2})}) + - x]_{\ln (\frac{1}{2})}^{4}$

Step 5.8.2

Evaluate $- x$ at $4$ and at $\ln (\frac{1}{2})$ .

$2 (e^{4} - e^{\ln (\frac{1}{2})}) + - 1 \cdot 4 + \ln (\frac{1}{2})$

Step 5.8.3

Simplify.

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Step 5.8.3.1

Exponentiation and log are inverse functions.

$2 (e^{4} - \frac{1}{2}) - 1 \cdot 4 + \ln (\frac{1}{2})$

Step 5.8.3.2

Multiply $- 1$ by $4$ .

$2 (e^{4} - \frac{1}{2}) - 4 + \ln (\frac{1}{2})$

Step 5.9

Simplify.

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Step 5.9.1

Simplify each term.

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Step 5.9.1.1

Apply the distributive property.

$2 e^{4} + 2 (- \frac{1}{2}) - 4 + \ln (\frac{1}{2})$

Step 5.9.1.2

Cancel the common factor of $2$ .

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Step 5.9.1.2.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$2 e^{4} + 2 (\frac{- 1}{2}) - 4 + \ln (\frac{1}{2})$

Step 5.9.1.2.2

Cancel the common factor.

$2 e^{4} + 2 (\frac{- 1}{2}) - 4 + \ln (\frac{1}{2})$

Step 5.9.1.2.3

Rewrite the expression.

$2 e^{4} - 1 - 4 + \ln (\frac{1}{2})$

Step 5.9.2

Subtract $4$ from $- 1$ .

$2 e^{4} - 5 + \ln (\frac{1}{2})$

Step 6

Add the areas $\frac{2}{e^{2}} + \ln (\frac{1}{2}) + 1 + 2 e^{4} - 5 + \ln (\frac{1}{2})$ .

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Step 6.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\frac{2}{e^{2}} + 1 + 2 e^{4} - 5 + \ln (\frac{1}{2} \cdot \frac{1}{2})$

Step 6.2

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

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Step 6.2.1

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

$\frac{2}{e^{2}} + 1 + 2 e^{4} - 5 + \ln (\frac{1}{2 \cdot 2})$

Step 6.2.2

Multiply $2$ by $2$ .

$\frac{2}{e^{2}} + 1 + 2 e^{4} - 5 + \ln (\frac{1}{4})$

Step 6.3

Subtract $5$ from $1$ .

$\frac{2}{e^{2}} + 2 e^{4} - 4 + \ln (\frac{1}{4})$

Step 7